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https://human.libretexts.org/Courses/University_of_Houston/Improvisations_-_2/06%3A_Unite_5 | 6: Unité 5 - Qu’est-ce qui ne va pas ?
- décrire son style vestimentaire et interviewer des partenaires sur leurs styles vestimentaires
- décrire son état de santé et répondre aux questions sur sa santé
- donner des conseils et des recommandations | libretexts | 2025-03-17T22:26:32.176825 | 2023-08-17T17:16:10 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://human.libretexts.org/Courses/University_of_Houston/Improvisations_-_2/06%3A_Unite_5",
"book_url": "https://commons.libretexts.org/book/human-195014",
"title": "6: Unité 5 - Qu’est-ce qui ne va pas ?",
"author": null
} |
https://human.libretexts.org/Courses/University_of_Houston/Improvisations_-_2/06%3A_Unite_5/6.01%3A_New_Page | 6.1: Vocabulaire - présentation
Ecoutez et répétez.
| un défilé de mode | fashion show |
| une marque | brand |
| un chapeau | hat |
| une casquette | cap |
| des vêtements (m) | clothes |
| un survêtement / un jogging | tracksuit |
| des sous-vêtements | underwear |
| un soutien-gorge | bra |
| un slip | briefs |
| une culotte | panties |
| un caleçon / boxer | boxers |
| une chemise | man's shirt |
| un chemisier | blouse |
| un polo | polo shirt |
| un tee-shirt | T-shirt |
| un pull | sweater |
| un pantalon | (a pair of) pants |
| un jean (sing.) | jeans |
| un short (sing.) | shorter shorts |
| un bermuda | longer shorts |
| une ceinture | belt |
| des chaussettes (f pl) | socks |
| des chaussures (f pl) | shoes |
| des baskets (f pl) | basketball shoes |
| des bottes (f pl) | boots |
| des sandales (f pl) | sandals |
| des tennis (f pl) | tennis shoes |
| des tongs | flip-flops |
| des talons (m) / des chaussures à talons | high-heels |
| une robe | dress |
| une jupe | skirt |
| un tailleur | woman's suit |
| un costume | man's suit |
| une cravate | tie |
| des lunettes (f) de soleil | sunglasses |
| un maillot de bain | swimsuit |
| un blouson | short jacket, leather jacket |
| une veste | jacket |
| un manteau | coat |
| un anorak | parka |
| un imperméable | raincoat |
| une écharpe | scarf |
| des gants (m) | gloves |
Ecoutez et répétez.
| la course à pied / faire de la course à pied | running / to go running |
| le footing / le jogging / faire du footing / du jogging | jogging / to go jogging |
| le roller / faire du roller | roller blading / to go roller blading |
| le cyclisme / faire du cyclisme | cycling / to go cycling |
| l'escalade (f) / faire de l'escalade | rock-climbing / to go rock-climbing |
| la musculation / faire de la musculation | weight training / to train with weights |
| la natation / faire de la natation | swimming / to go swimming |
| le canoë / faire du canoë | canoeing / to go canoeing |
| le kayak / faire du kayak | kayaking / to go kayaking |
| la peinture / faire de la peinture | painting / to paint (art) |
| les échecs (m) / jouer aux échecs | chess / to play chess |
| le yoga / faire du yoga | yoga |
| le surf / faire du surf | surfing |
| le cheval / faire du cheval | horseback riding |
Ecoutez et répétez.
| une allergie | allergy |
| la grippe | the flu |
| une migraine | migraine headache |
| un rhume | a cold |
| des médicaments (m) | medicine |
| des vitamines (f) | vitamins |
| un somnifère | sleeping pill |
Ecoutez et répétez.
| la gorge | throat |
| le bras | arm |
| le coude | elbow |
| le poignet | wrist |
| la main | hand |
| le doigt | finger |
| la poitrine | chest |
| le ventre | stomach |
| le dos | back |
| le derrière | rear, behind |
| les fesses (f) | buttocks |
| le genou / les genoux | knee / knees |
| la cheville | ankle |
| le pied | foot |
| les orteils (m) | toes |
Ecoutez et répétez.
| à la mode / branché(e) | in fashion |
| démodé(e) | out of style |
| décontracté(e) | casual |
| habillé(e) | dressy |
| discipliné(e) | disciplined |
| sportif / sportive | athletic |
| stressé(e) | stressed |
| malade | sick |
Ecoutez et répétez.
| s'habiller | to get dressed |
| mettre | to put (on) |
| porter | to wear |
| aller bien / aller mal | to fit well/poorly |
| grossir | to gain weight |
| être au régime / faire un régime | to be on a diet |
| faire du sport | to exercise |
| mincir | to lose weight |
| être en forme | to be in shape |
| tomber malade | to get sick |
| tousser | to cough |
| éternuer | to sneeze |
| guérir | to recover |
| avoir mal à... | to hurt (body part) |
| (avoir mal à la tête, au dos, aux pieds, etc) | (to have a headache, a backache, sore feet, etc) |
| se reposer | to rest |
| se détendre | to relax |
| fumer | to smoke |
| être en bonne santé | to be in good health | | libretexts | 2025-03-17T22:26:32.271979 | 2023-08-17T17:16:11 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://human.libretexts.org/Courses/University_of_Houston/Improvisations_-_2/06%3A_Unite_5/6.01%3A_New_Page",
"book_url": "https://commons.libretexts.org/book/human-195014",
"title": "6.1: Vocabulaire - présentation",
"author": null
} |
https://human.libretexts.org/Courses/University_of_Houston/Improvisations_-_2/06%3A_Unite_5/6.02%3A_New_Page | 6.2: Vocabulaire - activités
Instructors of LibreTexts and adopters can use the file impro2_ch5_vocabulaire.docx attached to this page (scroll down to see it).
Vrai ou Faux ? Dites si les phrases suivantes sont vraies ou fausses. Si une phrase est fausse, corrigez-la.
- Serena Williams est célèbre parce qu'elle fait du cheval.
- On peut faire du yoga sur le campus de UH.
- J'ai déjà fait de l'escalade au moins une fois dans ma vie.
- Michael Phelps est célèbre parce qu'il fait de la natation.
- Killiam Mbappé est célèbre parce qu'il fait de la peinture.
- Bobby Fischer est célèbre parce qu'il jouait aux échecs.
- On peut faire du surf au Texas.
Les charades. Votre prof va vous donner une petit bout de papier avec une expression française. Essayez de mimer cette expression. Votre partenaire doit deviner quelle expression vous mimez. Echangez vos expressions et trouvez un autre partenaire. Répétez plusieurs fois.
Vrai ou Faux ? Décidez si les phrases sont vraies ou fausses. Si une phrase est fausse, corrigez-la.
- Or regarde avec les yeux.
- On écoute avec les mains.
- Le poignet est une partie du visage.
- Le genou est une partie de la jambe.
- Les dents sont dans la bouche.
- Le ventre est sous les genoux.
- La poitrine est sous le cou.
Associations. Quelle(s) partie(s) du corps associez-vous avec les marques suivantes?
- Kleenex
- L'Oréal
- Head & Shoulders
- Nike
- Rolex
- Evian
- Bic
- Lululemon
Aïe ! J'ai mal ! Expliquez où on a mal selon la situation indiquée.
Modèle:
Quand on fait trop de planche à voile… => On a mal aux bras, au dos…
- Quand on mange trop de chocolat...
- Quand on boit trop...
- Quand on chante trop...
- Quand on fait trop de musculation…
- Quand on fait de la course à pied en portant des sandales…
- Quand on fait du cyclisme...
- Quand on fait du roller...
- Quand on fait du canoë...
- Quand on étudie trop...
- Quand on fait trop de natation ...
Les parties du corps. Connaissez-vous les parties du corps humain ? Montrez les parties du corps sur cette image.
Faites correspondre les vêtements suivants (colonne A) aux parties du corps correspondantes (colonne B).
| A | B |
| une casquette | les mains |
| des gants | les yeux |
| une écharpe | le ventre |
| des tongs | la tête |
| un pantalon | le cou |
| des lunettes de soleil | les pieds |
| une ceinture | les jambes |
Comment dit-on en français ? Votre prof va vous donner des images. Trouvez un partenaire, montrez-lui votre image, et dites : « Comment dit-on en français ? » Votre partenaire va répondre en français, par une phrase complète, par exemple : « C'est un pantalon » ou « Ce sont des sandales ». Vous faites la même chose pour l’image de votre partenaire. Vous échangez vos images, et vous trouvez un nouveau partenaire. Répétez plusieurs fois.
Découvrez votre style vestimentaire. Lisez les phrases suivantes et choisissez la réponse qui vous convient le mieux.
1. Pour aller en cours, je mets plutôt a. une chemise blanche et un pantalon / un chemisier blanc et une jupe
b. un tee-shirt noir et un jean noir c. un tee-shirt et un short d. une chemise à fleurs / une robe longue
2. Quand je sors en boite, je mets plutôt a. un costume et une cravate / une robe élégante
b. une chemise noire et un pantalon en cuir (leather) / une jupe en cuir c. un survêtement d. un tee-shirt et un jean / une robe à fleurs
3. Quand il fait froid, je mets plutôt a. un manteau b. un imperméable sombre c. un anorak d. une écharpe en laine
4. En été, je mets plutôt a. un bermuda et un polo b. un jean noir c. un short et un tee-shirt
d. un tee-shirt et short décontracté / un maillot de bain et une jupe
5. Dans un magasin de chaussures, j’achète a. des chaussures en cuir / des talons hauts b. des bottes noires c. des baskets / des tennis
d. des sandales / des tongs
6. Qu’est-ce que vous portez en hiver? a. un pull b. une veste en cuir c. un sweat / un corsaire d. un cardigan
7. Mon accessoire préféré : a. des lunettes de soleil b. une ceinture large c. une casquette d. un chapeau
8. Mes sous-vêtements préférés sont a. un caleçon à carreaux (plaid) b. un slip / un soutien-gorge noir
c. un boxer uni d. un caleçon / une culotte à pois (polka dot)
Calculez combien de a, b, c, d vous avez au total. Comparez vos résultats avec votre partenaire. Quel est le style de votre partenaire? Expliquez.
| Si vous avez une majorité de a | Si vous avez une majorité de b | Si vous avez une majorité de c | Si vous avez une majorité de d |
| Vous adoptez plutôt le style BCBG (Bon chic, bon genre = classic style) qui est classique, parfois habillé mais jamais de mauvais goût | Il est clair que vous aimez le noir et les couleurs sombres. Votre style est gothique et le cuir est un élément important dans votre look. | Très actif et toujours prêt à faire de l’exercice, vous êtes sportif et aimez les vêtements qui vous permettent de faire des activités physiques. | Vous préférez être à l’aise (comfortable) et choisissez des vêtements décontractés. Le style baba cool (hippy, flower child) vous va très bien. |
Découvrez votre état de santé. Avec un partenaire, répondez aux questions suivantes. Notez les réponses de votre partenaire.
| Souvent | Quelquefois | Jamais | |
| 1. Est-ce que vous prenez des vitamines ? | |||
| 2. Est-ce que vous dormez assez ? | |||
| 3. Est-ce que vous faites du sport ? | |||
| 4. Prenez-vous le temps de vous détendre ? | |||
| 5. Allez-vous chez le médecin ou le dentiste? | |||
| 6. Est-ce que vous faites attention à votre alimentation? | |||
| 7. Est-ce que vous buvez de l’eau minérale? | |||
| 8. Est-ce que vous consommez beaucoup d’alcool? | |||
| 9. Est-ce que vous fumez des cigarettes, des cigares, la pipe? | |||
| 10. Prenez-vous des somnifères? | |||
| 11. Est-ce que vous mangez beaucoup de fast food? | |||
| 12. Avez-vous des allergies? | |||
| 13. Est-ce que vous avez souvent des migraines et des maux de tête? | |||
| 14. Est-ce que vous bronzez (tan) sans protection? | |||
| 15. Etes-vous stressé(e)? |
Regardez les évaluations suivantes ( Félicitations, Pas mal, Il faut changer, Quelle horreur ) et choisissez le meilleur message pour votre partenaire. Est-ce que votre partenaire est d'accord avec votre conclusion ? Et vous, êtes-vous d'accord avec la conclusion de votre partenaire ?
| Félicitations ! | Pas mal | Il faut changer | Quelle horreur ! |
| Félicitations ! Vous devez être fier ! Ne changez pas vos habitudes. Vous allez probablement vivre très longtemps. | Pas mal. Vous avez peut-être de petits changements à faire mais vous êtes en assez bonne santé. | Si vous voulez changer vos habitudes, c’est le bon moment! | Quelle horreur ! Vous allez tomber malade si vous ne changez pas tout de suite votre style de vie ! |
Aknowledgment: some activites on this page are partially adopted from Francais Interactif . | libretexts | 2025-03-17T22:26:32.368175 | 2023-08-17T17:16:12 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://human.libretexts.org/Courses/University_of_Houston/Improvisations_-_2/06%3A_Unite_5/6.02%3A_New_Page",
"book_url": "https://commons.libretexts.org/book/human-195014",
"title": "6.2: Vocabulaire - activités",
"author": null
} |
https://human.libretexts.org/Courses/University_of_Houston/Improvisations_-_2/06%3A_Unite_5/6.03%3A_New_Page | 6.3: Grammaire - présentation
The verb mettre is irregular. Listen carefully to its forms in the present.
| mettre 'to place, put' | |
| je mets | nous mettons |
| tu mets | vous mettez |
| il/elle/on met | ils/elles mettent |
| past participle : mis |
Mettre literally means 'to place, to put.' It is also used in the following expressions:
| mettre la table (le couvert) , to set the table |
| mettre + article of clothing, to put on |
| mettre + electrical item (radio, light), to turn on |
Other verbs like mettre include:
| admettre , to admit |
| permettre , to permit |
| promettre , to promise |
| remettre , to turn in (a report), to postpone |
1. Les étudiants .... (mettre) des vêtements à la mode quand ils vont en boîte. 2. Le professeur .... (remettre) l'examen à la semaine prochaine. 3. Nous .... (admettre) Shasta à notre club.
- Answer
-
1. mettent 2. met 3. admettons
Qui
is used to ask questions about people. It may be the subject or the direct object in a sentence and thus, can mean either 'who?' or 'whom?' You may also choose to use the longer forms:
qui est-ce qui
to ask 'who?',
qui est-ce que
to ask 'whom?'
As the subject:
qui
,
qui est-ce qui
As the direct object:
qui
,
qui est-ce que
Examples: Qui vient à ma fête ? Qui est-ce qui vient à ma fête? (subject)
Qui vois-tu sur la photo ? Qui est-ce que tu vois sur le photo ? (direct object)
As the object of a preposition: prep + qui, prep + qui est-ce que
Examples: Avec qui vas-tu en France ? À qui est-ce que tu parles ?
1. Who likes strawberries ? 2. Whom did you go to the movies with ? 3. Whom do you respect ?
- Answer
-
1. Qui aime les fraises ? / Qui est-ce qui aime les fraises ? 2. Avec qui est-ce que tu vas au cinéma ? / Avec qui vas-tu au cinéma? 3. Qui est-ce que tu respectes ? Qui respectes-tu ?
Qu'est-ce qui
,
que
and its longer form
qu'est-ce que
are used to ask questions about things. They are used to ask 'what?'.
Que
and its longer form
qu'est-ce que
are direct objects in a sentence, whereas
qu'est-ce qui
may be used only as the subject. (There is no short form of
qu'est-ce qui
)
As the subject:
qu'est-ce qui
As the direct object:
que
,
qu'est-ce que
As the object of a preposition:
prep +
quoi, prep + quoi est-ce que
Examples: Qu'est-ce qui te fait peur ? (subject)
Que fais-tu demain ? Qu'est-ce que tu fais demain ? (direct object)
De quoi parles-tu ? De quoi est-ce que tu parles (object of a preposition)
1. What amuses you ? 2. What are you watching ? 3. What do you serve this salad with ?
- Answer
-
1. Qu'est-ce qui t'amuse ? 2. Que regardes-tu ? Qu'est-ce que tu regardes ? 3. Avec quoi sers-tu cette salade ? Avec quoi est-ce que tu sers cette salade?
The impersonal pronoun il ('it') is used in French when an action has no agent, that is, when there is no person or animate being responsible for the action. The conjugated verb is always in the third person singular, no matter what tense the impersonal verb takes.
Examples: Tiens, il pleut très fort!
The action of the verb ( pleut , 'is raining') is an impersonal, natural force. The impersonal pronoun il is often referred to as a 'dummy subject' because it fills the syntactic position of subject but doesn't have any real meaning.
Weather expressions in both French and English require impersonal subjects. The infinitive of 'weather verbs' can only be conjugated in the third person singular form (the il form).
| pleuvoir, to rain | Il pleut. | It's raining. |
| neiger, to snow | Il neige. | It's snowing. |
| grêler, to sleet | Il grêle. | It's sleeting. |
| geler, to freeze | Il gèle. | It's freezing. |
| bruiner, to drizzle | Il bruine. | It's drizzling. |
Weather conditions are also expressed in French using the verb faire followed by an adjective or noun. Of course, when the verb faire is used impersonally in such weather expressions, it can only be conjugated in the third person singular form ( il fait ).
| Il fait chaud. | It's hot. |
| Il fait du vent. | It's windy. |
| Il fait beau. | It's beautiful. |
The verb falloir only exists in the impersonal form ( il faut ). It always expresses the notion of necessity or obligation which is translated into English in various ways ('must,' 'should,' 'have to'). Falloir may be followed by an infinitive, by a noun, or by a subordinate clause introduced by que ; note that the verb in the subordinate clause in (a) requires the subjonctive mood.
Examples: Il faut parler français en classe. Pour ça, il faut du courage. Il faut que les étudiants soient courageux et persistants.
The impersonal subject
il
may appear with the verb
être
followed by an adjective and an infinitive. Note that the infinitive is always introduced by the prepostion
de
in such impersonal expressions:
Il est + [adjective] + de + [infinitive]
Examples: Il est amusant de faire des achats chez H & M.
Il est difficile de rester en bonne santé quand on mange beaucoup de sucre.
Other common impersonal expressions include:
Il y a
, there is, there are
Il est
+ clock time (
Il est deux heures
, It's two o'clock.)
Il s'agit de
, to be about, to be a matter of, to be a question of
Il vaut mieux
, to be better, to be advisable
1. ..... du vent aujourd'hui. 2. À UH .... environ 37,000 étudiants. 3. Ah, zut! .... déjà 23 heures, et je n'ai pas fini mes devoirs! 4. Pour obtenir un A, .... venir en classe et faire les devoirs.
- Answer
-
1. Il fait 2. Il y a 3. Il est 4. Il faut
Disjunctive pronouns (also known as tonic or stressed pronouns) refer to people whose names have already been mentioned or whose identity is obvious from context. They are used in a variety of situations in French, most often in short answers without verbs, for emphasis, or for contrast with subject pronouns. Here are all the disjunctive pronouns:
| disjunctive pronouns | |
| moi (I, me) | nous (we, us) |
| toi (you) | vous (you) |
|
lui
(he, him)
elle (she, her) soi (one) |
eux
(they, them; masc.)
elles (they, them; fem.) |
Disjunctive pronouns are used in the following cases:
1. after prepositions
Examples: Astérix, attends, je veux venir avec toi . Ne pars pas sans moi .
2. in short answers or exclamations when no verb is expressed
Examples: Qui va jouer avec Shasta ? - Moi !
3. with ni...ni, ne...que
Examples: Astérix, qui est-ce que tu admires : Lady Gaga ou Justin Bieber ? - Ni elle ni lui . Je n'admire que mon ami Obélix.
4. in a compound subject or object
Examples: Shasta et moi , nous sommes amateurs du foot.
5. in simple agreements or disagreements when no verb is expressed
Examples: Moi aussi! Pas lui . Lui non plus.
6. for emphasis
Examples: Eux , ils sont disciplinés, mais vous , vous êtes paresseux.
7. after c'est or ce sont
Examples: C'est moi le plus intelligent. Ce sont eux qui ne font jamais rien.
8. with même , to mean self
Examples: Fais toi -même les devoirs, ne compte pas sur Chat GPT.
9. in comparisons
Examples: Astérix est plus intelligent que moi . Obélix est plus fort que lui.
1. I am not sick. - Me neither. 2. You are stronger than him. 3. We took our warm clothes with us. 4. Did you take your green jacket with you ?
- Answer
-
1. Je ne suis pas malade. - Moi non plus. 2. Tu es plus fort que lui. 3. Nous avons pris nos vêtements chauds avec nous. 4. Est-ce que tu as pris ta veste verte avec toi ?
Aknowledgment: some parts of this page are partially adopted from Francais Interactif . | libretexts | 2025-03-17T22:26:32.535449 | 2023-08-17T17:16:15 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://human.libretexts.org/Courses/University_of_Houston/Improvisations_-_2/06%3A_Unite_5/6.03%3A_New_Page",
"book_url": "https://commons.libretexts.org/book/human-195014",
"title": "6.3: Grammaire - présentation",
"author": null
} |
https://human.libretexts.org/Courses/University_of_Houston/Improvisations_-_2/06%3A_Unite_5/6.04%3A_New_Page | 6.4: Grammaire - activités
Quelle est la conséquence ? Reliez les phrases de la colonne A aux phrases de la colonne B qui sont des conséquences logiques.
| A | B |
| Aujourd'hui c'est vendredi. | Le prof leur a permis de faire une pause de dix minutes. |
| Sasha a acheté une nouvelle robe. | Il lui a promis une soirée à Sweet Paris . |
| Les étudiants ont dit qu'ils étaient fatigués. | Les étudiants de UH mettent des T-shirts rouges. |
| Sasha avait envie de dîner avec Shasta. | Nous mettons de belles robes et des talons. |
| Nous sortons en boîte de nuit ce soir. | Elle l'a mise pour son anniversaire. |
Quels vêtements associez-vous avec :
- un look habillé ?
- un look branché ?
- un look décontracté ?
Maintenant, posez les questions suivantes à deux ou trois partenaires.
- Qu’est-ce que tu mets pour aller à un mariage?
- Qu’est-ce que tu mets quand tu sors avec des amis?
- Qu’est-ce que tu mets quand tu restes chez toi le week-end?
- Qu’est-ce que tu mets pour aller en cours?
Décidez si vos partenaires ont un look plutôt habillé, branché ou décontracté. Rapportez votre résultat à la classe, en écrivant deux ou trois phrases au tableau.
Modèle :
Mon partenaire ... a un look plutôt habillé parce qu'il met toujours un costume et une cravate pour aller en cours. Par contre, mes partenaires .... et .... sont plutôt décontractés parce qu'ils mettent des pyjamas et des tongs pour aller en cours.
Répondez aux questions suivantes avec votre partenaire. Ensuite, classez les questions en catégories dans le tableau.
- En général, qu’est-ce que tu fais le week-end ?
- Qui est-ce que tu trouves drôle dans la pop-culture américaine ?
- Avec qui est-ce que tu partages tes secrets ?
- Qui a gagné le championnat de football américain universitaire l’année dernière ?
- Qu’est-ce qui est rouge et liquide et que les Français adorent ?
- Sans quoi est-ce que tu ne peux pas étudier ?
| Sujet animé | Sujet inanimé | Objet direct animé | Objet direct inanimé | Objet de préposition animé | Objet de préposition inanimé |
Encore des questions. Répondez aux questions suivantes avec votre partenaire. Toutes les questions sont sur les célébrités. Décidez si les questions portent sur le sujet animé, l'objet direct animé, ou l'objet de préposition animé.
Modèle :
- Qui chante très bien ?
- A mon avis, Taylor Swift chante très bien.
- Qui s'habille très excentriquement ?
- Qui joue souvent les méchants dans les films?
- Qui est très généreux ?
- Qui a une mauvaise réputation ?
- Qui est-ce que tu admires pour son talent ?
- Qui est-ce que tu trouves moche ?
- Qui est-ce que tu ne connais pas bien ?
- Qui est-ce que tu voudrais rencontrer ?
- De qui est-ce que tu parles souvent avec tes amis ?
Encore des questions. Répondez aux questions suivantes avec votre partenaire. Toutes les questions sont sur toi et ta famille. Décidez si les questions portent sur le sujet inanimé, l'objet direct sujet inanimé ou l'objet de préposition inanimé.
- Qu'est-ce que tes parents aiment faire le weekend ?
- Qu'est-ce que tu aimes manger ?
- Qu'est-ce que tu aimes boire ?
- Qu'est-ce qui t'inspire dans la vie de tes parents ?
- Qu'est-ce qui te dérange dans ta maison ?
- Qu'est-ce qui te fait peur dans ton avenir ?
- De quoi est-ce que tu parles souvent avec tes parents ?
Pardon ? J'ai pas compris . Voici une conversation téléphonique entre Astérix et Obélix. Parfois, il y a des bruits sur la ligne et Astérix ne comprend pas les paroles de son ami. Quelle question doit-il poser ?
- Salut Astérix ! Ça va?
- Salut Obélix. Oui, bien, et toi ?
- Ça va. Ecoute, hier j'ai vu ...., tu le connais, n'est-ce pas ? C'est un prof très stricte.
Et tu sais, aujourd'hui je regarde .... C'est une émission très marrante!
Demain, il y a une fête chez .... Ils nous invitent, tu veux y aller? Moi, je veux bien y aller, parce que .... va être là. J'adore cette jeune fille. Quand je la vois, nous parlons toujours de ....., parce que nous adorons ce livre.
Ecoute, je te laisse, parce que je dois me préparer pour mon atelier. Faire .... m'amuse beaucoup, comme tu sais.
Interview. Choisissez une célébrité et imaginez que vous pouvez l’interviewer. Quelles questions est-ce que vous avez envie de lui poser? Utilisez obligatoirement : qui est-ce que, qui est-ce qui, qu’est-ce que, qu’est-ce qui, préposition+qui. préposition+quoi. Ecrivez 6 questions. Puis, comparez vos questions.
Vrai ou Faux ? Lisez les phrases suivantes et décidez si elles sont vraies ou fausses. Si une phrase est fausse, corrigez-la.
- Pour être en bonne santé, il faut boire un verre de vin rouge tous les jours.
- Il est amusant de jouer aux échecs.
- Pour mincir, il vaut mieux ne jamais manger de chocolat.
- Il est important de mettre un anorak quand il fait froid.
- Il est facile de faire du cheval tous les jours.
- Il ne faut pas fumer dans les avions.
- Pour se sentir bien, il est essentiel de faire de la musculation.
Vos recommandations aux nouveaux étudiants. Donnez des conseils aux nouveaux étudiants de UH. Complétez les phrases suivantes.
Modèle:
Pour être un bon étudiant à UH, il faut... => Pour être un bon étudiant à UH, il faut faire les devoirs et poser des questions aux professeurs.
- Pour combattre le stress, il ne faut pas ...
- Pour rester en bonne forme physique, il est utile de ...
- Pour venir aux cours à l'heure, il est nécessaire de ...
- Pour être à la mode, il faut ...
- Pour choisir de bons cours, il est préférable de ...
- Pour être ami avec Shasta, il est bon de ...
- Pour recevoir un A dans le cours de français, il vaut mieux ...
Devinettes. Devinez à qui se réfère le pronom souligné.
Modèle:
Le professeur parle français avec eux . = > Le prof parle français avec les étudiants .
- Quand j'étais petit, je partais en vacances avec eux .
- Brad Pitt a quitté Jeniffer Aniston pour elle .
- Lui , il est le mascot de UH.
- Elle , elle est la mascotte de UH.
- Eux , ils détestent les devoirs.
- Lui , il est le chien d'Obélix.
- Les enfants achètent des cadeaux pour elles en mai.
Et vous ? Posez les questions suivantes à un partenaire et comparez vos réponses.
Modèle:
- Je fume. Et toi? - Moi aussi, je fume. OU Pas moi, je ne fume pas.
- Je ne fume pas. Et toi? - Moi non plus, je ne fume pas. OU Moi si, je fume.
- J’aime voyager / Je n’aime pas voyager. Et toi ?
- Je danse chez moi / Je ne danse jamais chez moi. Et toi ?
- Je suis sorti(e) hier soir / Je ne suis pas sorti(e) hier soir. Et toi ?
- Je fais un sport extrême / Je ne fais pas de sport extrême. Et toi ?
- Je mange beaucoup de sucre / Je ne mange jamais de sucre. Et toi ?
- Imaginez une autre situation.
Aknowledgment: some activites on this page are partially adopted from Francais Interactif . | libretexts | 2025-03-17T22:26:32.621340 | 2023-08-17T17:16:15 | {
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"title": "6.4: Grammaire - activités",
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https://human.libretexts.org/Courses/University_of_Houston/Improvisations_-_2/06%3A_Unite_5/6.05%3A_New_Page | 6.5: Phonétique
In addition to oral vowels, French also has three nasal vowels (however, in some Francophone areas, there are four nasal vowels). Oral vowels are produced mainly within the oral cavity. Nasal vowels are produced when air passes through the nose as well as the mouth.
English has nasal-like vowels in words such as sing and impossible, but the nasal consonants /n/ and /m/ are still pronounced. These consonants are not pronounced in French when following a nasal vowel. The consonant is totally assimilated into the vowel pronunciation.
|
| ã | centre, pantalon, sandales, manteau, planche, randonnée, médicaments, ventre, jambe, branché | |
| ɛ̃ | moins, ceinture, bain, imperméable, alpin, peinture, main, mincir | |
| õ | talon, décontracté, tomber, bon, longue, son, pont, onze, content, maison |
ã
Lisez les paires de mots suivants:
gras/grand
chat/chanter
Jeanne/Jean
âne/an
ɛ̃
Lisez les paires de mots suivants:
certaine/certain
ancienne/ancien
aucune/aucun
õ
Lisez les paires de mots suivants:
peau/pont
mot/mont
beau/bon
Aknowledgment: some parts of this page are partially adopted from Francais Interactif . | libretexts | 2025-03-17T22:26:32.681681 | 2023-08-17T17:16:17 | {
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"title": "6.5: Phonétique",
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https://human.libretexts.org/Courses/University_of_Houston/Improvisations_-_2/06%3A_Unite_5/6.06%3A_New_Page | 6.6: Culture
Avant de visionner une vidéo culturelle, répondez aux questions suivantes :
- Qui sont des mannequins?
- Quels mannequins célèbres connaissez-vous ?
- Que pouvez-vous dire sur leur physique ? Comment sont-ils /elles ?
- Est-ce que vous rêviez de devenir mannequin quand vous étiez plus jeunes ?
Regardez maintenant la vidéo en prenant des notes, pour répondre aux questions suivantes:
1. Depuis quand existe la loi mannequins en France ?
A. Depuis 2016 B. Depuis 2007 C. Depuis 2017
2. Combien de fois par an les mannequins doivent avoir un rendez-vous avec le docteur ?
A. Une fois par an B. Deux fois par an C. Trois fois par an
3. Que veut dire être maigre ?
A. être très mince B. être très gros(se) C. être très beau (belle)
4. La mode, influence-t-elle uniquement les mannequins ?
A. Oui, elle ne concerne que les mannequins B. Non, elle concerne tout le monde
5. À quoi ressemblent les corps fabriqués par les marques ?
A. À des corps réels B. À des corps irréels C. À des corps réels, mais plus beaux qu'un corps moyen
6. Quelle taille a disparu en septembre 2017 ?
A. La taille 36 B. La taille 34 C. La taille 32
7. Est-ce qu’il existe le corps idéal, d'après cette vidéo ?
A. Oui, le corps idéal est un corps mince B. Non, la beauté a mille façons de s’exprimer. | libretexts | 2025-03-17T22:26:32.740103 | 2023-08-17T17:16:18 | {
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"title": "6.6: Culture",
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https://human.libretexts.org/Courses/University_of_Houston/Improvisations_-_2/06%3A_Unite_5/6.07%3A_New_Page | 6.7: Chansons et activités d'écoute
« Miss France », interprétée par Helmut Fritz.
Discutez les questions suivantes avec un partenaire:
1. Quelle est votre opinion sur les concours de beauté, comme Miss France , Miss USA , Miss Monde ? Est-ce que vous regardez ces concours à la télé?
2. Comment devrait être une reine de beauté à votre avis ? Décrivez son portrait physique et moral.
Maintenant, écoutez attentivement les belles paroles de cette chanson et écrivez les mots ou les expressions qui manquent.
Que 1. ........ faire plus tard ?
Le bien autour de moi, Jean-Pierre
Tu 2. ....... la foire agricole
Tu 3. ......... dans les écoles
T’as 4. ....... qui touchent plus le sol
Tu iras le dire sur Canal
Quoi de mieux qu’un Grand Journal
Pour crier « la guerre c’est mal »
Faut qu’tu sois 5. ..... mais y’a aussi c’que tu 6. .....
J’veux être Miss France
En 7. ......, tu parles de la crise en France
J’veux être Miss France
Que 8. ...... faire plus tard ?
Le bien autour de moi, Jean-Pierre
Tu feras Vogue et Marie-Claire
Des dédicaces dans les hypers,
9. ......... et 10. .......... seront fiers
T’iras au concours de Miss Monde
Tu seras 11. ........ parmi les blondes
Eliminée en vingt secondes
« And the winner is….. Miss America »
Faut qu’tu sois 12. ...... mais y’a aussi c’que tu 13. ......
J’veux être Miss France
En 14. ........ , tu parles de la crise en France
J’veux être Miss France
En 15. ......... du soir, tu prônes la tolérance
J’veux être Miss France
Les gens qui appellent c’est peut-être pas pour ta science
Non peut-être pas…
Un jour tu seras moitié nue
Sur la couverture d’Entrevue
C’est Geneviève qui sera déçue, la pauvre
Et tu te foutras comme il faut
D’Yves Rocher, de Frank Provost
Et de ta 307 Peugeot
Faut qu’tu sois 16. ...... mais y’a aussi c’que tu 17. .....
J’veux être Miss France
En 18. ...... du soir, tu prônes la tolérance
Je suis actuellement en BTS Esthéticienne dans le but de devenir…
… esthéticienne !
Discutez les questions suivantes avec un partenaire.
- Que portent les filles dans la vidéo ? Comment sont les filles : belles? minces ? intelligentes ? timides ?
- Que porte Helmut Fritz dans la vidéo ? Comment est Helmut Fritz: beau ? intelligent ? arrogant ? timide ?
- Quelle est l'idée principale de cette chanson, à votre avis ? Est-ce que vous pensez qu'être belle n'est pas suffisant pour une reine de beauté ? | libretexts | 2025-03-17T22:26:32.799214 | 2023-08-17T17:16:19 | {
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"title": "6.7: Chansons et activités d'écoute",
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https://human.libretexts.org/Courses/University_of_Houston/Improvisations_-_2/06%3A_Unite_5/6.08%3A_New_Page | 6.8: Projet
Travaillez en groupes de deux et préparez un jeu de rôle entre un médecin et un patient. Le médecin doit poser des questions au patient (au moins 8 questions) pour le diagnostiquer correctement et lui conseiller des médicaments appropriés. Le patient doit répondre aux questions en donnant beaucoup de détails sur sa santé. Utilisez le vocabulaire actif, mais aussi votre imagination et votre sens de l'humour.
Modèle:
- Bonjour Mademoiselle. Qu'est-ce qui ne va pas ?
- Bonjour docteur. Je suis très malade. J'ai mal partout: à la tête, à la gorge, dans la poitrine.
- Est-ce que vous avez de la fièvre?
- Oui, docteur. J’ai pris ma température ce matin et j’avais 38 degrès de fièvre.
- Est-ce que vous toussez beaucoup?
- Oui docteur. Et parfois j'éternue aussi.
- Très bien, mademoiselle. Vous avez les symptômes de la grippe. Vous devez vous reposer et prendre de l'aspirine pendant trois jours. Aussi, n'oubliez pas de boire beaucoup de thé au citron avec du miel.
- Merci, docteur. Est-ce que j'ai besoin d'une ordonnance pour l'aspirine ?
- Pas du tout, mademoiselle. Vous pouvez en acheter dans une pharmacie sans ordonnance.
- Merci, docteur.
- Je vous en prie. Bon rétablissement, portez-vous bien !
Aknowledgment: some parts of this page are partially adopted from Francais Interactif . | libretexts | 2025-03-17T22:26:32.854898 | 2023-08-17T17:16:20 | {
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"book_url": "https://commons.libretexts.org/book/human-195014",
"title": "6.8: Projet",
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https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/01%3A_The_Dawn_of_the_Quantum_Theory/1.01%3A_Blackbody_Radiation_Cannot_Be_Explained_Classically | 1.1: Blackbody Radiation Cannot Be Explained Classically
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Learning Objectives
One experimental phenomenon that could not be adequately explained by classical physics was blackbody radiation. Objectives for this section include
- Be familiar with black-body radiators
- Apply Stefan-Boltzmann’s Law to estimate total light output from a radiator
- Apply Wien’s Displacement Law to estimate the peak wavelength (or frequency) of the output from a black body radiator
- Understand the Rayleigh-Jeans Law and how it fails to properly model black-body radiation
All normal matter at temperatures above absolute zero emits electromagnetic radiation, which represents a conversion of a body's internal thermal energy into electromagnetic energy, and is therefore called thermal radiation . Conversely, all normal matter absorbs electromagnetic radiation to some degree. An object that absorbs ALL radiation falling on it, at all wavelengths, is called a blackbody. When a blackbody is at a uniform temperature, its emission has a characteristic frequency distribution that depends on the temperature. This emission is called blackbody radiation .
A room temperature blackbody appears black, as most of the energy it radiates is infra-red and cannot be perceived by the human eye. Because the human eye cannot perceive light waves at lower frequencies, a black body, viewed in the dark at the lowest just faintly visible temperature, subjectively appears grey, even though its objective physical spectrum peaks in the infrared range. When it becomes a little hotter, it appears dull red. As its temperature increases further it becomes yellow, white, and ultimately blue-white.
Blackbody radiation has a characteristic, continuous frequency spectrum that experimentally depends only on the body's temperature. In fact, we can be much more precise:
A body emits radiation at a given temperature and frequency exactly as well as it absorbs the same radiation.
This statement was proved by Gustav Kirchhoff: the essential point is that if we instead suppose a particular body can absorb better than it emits, then in a room full of objects all at the same temperature, it will absorb radiation from the other bodies better than it radiates energy back to them. This means it will get hotter, and the rest of the room will grow colder, contradicting the second law of thermodynamics. Thus, a body must emit radiation exactly as well as it absorbs the same radiation at a given temperature and frequency in order to not violate the second law of thermodynamics.
Any body at any temperature above absolute zero will radiate to some extent, the intensity and frequency distribution of the radiation depending on the detailed structure of the body. To begin analyzing heat radiation, we need to be specific about the body doing the radiating: the simplest possible case is an idealized body which is a perfect absorber, and therefore also (from the above argument) a perfect emitter. So how do we construct a perfect absorber in the laboratory? In 1859 Kirchhoff had a good idea: a small hole in the side of a large box is an excellent absorber, since any radiation that goes through the hole bounces around inside, a lot getting absorbed on each bounce, and has little chance of ever getting out again. So, we can do this in reverse : have an oven with a tiny hole in the side, and presumably the radiation coming out the hole is as good a representation of a perfect emitter as we’re going to find (Figure 1.1.2 ).
By the 1890’s, experimental techniques had improved sufficiently that it was possible to make fairly precise measurements of the energy distribution of blackbody radiation. In 1895, at the University of Berlin, Wien and Lummer punched a small hole in the side of an otherwise completely closed oven, and began to measure the radiation coming out. The beam coming out of the hole was passed through a diffraction grating, which sent the different wavelengths/frequencies in different directions, all towards a screen. A detector was moved up and down along the screen to find how much radiant energy was being emitted in each frequency range. They found a radiation intensity/frequency curve close to the distributions in Figure 1.1.3 .
By measuring the blackbody emission curves at different temperatures (Figure 1.1.3 ), they were also able to construct two important phenomenological Laws (i.e., formulated from experimental observations, not from basic principles of nature): Stefan-Boltzmann’s Law and Wien’s Displacement Law .
Not all radiators are blackbody radiators
The radiation of a blackbody radiator is produced by the thermal activity of the material, not the nature of the material, nor how it got thermally excited. Some examples of blackbodies include incandescent light bulbs, stars, and hot stove tops. The emission appears as a continuous spectrum (Figure 1.1.3 ) with multiple coexisting colors. However, not every radiator is a blackbody radiator. For example, the emission of a fluorescence bulb is not one. The following spectrum show the distribution of light from a fluoresce light tube and is a mixture of discrete bands at different wavelengths of light in contrast to the continuous spectra in Figure 1.1.3 for blackbody radiators.
Fluorescent light bulbs contain a mixture of inert gases (usually argon and neon) together with a drop of mercury at low pressure. A different mix of visible colors blend to produce a light that appears to us white with different shadings.
The Stefan-Boltzmann Law
The first quantitative conjecture based on experimental observations was the Stefan-Boltzmann Law (1879) which states the total power (i.e., integrated over all emitting frequencies in Figure 1.1.3 ) radiated from one square meter of black surface goes as the fourth power of the absolute temperature (Figure 1.1.4 ):
\[P = \sigma T^4 \label{Eq1} \]
where
- \(P\) is the total amount of radiation emitted by an object per square meter (\(Watts/ m^{2}\))
- \(\sigma\) is a constant called the Stefan-Boltzman constant (\(5.67 \times 10^{-8}\, Watts\; m^{-2} K^{-4}\))
- \(T\) is the absolute temperature of the object (in K)
The Stefan-Boltzmann Law is easily observed by comparing the integrated value (i.e., under the curves) of the experimental black-body radiation distribution in Figure 1.1.3 at different temperatures. In 1884, Boltzmann derived this \(T^4\) behavior from theory by applying classical thermodynamic reasoning to a box filled with electromagnetic radiation, using Maxwell’s equations to relate pressure to energy density. That is, the tiny amount of energy coming out of the hole (Figure 1.1.2 ) would of course have the same temperature dependence as the radiation intensity inside.
Example 1.1.1
The sun’s surface temperature is 5700 K.
- How much power is radiated by the sun?
- Given that the distance to earth is about 200 sun radii, what is the maximum power possible from a one square kilometer solar energy installation?
Solution
(a) First, we calculate the area of the sun followed by the flux (power). The sun has a radius of \( 6.96 \times 10^{8} m \)
The area of the sun is \( A = 4 \pi R^{2} \).
\[ \begin{align*} A &= 4 (3.1416)(6.96 \times 10^{8} m)^{2} \\[4pt] &= 6.08 \times 10^{18} m^2 \end{align*} \]
The power radiated from the sun (via Stefan-Boltzmann Law) is \(P = \sigma T^{4}\) (Equation \ref{Eq1}).
\[ \begin{align*} P &= (5.67 \times 10^{-8}\, Watts\; m^{-2} K^{-4})(5700 K)^{4} \\[4pt] &= 5.98 \times 10^{7} Watts/m^2 \end{align*} \]
This value is per square meter.
(b) To calculate the total power radiated by the sun is thus:
\[ \begin{align*} P_{total} &= P A = (5.98 \times 10^{7} Watts/m^2)( 6.08 \times 10^{18} m^2) \\[4pt] &= 3.6 \times 10^{26} Watts \end{align*} \]
Wien’s Displacement Law
The second phenomenological observation from experiment was Wien’s Displacement Law . Wien's law identifies the dominant (peak) wavelength, or color, of light coming from a body at a given temperature. As the oven temperature varies, so does the frequency at which the emitted radiation is most intense (Figure 1.1.3 ). In fact, that frequency is directly proportional to the absolute temperature:
\[\nu_{max} \propto T \label{Eq2} \]
where the proportionality constant is \(5.879 \times 10^{10} Hz/K\).
Wien himself deduced this law theoretically in 1893, following Boltzmann’s thermodynamic reasoning. It had previously been observed, at least semi-quantitatively, by an American astronomer, Langley . This upward shift in \(\nu_{max}\) with \(T\) is familiar to everyone—when an iron is heated in a fire (Figure 1.1.1 ), the first visible radiation (at around 900 K) is deep red, the lowest frequency visible light. Further increase in \(T\) causes the color to change to orange then yellow, and finally blue at very high temperatures (10,000 K or more) for which the peak in radiation intensity has moved beyond the visible into the ultraviolet.
Another representation of Wien's Law (Equation \(\ref{Eq2}\)) in terms of the peak wavelength of light is
\[\lambda_{max} = \dfrac{b}{T} \label{Eq2a} \]
where \(T\) is the absolute temperature in kelvin and \(b\) is a constant of proportionality called Wien's displacement constant , equal to \(2.89 \times 10^{−3} m\, K\), or more conveniently to obtain wavelength in micrometers, \(b≈2900\; μm \cdot K\). This is an inverse relationship between wavelength and temperature. So the higher the temperature, the shorter or smaller the wavelength of the thermal radiation. The lower the temperature, the longer or larger the wavelength of the thermal radiation. For visible radiation, hot objects emit bluer light than cool objects.
Example 1.1.2
If surface body temperature is 90 °F.
- How much radiant energy in \(W\, m^{-2}\) would your body emit?
- What is the peak wavelength of emitted radiation?
- What is the total radiant energy emitted by your body in Watts? Note: The average adult human male has a body surface area of about 1.9 \(m^2\) and the average body surface area for a woman is about 1.6 \(m^2\).
Solution
(a) 90 °F is 305 K. We use Stefan-Boltzmann Law (Equation \ref{Eq1}). The total amount of radiation emitted will be \( P = \sigma T^4 \).
\[ \begin{align*} P &= (5.67 \times 10^{-8}\, Watts\; m^{-2} K^{-4})(305 K)^4 \\[4pt] &= 491 W\, m^{-2} \end{align*} \]
The peak wavelength of emitted radiation is found using Wien's Law:
\[ \begin{align*} \lambda_{max} &= \frac{ 2.898 \times 10^{-3} m \cdot K}{T} \\[4pt] &= \frac{ 2.898 \times 10^{-3} m \cdot K}{305 K} \\[4pt] &= 9.5 \times 10^{-6} m = 9.5 \mu m\end{align*} \]
The total radiant energy density in Watts is :
\[ \begin{align*} \text{Energy}_{\text{male}} &= (491 W\, m^{-2})(1.9 m^{2}) = 933 W \\[4pt] \text{Energy}_{\text{female}} &= (491 W\, m^{-2})(1.6 m^{2}) = 786 W \end{align*} \]
Example 1.1.3 : The Temperature of the Sun
For example, if the Sun has a surface temperature of 5700 K, what is the wavelength of maximum intensity of solar radiation?
Solution
If we substitute 5700 K for \(T\) in Equation \(\ref{Eq2a}\), we have
\[\begin{align*} λ_{max} &= \dfrac{0.0029}{5700} \\[4pt] &= 5.1 \times 10^{-7} \, m \end{align*} \]
Knowing that violet light has a wavelength of about \(4.0 \times 10^{-7}\) meters, yellow about \(5.6 \times 10^{-7}\) meters, and red about \(6.6 \times 10^{-7}\) meters, what can we say about the color of the Sun's peak radiation? The peak wavelength of the Sun's radiation is at a slightly shorter wavelength than the color yellow, so it is a slightly greenish yellow. To see this greenish tinge to the Sun, you would have to look at it from space. It turns out that the Earth's atmosphere scatters some of the shorter waves of sunlight, which shifts its peak wavelength to pure yellow.
Remember that thermal radiation always spans a wide range of wavelengths (Figure 1.1.2 ) and Equation \ref{Eq2a} only specifies the single wavelength that is the peak of the spectrum. So although the Sun appears yellowish-white, when you disperse sunlight with a prism you see radiation with all the colors of the rainbow. Yellow just represents a characteristic wavelength of the emission.
Exercise 1.1.1
- At what wavelength does the sun emit most of its radiation if it has a temperature of 5,778 K?
- At what wavelength does the earth emit most of its radiation if it has a temperature of 288 K?
- Answer a
-
500 nm
- Answer b
-
10.0 microns
The Rayleigh-Jeans Law
Lord Rayleigh and J. H. Jeans developed an equation which explained blackbody radiation at low frequencies. The equation which seemed to express blackbody radiation was built upon all the known assumptions of physics at the time. The big assumption which Rayleigh and Jean implied was that infinitesimal amounts of energy were continuously added to the system when the frequency was increased . Classical physics assumed that energy emitted by atomic oscillations could have any continuous value. This was true for anything that had been studied up until that point, including things like acceleration, position, or energy. Their resulting Rayleigh-Jeans Law was
\[ \begin{align} d\rho \left( \nu ,T \right) &= \rho_{\nu} \left( T \right) d\nu \\[4pt] &= \dfrac{8 \pi k_B T}{c^3} \nu^2 d\nu \label{Eq3} \end{align} \]
Experimental data performed on the black box showed slightly different results than what was expected by the Rayleigh-Jeans law (Figure 1.1.5 ). The law had been studied and widely accepted by many physicists of the day, but the experimental results did not lie, something was different between what was theorized and what actually happens. The experimental results showed a bell type of curve, but according to the Rayleigh-Jeans law the frequency diverged as it neared the ultraviolet region (Equation \(\ref{Eq3}\)). Ehrenfest later dubbed this the "ultraviolet catastrophe". It is important to emphasizing that Equation \(\ref{Eq3}\) is a classical result: the only inputs are classical dynamics and Maxwell’s electromagnetic theory .
Differential vs. Integral Representation of the Distribution
Radiation is understood as a continuous distribution of amplitude vs. wavelength or, equivalently, amplitude vs. frequency (Figure 1.1.5 ). According to Rayleigh-Jeans law, the intensity at a specific frequency \(\nu\) and temperature is
\[\rho(\nu,T) = \dfrac{8\pi k_bT\nu^2}{c^3} \text{.} \nonumber \]
However, in practice, we are more interested in frequency intervals. An exact frequency is the limit of a sequence of smaller and smaller intervals. If we make the assumption that, for a sufficiently small interval, \(ρ(\nu,T)\) does not vary, we get your definition for the differential \(dρ(ν,T)\) in Equation \ref{Eq3}:
The assumption is fair due to the continuity of \(ρ(\nu,T)\). This is the approximation of an integral on a very small interval \(d\nu\) by the height of a point inside this interval (\(\frac{8\pi k_bT\nu^2}{c^3}\)) times its length (\(d\nu\)). So, if we sum an infinite amount of small intervals like the one above we get an integral. The total radiation between \(\nu_1\) and \(\nu_2\) will be:
\[ \begin{align*} \int_{\nu_1}^{\nu_2}\operatorname{d}\!\rho(\nu,T) &= \int_{\nu_1}^{\nu_2}\rho(\nu, T)\operatorname{d}\!\nu \\[4pt] &= \int_{\nu_1}^{\nu_2}\frac{8\pi k_bT\nu^2}{c^3}\operatorname{d}\!\nu \\[4pt] &= 8 \pi k_b T \frac{v_2^3 - v_1^3}{3 c^3}\text{.} \end{align*} \]
Observe that \(ρ(\nu,T)\) is quadratic in \(\nu\).
Example 1.1.4 : The Ultraviolet Catastrophe
What is the total spectral radiance of a radiator that follows the Rayleigh-Jeans law for its emission spectrum?
Solution
The total spectral radiance \(\rho_{tot}(T)\) is the combined emission over all possible wavelengths (or equivalently, frequencies), which is an integral over the relevant distribution (Equation \ref{Eq3} for the Rayleigh-Jeans Law).
\[ \begin{align*} \rho_{tot}(T) &= \int_0^\infty d\rho \left( \nu ,T \right) \\[4pt] &= \int_0^\infty \dfrac{8 \pi k_B T}{c^3} \nu^2 d\nu \end{align*} \]
but the integral
\[\int_0^\infty x^2\mathrm{d}x \nonumber \]
does not converge. Worse, it is infinite,
\[ \lim_{k\to\infty}\int_0^k x^2\mathrm{d}x = \infty \nonumber \]
Hence, the classically derived Rayleigh-Jeans law predicts that the radiance of a blackbody is infinite . Since radiance is power per angle and unit area, this also implies that the total power and hence the energy a blackbody emitter gives off is infinite, which is patently absurd. This is called the ultraviolet catastrophe because the absurd prediction is caused by the classical law not predicting the behavior at high frequencies/small wavelengths correctly (Figure 1.1.5 ).
Contributors and Attributions
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Michael Fowler (Beams Professor, Department of Physics , University of Virginia)
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David M. Hanson, Erica Harvey, Robert Sweeney, Theresa Julia Zielinski (" Quantum States of Atoms and Molecules ")
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Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).
- ACuriousMind (StackExchange) | libretexts | 2025-03-17T22:26:38.506152 | 2023-01-07T00:42:19 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/01%3A_The_Dawn_of_the_Quantum_Theory/1.01%3A_Blackbody_Radiation_Cannot_Be_Explained_Classically",
"book_url": "https://commons.libretexts.org/book/chem-424908",
"title": "1.1: Blackbody Radiation Cannot Be Explained Classically",
"author": null
} |
https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/01%3A_The_Dawn_of_the_Quantum_Theory/1.09%3A_The_Heisenberg_Uncertainty_Principle | 1.9: The Heisenberg Uncertainty Principle
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Learning Objectives
- To understand that sometime you cannot know everything about a quantum system as demonstrated by the Heisenberg uncertainly principle.
In classical physics, studying the behavior of a physical system is often a simple task due to the fact that several physical qualities can be measured simultaneously. However, this possibility is absent in the quantum world. In 1927 the German physicist Werner Heisenberg described such limitations as the Heisenberg Uncertainty Principle, or simply the Uncertainty Principle, stating that it is not possible to measure both the momentum and position of a particle simultaneously.
The Heisenberg Uncertainty Principle is a fundamental theory in quantum mechanics that defines why a scientist cannot measure multiple quantum variables simultaneously. Until the dawn of quantum mechanics, it was held as a fact that all variables of an object could be known to exact precision simultaneously for a given moment. Newtonian physics placed no limits on how better procedures and techniques could reduce measurement uncertainty so that it was conceivable that with proper care and accuracy all information could be defined. Heisenberg made the bold proposition that there is a lower limit to this precision making our knowledge of a particle inherently uncertain.
Probability
Matter and photons are waves, implying they are spread out over some distance. What is the position of a particle, such as an electron? Is it at the center of the wave? The answer lies in how you measure the position of an electron. Experiments show that you will find the electron at some definite location, unlike a wave. But if you set up exactly the same situation and measure it again, you will find the electron in a different location, often far outside any experimental uncertainty in your measurement. Repeated measurements will display a statistical distribution of locations that appears wavelike (Figure 1.9.1 ).
After de Broglie proposed the wave nature of matter, many physicists, including Schrödinger and Heisenberg, explored the consequences. The idea quickly emerged that, because of its wave character, a particle’s trajectory and destination cannot be precisely predicted for each particle individually . However, each particle goes to a definite place (Figure 1.9.1 ). After compiling enough data, you get a distribution related to the particle’s wavelength and diffraction pattern. There is a certain probability of finding the particle at a given location, and the overall pattern is called a probability distribution . Those who developed quantum mechanics devised equations that predicted the probability distribution in various circumstances.
It is somewhat disquieting to think that you cannot predict exactly where an individual particle will go, or even follow it to its destination. Let us explore what happens if we try to follow a particle. Consider the double-slit patterns obtained for electrons and photons in Figure 1.9.2 . The interferrence patterns build up statistically as individual particles fall on the detector. This can be observed for photons or electrons—for now, let us concentrate on electrons. You might imagine that the electrons are interfering with one another as any waves do. To test this, you can lower the intensity until there is never more than one electron between the slits and the screen. The same interference pattern builds up!
This implies that a particle’s probability distribution spans both slits, and the particles actually interfere with themselves. Does this also mean that the electron goes through both slits? An electron is a basic unit of matter that is not divisible. But it is a fair question, and so we should look to see if the electron traverses one slit or the other, or both. One possibility is to have coils around the slits that detect charges moving through them. What is observed is that an electron always goes through one slit or the other; it does not split to go through both.
But there is a catch. If you determine that the electron went through one of the slits, you no longer get a double slit pattern—instead, you get single slit interference. There is no escape by using another method of determining which slit the electron went through. Knowing the particle went through one slit forces a single-slit pattern. If you do not observe which slit the electron goes through, you obtain a double-slit pattern. How does knowing which slit the electron passed through change the pattern? The answer is fundamentally important— measurement affects the system being observed . Information can be lost, and in some cases it is impossible to measure two physical quantities simultaneously to exact precision. For example, you can measure the position of a moving electron by scattering light or other electrons from it. Those probes have momentum themselves, and by scattering from the electron, they change its momentum in a manner that loses information . There is a limit to absolute knowledge, even in principle.
Heisenberg’s Uncertainty Principle
It is mathematically possible to express the uncertainty that, Heisenberg concluded, always exists if one attempts to measure the momentum and position of particles. First, we must define the variable “x” as the position of the particle, and define “p” as the momentum of the particle. The momentum of a photon of light is known to simply be its frequency, expressed by the ratio \(h/λ\), where h represents Planck’s constant and \(\lambda\) represents the wavelength of the photon. The position of a photon of light is simply its wavelength (\(\lambda\)). To represent finite change in quantities, the Greek uppercase letter delta, or Δ, is placed in front of the quantity. Therefore,
\[\Delta{p}=\dfrac{h}{\lambda} \label{1.9.1} \]
\[\Delta{x}= \lambda \label{1.9.2} \]
By substituting \(\Delta{x}\) for \(\lambda\) into Equation \(\ref{1.9.1}\), we derive
\[\Delta{p}=\dfrac{h}{\Delta{x}} \label{1.9.3} \]
or,
\[\underset{\text{early form of uncertainty principle }}{\Delta{p}\Delta{x}=h} \label{1.9.4} \]
A Common Trend in Quantum Systems
Equation \(\ref{1.9.4}\) can be derived by assuming the particle of interest is behaving as a particle, and not as a wave. Simply let \(\Delta p=mv\), and \(Δx=h/(m v)\) (from De Broglie’s expression for the wavelength of a particle). Substituting in \(Δp\) for \(mv\) in the second equation leads to Equation \(\ref{1.9.4}\).
Equation \ref{1.9.4} was further refined by Heisenberg and his colleague Niels Bohr, and was eventually rewritten as
\[\Delta{p_x}\Delta{x} \ge \dfrac{h}{4\pi} = \dfrac{\hbar}{2} \label{1.9.5} \]
with \(\hbar = \dfrac{h}{2\pi}= 1.0545718 \times 10^{-34}\; m^2 \cdot kg / s\).
Equation \(\ref{1.9.5}\) reveals that the more accurately a particle’s position is known (the smaller \(Δx\) is), the less accurately the momentum of the particle in the x direction (\(Δp_x\)) is known. Mathematically, this occurs because the smaller \(Δx\) becomes, the larger \(Δp_x\) must become in order to satisfy the inequality. However, the more accurately momentum is known the less accurately position is known (Figure 1.9.2 ).
What is the Proper Definition of Uncertainty?
Equation \(\ref{1.9.5}\) relates the uncertainty of momentum and position. An immediate questions that arise is if \(\Delta x\) represents the full range of possible \(x\) values or if it is half (e.g., \(\langle x \rangle \pm \Delta x\)). \(\Delta x\) is the standard deviation and is a statistic measure of the spread of \(x\) values. The use of half the possible range is more accurate estimate of \(\Delta x\). As we will demonstrated later, once we construct a wavefunction to describe the system, then both \(x\) and \(\Delta x\) can be explicitly derived. However for now, Equation \ref{1.9.5} will work.
For example: If a problem argues a particle is trapped in a box of length, \(L\), then the uncertainly of it position is \(\pm L/2\). So the value of \(\Delta x\) used in Equation \(\ref{1.9.5}\) should be \(L/2\), not \(L\).
Example 1.9.1
An electron is confined to the size of a magnesium atom with a 150 pm radius. What is the minimum uncertainty in its velocity?
Solution
The uncertainty principle (Equation \(\ref{1.9.5}\)):
\[\Delta{p}\Delta{x} \ge \dfrac{\hbar}{2} \nonumber \]
can be written
\[\Delta{p} \ge \dfrac{\hbar}{2 \Delta{x}} \nonumber \]
and substituting \(\Delta p=m \Delta v \) since the mass is not uncertain.
\[\Delta{v} \ge \dfrac{\hbar}{2\; m\; \Delta{x}} \nonumber \]
the relevant parameters are
- mass of electron \(m=m_e= 9.109383 \times 10^{-31}\; kg\)
- uncertainty in position: \(\Delta x=150 \times 10^{-12} m \)
\[ \begin{align*} \Delta{v} &\ge \dfrac{1.0545718 \times 10^{-34} \cancel{kg} m^{\cancel{2}} / s}{(2)\;( 9.109383 \times 10^{-31} \; \cancel{kg}) \; (150 \times 10^{-12} \; \cancel{m}) } \\[4pt] &= 3.9 \times 10^5\; m/s \end{align*} \nonumber \]
Exercise 1.9.1
What is the maximum uncertainty of velocity the electron described in Example 1.9.1 ?
- Answer
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Infinity. There is no limit in the maximum uncertainty, just the minimum uncertainty.
Understanding the Uncertainty Principle through Wave Packets and the Slit Experiment
It is hard for most people to accept the uncertainty principle, because in classical physics the velocity and position of an object can be calculated with certainty and accuracy. However, in quantum mechanics, the wave-particle duality of electrons does not allow us to accurately calculate both the momentum and position because the wave is not in one exact location but is spread out over space. A "wave packet" can be used to demonstrate how either the momentum or position of a particle can be precisely calculated, but not both of them simultaneously. An accumulation of waves of varying wavelengths can be combined to create an average wavelength through an interference pattern: this average wavelength is called the "wave packet". The more waves that are combined in the "wave packet", the more precise the position of the particle becomes and the more uncertain the momentum becomes because more wavelengths of varying momenta are added. Conversely, if we want a more precise momentum, we would add less wavelengths to the "wave packet" and then the position would become more uncertain. Therefore, there is no way to find both the position and momentum of a particle simultaneously.
Several scientists have debated the Uncertainty Principle, including Einstein. Einstein created a slit experiment to try and disprove the Uncertainty Principle. He had light passing through a slit, which causes an uncertainty of momentum because the light behaves like a particle and a wave as it passes through the slit. Therefore, the momentum is unknown, but the initial position of the particle is known. Here is a video that demonstrates particles of light passing through a slit and as the slit becomes smaller, the final possible array of directions of the particles becomes wider. As the position of the particle becomes more precise when the slit is narrowed, the direction, or therefore the momentum, of the particle becomes less known as seen by a wider horizontal distribution of the light.
Example 1.9.2
The speed of a 1.0 g projectile is known to within \(10^{-6}\;m/s\).
- Calculate the minimum uncertainty in its position.
- What is the maximum uncertainty of its position?
Solution a
From Equation \(\ref{1.9.5}\), the \(\Delta{p_x} = m \Delta v_x\) with \(m=1.0\;g\). Solving for \(\Delta{x}\) to get
\[ \begin{align*} \Delta{x} &= \dfrac{\hbar}{2m\Delta v} \\[4pt] &= \dfrac{1.0545718 \times 10^{-34} \; m^2 \cdot kg / s}{(2)(0.001 \; kg)(10^{-6} \;m/s)} \\[4pt] &= 5.3 \times 10^{-26} \,m \end{align*} \nonumber \]
This negligible for all intents and purpose as expected for any macroscopic object.
Solution b
Unlimited (or the size of the universe). The Heisenberg uncertainty principles does not quantify the maximum uncertainty.
Exercise 1.9.2
Estimate the minimum uncertainty in the speed of an electron confined to a hydrogen atom within a diameter of \(1 \times 10^{-10} m\)?
- Answer
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We need to quantify the uncertainty of the electron in position. We can estimate that as \(\pm 5 \times 10^{-10} m\). Hence, substituting the relavant numbers into Equation \ref{1.9.5} and solving for \(\Delta v\) we get
\[\Delta v= 1.15 \times 10^6\, km/s \nonumber \]
Notice that the uncertainty is significantly greater for the electron in a hydrogen atom than in the magnesium atom (Example 1.9.1 ) as expected since the magnesium atom is appreciably bigger.
Heisenberg’s Uncertainty Principle not only helped shape the new school of thought known today as quantum mechanics, but it also helped discredit older theories. Most importantly, the Heisenberg Uncertainty Principle made it obvious that there was a fundamental error in the Bohr model of the atom. Since the position and momentum of a particle cannot be known simultaneously, Bohr’s theory that the electron traveled in a circular path of a fixed radius orbiting the nucleus was obsolete. Furthermore, Heisenberg’s uncertainty principle, when combined with other revolutionary theories in quantum mechanics, helped shape wave mechanics and the current scientific understanding of the atom.
Humor: Heisenberg and the Police
- Heisenberg get pulled over for speeding by the police. The officer asks him "Do you know how fast you were going?"
- Heisenberg replies, "No, but we know exactly where we are!"
- The officer looks at him confused and says "you were going 108 miles per hour!"
- Heisenberg throws his arms up and cries, "Great! Now we're lost!" | libretexts | 2025-03-17T22:26:38.915334 | 2023-01-07T00:42:36 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/01%3A_The_Dawn_of_the_Quantum_Theory/1.09%3A_The_Heisenberg_Uncertainty_Principle",
"book_url": "https://commons.libretexts.org/book/chem-424908",
"title": "1.9: The Heisenberg Uncertainty Principle",
"author": null
} |
https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.03%3A_The_Harmonic_Oscillator_Approximates_Molecular_Vibrations | 5.3: The Harmonic Oscillator Approximates Molecular Vibrations
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Learning Objectives
- Understand the utility and limits of using the quantum harmonic oscillator as a model for molecular vibrations
The quantum harmonic oscillator is one of the most important model systems in quantum mechanics. This is due in partially to the fact that an arbitrary potential curve \(V(x)\) can usually be approximated as a harmonic potential at the vicinity of a stable equilibrium point. Furthermore, it is one of the few quantum-mechanical systems for which an exact, analytical solution exists. Solving other potentials typically require either approximations or numerical approaches to identify the corresponding eigenstates and eigenvalues (i.e., wavefunctions and energies).
A general potential energy (\(V(x)\)) curve for a molecular vibration can be expanded as a Taylor series
\[V(x) = V(x_0) + \left. \dfrac {d V(x)}{d x} \right|_{x_0}^{x} (x - x_0) + \left. \dfrac {1}{2!} \dfrac {d^2 V(x)}{d x^2} \right|_{x_0}^{x} (x - x_0)^2 + \ldots + \left. \dfrac {1}{n!} \dfrac {d^n V(x)}{d x^n} \right|_{x_0}^{x} (x - x_0)^n \label{5.3.1} \]
It is important to note that this approximation is only good for \(x\) near \(x_0\), and that \(x_0\) stands for the equilibrium bond distance. \(V(x)\) is often (but not always) shortened to the cubic term and can be rewritten as
\[V(x) = \dfrac {1}{2} kx^2 + \dfrac {1}{6} \gamma x^3 \label{5.3.2} \]
where \(V(x_0) = 0\), \(k\) is the harmonic force constant (harmonic term), and \(\gamma\) is the first anharmonic term (i.e., cubic) . As Figure 5.3.2 demonstrates, the harmonic oscillator (red curve) is a good approximation for the exact potential energy of a vibration (blue curve).
Adding anharmonic perturbations to the harmonic oscillator (Equation \(\ref{5.3.2}\)) better describes molecular vibrations. Anharmonic oscillation is defined as the deviation of a system from harmonic oscillation, or an oscillator not oscillating in simple harmonic motion. Anharmonic oscillation is described as the restoring force is no longer proportional to the displacement. Adding the cubic term (Figure 5.3.2 ; green curve) improves the harmonic oscillation approximation especially under greater displacement from equilibrium.
Naturally, adding higher order anharmonic terms, like quartic terms (Figure \(\PageIndex{2; right}\)) improve the approximation. Almost all diatomics have experimentally determined potentials for their lowest energy states. \(\ce{H2}\), \(\ce{Li2}\), \(\ce{O2}\), \(\ce{N2}\), and \(\ce{F2}\) with terms up to \(n < 10\) determined of Equation \(\ref{5.3.1}\).
Figure 5.3.1 shows the the general potential with (numerically) calculated energy levels (\(E_0\), \(E_1\) etc.). \(D_o\) is the dissociation energy, which is different from the well depth \(D_e\). These vibrational energy levels of this plot can be calculated using the harmonic oscillator model (i.e., Equation 5.3.1 with the Schrödinger equation) and have the general form
\[ E_v = \left(v + \dfrac{1}{2}\right) v_e - \left(v + \dfrac{1}{2}\right)^2 v_e x_e + \left(v + \dfrac{1}{2}\right)^3 v_e y_e + \text{higher terms} \nonumber\]
where \( v \) is the vibrational quantum number and \( x_e\) and \( y_e\) are the first and second anharmonicity constants, respectively.
The \(v = 0\) level is the vibrational ground state. Because this potential is less confining than a parabola used in the harmonic oscillator, the energy levels become less widely spaced at high excitation (Figure 5.3.1 ; top of potential).
Limitations of the Harmonic Oscillator Model for Molecular Vibrations
The harmonic oscillation is a great approximation of a molecular vibration, but has key limitations:
- Due to equal spacing of energy, all transitions occur at the same frequency (i.e. single line spectrum). However experimentally many lines are often observed (called overtones).
- The harmonic oscillator does not predict bond dissociation; you cannot break it no matter how much energy is introduced.
Morse Potentials are better Approximations of Vibrational Motion
A more powerful approach than just "patching up" the harmonic oscillator solution with anharmonic corrections is to adopt a different potential (\(V(x)\)). One such approach is the Morse potential, named after physicist Philip M. Morse, and a better approximation for the vibrational structure of the molecule than the harmonic oscillator because it explicitly includes the effects of bond breaking and accounts for the anharmonicity of real bonds (Figure 5.3.4 ).
The Morse Potential is a good approximation to \(V(x)\) and is best when looking for a general formula for all \(x\) from 0 to \(\infty\), not just applicable for the local region around the \(x_o\):
\[V(x) = D(1-e^{-\beta (x - x_0)})^2 \label{5.3.8} \]
with \(V(x = x_0) = 0\) and \(V(x = \infty) = D\).
The Morse Potential (Figure 5.3.4 ) approaches zero at infinite \(r_e\) and equals \(-D_e\) at its minimum (i.e. \(r=r_e\)). It clearly shows that the Morse potential is the combination of a short-range repulsion term (small \(r\) values) and a long-range attractive term (large \(r\) values).
Solving the Schrödinger Equation with the Morse Potential (Equation \ref{5.3.8}) is not trivial, but can be done analytically.
\[ \hat{H}|\psi \rangle = E_n | \psi \rangle \nonumber \]
with
\[ \begin{align*} \hat{H} &= \hat{T} + \hat{V} \\[4pt] &= \dfrac{- \hbar ^2 d^2}{2m \;dx^2} + D(1-e^{-\beta (x - x_0)})^2 \end{align*} \]
The solutions and energies for the Morse potential will not be used in this course and will not be discussed in more detail. | libretexts | 2025-03-17T22:26:39.725495 | 2023-01-07T00:43:12 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.03%3A_The_Harmonic_Oscillator_Approximates_Molecular_Vibrations",
"book_url": "https://commons.libretexts.org/book/chem-424908",
"title": "5.3: The Harmonic Oscillator Approximates Molecular Vibrations",
"author": null
} |
https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_Hermite_Polynomials_are_either_Even_or_Odd_Functions | 5.6: Hermite Polynomials are either Even or Odd Functions
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Learning Objectives
- Understand key properties of the Hermite polynomials including orthogonality and symmetry.
- Be proficient at using symmetries of integrands to quickly solve integrals.
Hermite polynomials were defined by Laplace (1810) though in scarcely recognizable form, and studied in detail by Chebyshev (1859). Chebyshev's work was overlooked and they were named later after Charles Hermite who wrote on the polynomials in 1864 describing them as new. They were consequently not new although in later 1865 papers Hermite was the first to define the multidimensional polynomials. The first six Hermite polynomial are plotted in Figure 5.7.1 .
Generating Formula
Any Hermite polynomial \(H_n(x)\) can be generated from a previous one \(H_{n-1}(x)\) via the following using the recurrence relation
\[ H_{n+1} (x)=2xH_n (x)-2nH_{n-1} (x). \label{5.7.2} \]
Hermite Polynomials are Symmetric
Let \(f(x)\) be a real-valued function of a real variable.
- Then \(f\) is even if the following equation holds for all x and -x in the domain of f \[f(x) = f(-x) \nonumber \]
- Then \(f\) is odd if the following equation holds for all x and -x in the domain of f \[-f(x) = f(-x) \nonumber \]
Even and odd are terms used to describe particularly well-behaved functions. An even function is symmetric about the y-axis (Figure 5.7.2 ; left). That is, if we reflect the graph of the function in the \(y\)-axis, then it does not change. Formally, we say that \(f\) is even if, for all \(x\) and \(−x\) in the domain of \(f\), we have
\[f(-x)=f(x) \nonumber \]
Two examples of even functions are \(f(x)=x^2\) and \(f(x)=\cos x\).
An odd function has rotational symmetry of order two about the origin (Figure 5.7.2 ; middle). That is, if we rotate the graph of the function 180° about the origin, then it does not change. Formally, we say that ff is odd if, for all \(x\) and \(−x\) in the domain of \(f\), we have
\[f(-x)=-f(x) \nonumber \]
Examples of odd functions are \(f(x)=x^3\) and \(f(x)=\sin x\).
Naturally, not all functions can be classified as even or odd. For example \(f=x^3+1\) shown in the right side of Figure 5.7.2 , is neither.
You can also think of these properties as symmetry conditions at the origin. More symmetries in 3D space are discussed in Group Theory.
Without proof, we can identify several key features involving multiplication properties of even and odd functions:
- The product of two even functions is an even function.
- The product of two odd functions is an even function.
- The product of an even function and an odd function is an odd function.
This can be shown graphically as a product table like that in Table 5.7.1 .
| Product table | Odd Function (anti-symmetric) | Even Function (symmetric) | No symmetry (neither) |
|---|---|---|---|
| Odd Function (anti-symmetric) | Even Function (symmetric) | Odd Function (anti-symmetric) | who knows |
| Even Function (symmetric) | Odd Function (anti-symmetric) | Even Function (symmetric) | who knows |
| No symmetry (neither) | who knows | who knows | who knows |
Notice that the Hermite polynomials in Figure 5.7.1 oscillate from even to odd. We can take advantage of that aspect in our calculation of Harmonic Oscillator wavefunctions. Hermite Polynomial is an even or odd function depends on its degree \(n\). Based on
\[H_n(-x) = (-1)^n H_n(x) \label{5.7.3} \]
- \(H_n(x)\) is an even function, when \(n\) is even.
- \(H_n(x)\) is an odd function, when \(n\) is odd.
Integration over Symmetric Functions
You often consider integrals of the form
\[I=\int_{-a}^a f(x)\,\mathrm{d}x \nonumber \]
If \(f\) is odd or even, then sometimes you can make solving this integral easier. For example, we can rewrite that integral in the following way:
\[\begin{align} I=\int_{-a}^a f(x)\,\mathrm{d}x &= \int_{-a}^0 f(x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \\ &= \int_0^a f(-x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \label{intsym} \end{align} \]
For an even function, we have \(f(-x)=f(x)\) and Equation \ref{intsym} can be simplified
\[\begin{align*} I &= \int_0^a f(-x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \\[4pt] &=2\int_0^a f(x)\,\mathrm{d}x \end{align*} \]
For an odd function, we have \(f(-x)=-f(x)\) and Equation \ref{intsym} can be simplified
\[\begin{align*} I &= -\int_0^a f(x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \\[4pt] &=0 \end{align*} \]
That’s what it means to simplify the integration: the integral of an odd or even function over the interval \([−L,L]\) can be put into a nicer form (and sometimes we can see that it vanishes without ever computing an integral).
Example 5.7.1
Technically, evaluating the orthogonality of Hermite polynomials requires integrating over the \(\exp(-x^2)\) weight function (Equations \(\ref{1}\) and \(\ref{2}\)).
Solution
For the Hermite polynomials \(H_n(x)\), the relevant inner product (using Dirac Notation)
\[ \langle f,g \rangle=\int_{-\infty}^\infty f(x)g(x)\color{red}{\exp(-x^2)}\,\mathrm dx \nonumber \]
While the \(H_2(x)H_3(x)\) product is indeed an odd function (Table 5.7.1 ), while \(exp(−x^2)\) is even. Their product is odd, and thus \(\langle f,g \rangle\) certainly ought to be zero.
Symmetry is an important aspect of quantum mechanics and mathematics, especially in calculating integrals. Using this symmetry, integrals can be identified to be equal to zero without explicitly solving them. For example, the integral of an odd integrand over all possible values will always be zero irrespective of the exact nature of the function:
\[ \int_{-\infty}^{\infty} f(x) \, dx= 0 \nonumber \]
This simplifies calculations greatly as demonstrated in the following chapters.
Hermite Polynomials are Orthogonal
Hermite polynomials \(H_n(x)\) are n th-degree polynomials for n = 0, 1, 2, 3 and form an orthogonal set of functions for the weight function \(e^{-x^2/2}\). The exact relation is:
\[ \int_{-\infty}^{\infty} H_m(x)H_n(x) e^{-x^2/2} dx = 0 \label{1} \]
if \(m \neq n\) and
\[ \int_{-\infty}^{\infty} H_m(x)H_n(x) e^{-x^2/2} dx = 2^n n! \sqrt{\pi} \label{2} \]
if \(m = n\).
This will not be proved, but can the demonstrated using any of the Hermite polynomials listed in the previous section. The orthogonality property becomes important when solving the Harmonic oscillator problems. Note that the integral of Equation \ref{2} is important for normalizing the quantum harmonic oscillator wavefunctions discussed in last Section.
Example 5.7.2 : Hermite Polynomials are Orthogonal
Demonstrate that \(H_2(x)\) and \(H_3(x)\) are orthogonal.
Solution
We need to confirm
\[\int_{-\infty}^{\infty}H_2(x)H_3(x) dx=0 \nonumber \]
or when substituted
\[\int_{-\infty}^{\infty} (4x^2-2)(8x^3-12x) dx=0 \nonumber \]
because it says I need to show it's orthogonal on \( [ -\infty, \infty ] \) or we can just evaluate it on a finite interval \([−L,L]\), where \(L\) is a constant.
\[ \begin{align*} \int_{-L}^{L} (4x^2-2)(8x^3-12x) dx &=\left. 8 \left(\dfrac{2 x^6}{3}-2 x^4+\dfrac{3 x^2}{2}\right)\right|_{-L}^{L} \\[4pt] &= 8 \left(\dfrac{2 L^6}{3}-2 L^4+\frac{3 L^2}{2}\right)-8 \left(\dfrac{2 (-L)^6}{3}-2 (-L)^4+\dfrac{3 (-L)^2}{2}\right) \\[4pt] &=0. \end{align*} \]
Concluding
Hermite polynomials are a component in the harmonic oscillator wavefunction that dictates the symmetry of the wavefunctions. If your integration interval is symmetric around 0, then the integral over any integrable odd function is zero, no exception. Therefore as soon as you've found that your integrand is odd and your integration interval is symmetric, you're done. Also, for general functions, if you can easily split them into even and odd parts, you only have to consider the integral over the even part for symmetric integration intervals.
Another important property is that the product of two even or of two odd functions is even, and the product of an even and an odd function is odd. For example, if ff is even, \(x↦f(x)\sin(x)\) is odd, and therefore the integral over it is zero (provided it is well defined). | libretexts | 2025-03-17T22:26:39.871365 | 2023-01-07T00:43:20 | {
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"url": "https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_Hermite_Polynomials_are_either_Even_or_Odd_Functions",
"book_url": "https://commons.libretexts.org/book/chem-424908",
"title": "5.6: Hermite Polynomials are either Even or Odd Functions",
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https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/07%3A_Multielectron_Atoms | 7: Multielectron Atoms
Electrons with more than one atom, such as Helium (He), and Nitrogen (N), are referred to as multi-electron atoms. Hydrogen is the only atom in the periodic table that has one electron in the orbitals under ground state. We will learn how additional electrons behave and affect a certain atom.
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- 7.1: Perturbation Theory and the Variational Method for Helium
- Both perturbation theory and variation method (especially the linear variational method) provide good results in approximating the energy and wavefunctions of multi-electron atoms. We address both approximations with respect to the helium atom.
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- 7.2: Hartree-Fock Equations are Solved by the Self-Consistent Field Method
- The Hartree method is used to determined the wavefunction and the energy of a quantum multi-electron system in a stationary state. The Hartree method often assumes that the exact, N-body wave function of the system can be approximated by a product of single-electron wavefucntions. By invoking the variational method, one can derive a set of N-coupled equations for the N spin orbitals.
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- 7.3: An Electron has an Intrinsic Spin Angular Momentum
- Spin is one of two types of angular momentum in quantum mechanics, the other being orbital angular momentum. The orbital angular momentum operator is the quantum-mechanical counterpart to the classical angular momentum of orbital revolution. The existence of spin angular momentum is inferred from the Stern–Gerlach experiment, in which particles are observed to possess angular momentum that cannot be accounted for by orbital angular momentum alone.
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- 7.4: Wavefunctions must be Antisymmetric to Interchange of any Two Electrons
- The probability |Ψ(r1, r2)|² should be identical to the probability |Ψ(r2, r1)|² because the electrons have no label and they cannot be told apart because of Heisenberg principle. You can naively think that Ψ(r1, r2)=±Ψ(r2, r1) but it turns out that the sign must always be minus for the electrons. This is an additional postulate of quantum mechanics.
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- 7.5: Hartree-Fock Calculations Give Good Agreement with Experimental Data
- The Hartree–Fock method is a method of approximation for the determination of the wave function and the energy of quantum many-body systems. The Hartree–Fock method often assumes that the exact, N-body wave function of the system can be approximated by a single Slater determinant of N spin-orbitals. By invoking the variational method, one can derive a set of N-coupled equations for the N spin orbitals. A solution of these equations yields the Hartree–Fock wave function and energy of the system.
Thumbnail: Neon Atom. (CC BY 3.0 Unported; BruceBlaus via Wikipedia ) | libretexts | 2025-03-17T22:26:40.456010 | 2023-01-07T00:43:36 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/07%3A_Multielectron_Atoms",
"book_url": "https://commons.libretexts.org/book/chem-424908",
"title": "7: Multielectron Atoms",
"author": null
} |
https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/07%3A_Multielectron_Atoms/7.01%3A_Perturbation_Theory_and_the_Variational_Method_for_Helium | 7.1: Perturbation Theory and the Variational Method for Helium
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Learning Objectives
- Demonstrate that both perturbation theory and variational methods can be used to solve the electron structure of the helium atom.
Both perturbation theory and variation method (especially the linear variational method) provide good results in approximating the energy and wavefunctions of multi-electron atoms. Below we address both approximations with respect to the helium atom.
Perturbation Theory of the Helium Atom
We use perturbation theory to approach the analytically unsolvable helium atom Schrödinger equation by focusing on the Coulomb repulsion term that makes it different from the simplified Schrödinger equation that we have just solved analytically. The electron-electron repulsion term is conceptualized as a correction, or perturbation, to the Hamiltonian that can be solved exactly, which is called a zero-order Hamiltonian . The perturbation term corrects the previous Hamiltonian to make it fit the new problem. In this way the Hamiltonian is built as a sum of terms, and each term is given a name. For example, we call the simplified or starting Hamiltonian, \(\hat {H} ^0\), the zero order term, and the correction term \(\hat {H} ^1\).
\[ \hat {H} = \hat {H} ^0 + \hat {H} ^1 \label {9-17} \]
The Hamilonian for the helium atom (in atomic units) is:
\[\begin{align} \hat {H} ^0 &= \underbrace{-\dfrac {1}{2} \nabla ^2_1 - \dfrac {2}{r_1}}_{\text{H atom Hamiltonian}} - \underbrace{\dfrac {1}{2} \nabla ^2_2 - \dfrac {2}{r_2}}_{\text{H atom Hamiltonian}} \label {9-18} \\[4pt] \hat {H} ^1 &= \dfrac {1}{r_{12}} = \dfrac{1}{|r_1-r_2|} \label {9-19} \end{align} \]
The expression for the first-order correction to the energy is
\[ \begin{align} E^1 &= \langle \psi ^{0} | \hat {H} ^1 | \psi ^{0} \rangle \nonumber \\[4pt] &= \int \psi ^{0*} \hat {H} ^1 \psi ^0 \,d\tau \nonumber \end{align} \label {9-28} \]
Equation \(\ref{9-28}\) is a general expression for the first-order perturbation energy, which provides an improvement or correction to the zero-order energy we already obtained. Hence, \(E^1\) is the average interaction energy of the two electrons calculated using wavefunctions that assume there is no interaction.
The solution to \(\hat{H}^{0}\) (Equation \ref{9-18}) is the product of two single-electron hydrogen wavefunctions (scaled by the increased nuclear charge) since \(\hat{H}^{0}\) can be separated into independent functions of each electron (i.e., Separation of Variables ).
\[ | \psi ^{0} \rangle = | \varphi _{1s} (r_1) \varphi _{1s} (r_2) \rangle \nonumber \]
So the integral in Equation \(\ref{9-28}\) is
\[ E^1 = \iint \varphi _{1s} (r_1) \varphi _{1s} (r_2) \dfrac {1}{r_{12}} \varphi _{1s} (r_1) \varphi _{1s} (r_2)\, d\tau _1 d\tau _2 \label {9-29} \]
where the double integration symbol represents integration over all the spherical polar coordinates of both electrons \(r_1, \theta _1, \varphi _1 , r_2 , \theta _2 , \varphi _2\). The evaluation of these six integrals is lengthy. When the integrals are done, the result is \(E^1\) = +34.0 eV so that the total energy calculated using our second approximation method, first-order perturbation theory, is
\[ E_{approx2} = E^0 + E^1 = - 74.8 eV \label {9-30} \]
The new approximate value for the binding energy represents a substantial (~30%) improvement over the zero-order energy:
\[E^{0} = \dfrac{2}{n^2} + \dfrac{2}{n^2} = 4\, \underbrace{E_h}_{hartrees} = 108.8\, eV \nonumber \]
so the interaction of the two electrons is an important part of the total energy of the helium atom. We can continue with perturbation theory and find the additional corrections, \(E^2\), \(E^3\), etc. For example,
\[E^0 + E^1 + E^2 = -79.2\, eV. \nonumber \]
So with two corrections to the energy, the calculated result is within 0.3% of the experimental value of -79.01 eV. It takes thirteenth-order perturbation theory (adding \(E^1\) through \(E^{13}\) to \(E^0\)) to compute an energy for helium that agrees with experiment to within the experimental uncertainty. Interestingly, while we have improved the calculated energy so that it is much closer to the experimental value, we learn nothing new about the helium atom wavefunction by applying the first-order perturbation theory to the energy above. He need to expand the wavefunctions to first order perturbation theory, which requires more effort. Below, we will employ the variational method approximation to modify zero-order wavefunctions to address one of the ways that electrons are expected to interact with each other.
The Hartree Unit of Energy
The hartree is the atomic unit of energy (named after the British physicist Douglas Hartree) and is defined as
\[E_h=2R_Hhc \nonumber \]
where \(R_H\) is the Rydberg constant, \(h\) is the Planck constant and \(c\) is the speed of light.
\[\begin{align} E_h &= 4.359 \times 10^{−18} \,J \nonumber \\[4pt] &= 27.21\, eV. \nonumber \end{align} \nonumber \]
The hartree is usually used as a unit of energy in atomic physics and computational chemistry. As discussed before for hydrogen emission, IR, and microwave spectroscopies, experimental measurements prefer the electronvolt (\(eV\)) or the wavenumber (\(cm^{−1}\)).
Variational Method Applied to the Helium Method
As discussed in Section 6.7, because of the electron-electron interactions, the Schrödinger's Equation cannot be solved exactly for the helium atom or more complicated atomic or ionic species. However, the ground-state energy of the helium atom can be estimated using approximate methods. One of these is the variational method which requires the minimizing of the following variational integral.
\[\begin{align} E_{trial} &= \dfrac{\langle \psi_{trial}| \hat{H} | \psi_{trial} \rangle }{\langle \psi_{trial}| \psi_{trial} \rangle}\label{7.3.1b} \\[4pt] &= \dfrac{\displaystyle \int_0^{\infty} \psi_{trial}^* \hat{H} \psi_{trial} d\tau}{\displaystyle \int_0^{\infty} \psi_{trial}^2 \, d\tau} \label{7.3.1a} \end{align} \]
The five trial wavefunctions discussions below are equally "valid" trial wavefunctions that describe the probability of finding each electron (technically the wavefunction squared). What separates the "poor" approximations from the "good" approximation is whether the trial wavefunction predicts experimental results. Consequently, for all the approximations used for the rest of this TextMap, it is important to compare the theoretical results to the "true" (i.e., experimental) results. No matter how complicated an approximation is, it is only as good as the accuracy of its predicted values to experimental values.
Trial Wavefunction #1: Simple Orbital Approximation with One Parameter
As is clear from Equation \(\ref{7.3.1b}\), the variational method approximation requires that a trial wavefunction with one or more adjustable parameters be chosen. A logical first choice for such a multi-electron wavefunction would be to assume that the electrons in the helium atom occupy two identical, but scaled, hydrogen 1s orbitals.
\[ \begin{align} | \psi (1,2) \rangle_{trial} &= \phi (1) \phi (2) \\[4pt] &= \exp\left[- \alpha (r_1 +r_2)\right] \label{7.3.2} \end{align} \]
The variational energy obtained after minimizing Equation \(\ref{7.3.1a}\) after substituting the trial wavefunction (Equation \ref{7.3.2}) by varying \(\alpha\) is
\[E_{trial} = -2.84766 \;E_h \nonumber \]
and the experimentally determined ground-state energy for the helium atom is the sum of first and second ionization energies
\[E_{\exp}= I_1 +I_2 = -2.90372 \;E_h \label{exp} \]
The deviation of energy for the optimized trial wavefunction from the experimental value is
\[ \begin{align} \left| \dfrac{E_{trial}(\alpha)-E_{\exp}}{E_{\exp}} \right| &= \left| \dfrac{-2.84766 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right| \\[4pt] &= 1.93 \% \label{trial1} \end{align} \]
The value of -2.8477 hartrees is within 2% of the known ground-state energy of the helium atom. The error in the calculation is attributed to the fact that the wavefunction is based on the orbital approximation and, therefore, does not adequately take electron-electron interactions into account. In other words, this wavefunction gives the electrons too much independence, given that they have like charges and tend to avoid one another.
Trial Wavefunction #2: Orbital Approximation with Two Parameters
Some electron-electron interactions can be built into the multi-electron wavefunction by assuming that each electron is in an orbital which is a linear combination of two different and scaled hydrogen 1s orbitals.
\[\phi(r_1) = \exp(- \alpha r_1) + \exp(- \beta r_1) \label{7.3.3} \]
Under the orbital approximation this assumption gives a trial wavefunction of the form
\[ \begin{align} | \psi (1,2) \rangle_{trial} &= \phi (1) \phi (2) \label{7.3.4a} \\[4pt] &= {\exp(- \alpha r_1 )\exp(- \alpha r_2)}+\exp(- \alpha r_1 )\exp(- \beta r_2)+\exp(- \beta r_1 )\exp(- \alpha r_2 )+\exp(- \beta r_1 )\exp(- \beta r_2 ) \label{7.3.4b} \end{align} \]
Inspection of this trial wavefunction indicates that 50% of the time the electrons are in different orbitals, while for the first trial wavefunction the electrons were in the same orbital 100% of the time. Notice the enormous increase in the complexity of the variational expression for the energy for this trial wavefunction (Equation \(\ref{7.3.1a}\)). However, the calculation is very similar to that using the previous trial wavefunction. The differences are that in this case the expression for the energy is more complex and that it is being minimized simultaneously with respect to two parameters (\(\alpha\) and \(\beta\)) rather than just one (\(\alpha\)).
The variational energy obtained after minimizing Equation \(\ref{7.3.1a}\) after substituting the trial wavefunction (Equation \ref{7.3.4b}) by varying \(\alpha\) and \(\beta\) is
\[E_{trial} =-2.86035 \;E_h \nonumber \]
The deviation of energy for the optimized trial wavefunction from the experimental value (Equation \ref{exp}) is
\[ \begin{align} \left| \dfrac{E_{trial}(\alpha, \beta)-E_{\exp}}{E_{\exp}} \right| &= \left| \dfrac{-2.86035 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right| \\[4pt] &= 1.49 \% \label{trial2} \end{align} \]
Clearly introducing some electron-electron interactions into the trial wavefunction has improved the agreement between theory and experiment (Equation \ref{trial1} vs. \ref{trial2}).
Trial Wavefunction #3: Orbital Approximation with Two Parameters
The extent of electron-electron interactions can be increased further by eliminating the first and last term in the second trial wavefunction (Equation \(\ref{7.3.4b}\)). This yields a multi-electron wavefunction of the form,
\[ | \psi (1,2) \rangle_{trial} = \exp(- \alpha r_1 )\exp(- \beta r_2 ) + \exp(- \beta r_1 )\exp(- \alpha r_2 ) \label{7.3.5} \]
This trial wavefunction places the electrons in different scaled hydrogen 1s orbitals 100% of the time this adds further improvement in the agreement with the literature value of the ground-state energy is obtained. The variational energy obtained after minimizing Equation \(\ref{7.3.1a}\) after substituting the trial wavefunction (Equation \ref{7.3.5}) by varying \(\alpha\) and \(\beta\) is
\[ E_{trial} = -2.87566 \;E_h \nonumber \]
The deviation of energy for the optimized trial wavefunction from the experimental value (Equation \ref{exp}) is
\[ \begin{align} \left| \dfrac{ E_{trial} (\alpha,\beta)-E_{\exp}}{E_{\exp}} \right| &= \left| \dfrac{-2.87566 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right| \\[4pt] &= 0.97 \%\label{trial3} \end{align} \]
This result is within 1% of the actual ground-state energy of the helium atom.
Trial Wavefunction #4: Approximation with Two Parameters
The third trial wavefunction, however, still rests on the orbital approximation and, therefore, does not treat electron-electron interactions adequately. Hylleraas took the calculation a step further by introducing electron-electron interactions directly into the first trial wavefunction by adding a term, \(r_{12}\), involving the inter-electron separation.
\[| \psi_{trial} (1,2) \rangle = \left(\exp[- \alpha ( r_1 + r_2 )]\right) \left(1 + \beta r_{12} \right) \label{7.3.6} \]
In the trial multi-electron wavefunction of Equation \ref{7.3.6}, if the electrons are far apart, then \(r_{12}\) is large and the magnitude of the wavefunction increases to favor that configuration. The variational energy obtained after minimizing Equation \(\ref{7.3.1a}\) after substituting the trial wavefunction (Equation \ref{7.3.6}) by varying \(\alpha\) and \(\beta\) is
\[ E_{trial} = - 2.89112\; E_h \nonumber \]
The deviation of energy for the optimized trial wavefunction from the experimental value (Equation \ref{exp}) is
\[ \begin{align} \left| \dfrac{ E_{trial} (\alpha,\beta)-E_{\exp}}{E_{\exp}} \right| &= \left| \dfrac{- 2.89112 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right| \\[4pt] &= 0.43 \% \label{trial4} \end{align} \]
This modification of the trial wavefunction has further improved the agreement between theory and experiment to within 0.5%.
Fifth Trial Wavefunction #5: Approximation with Three Parameters
Chandrasakar brought about further improvement by adding Hylleraas's \(r_{12}\) term to the third trial wavefunction (Equation \(\ref{7.3.5}\)) as shown here.
\[| \psi (1,2) \rangle_{trial} = \left[\exp(- \alpha r_1 )\exp(- \beta r_2 ) + \exp(- \beta r_1 )\exp(- \alpha r_2 ) \right][1 + \gamma r _{12} ] \label{7.3.7} \]
Chandrasakar's three parameter wavefunction gives rise to a fairly complicated variational expression for ground-state energy. The variational energy obtained after minimizing Equation \(\ref{7.3.1a}\) after substituting the trial wavefunction (Equation \ref{7.3.7}) by varying \(\alpha\), \(\beta\) and \(\gamma\) is
\[ E_{trial} = -2.90143 \;E_h \nonumber \]
The deviation of energy for the optimized trial wavefunction from the experimental value (Equation \ref{exp}) is
\[ \begin{align} \left| \dfrac{ E_{trial} (\alpha, \beta, \gamma)-E_{\exp}}{E_{\exp}} \right| &= \left| \dfrac{-2.90143 + 2.90372 \;E_h}{-2.90372 \;E_h} \right| \\[4pt] &= 0.0789 \% \label{trial5} \end{align} \]
Chandrasakar's wavefunction gives a result for helium that is within 0.07% of the experimental value for the ground-state energy.
Summary
The purpose of this section is to examine five trial wavefunctions for the helium atom used within the Perturbation Theory and Variational method approximation. For the Variational method approximation, the calculations begin with an uncorrelated wavefunction in which both electrons are placed in a hydrogenic orbital with scale factor \(\alpha\). The next four trial functions use several methods to increase the amount of electron-electron interactions in the wavefunction. As the summary of results that is appended shows this gives increasingly more favorable agreement with the experimentally determined value for the ground-state energy of the species under study. The detailed calculations show that the reason for this improved agreement with experiment is due to a reduction in electron-electron repulsion.
Five variational method calculations that have been outlined above for the helium atom (\(Z=2\)) can be repeated for two-electron atoms (e.g., \(\ce{H^-}\), \(\ce{Li^+}\), \(\ce{Be^{2+}}\), etc). The hydride anion is a particularly interesting case because the first two trial wavefunctions do not predict a stable ion (i.e., they are poor approximations). This indicates that electron-electron interactions is an especially important issue for atoms and ions with small nuclear charge. | libretexts | 2025-03-17T22:26:40.534623 | 2023-01-07T00:43:38 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/07%3A_Multielectron_Atoms/7.01%3A_Perturbation_Theory_and_the_Variational_Method_for_Helium",
"book_url": "https://commons.libretexts.org/book/chem-424908",
"title": "7.1: Perturbation Theory and the Variational Method for Helium",
"author": "Frank Rioux"
} |
https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/09%3A_Molecular_Spectroscopy/9.01%3A_The_Electromagnetic_Spectrum | 9.1: The Electromagnetic Spectrum
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An important aspect of studying Physical Chemistry is to be able to recognize the interaction of molecules to the surroundings. Molecular Spectroscopy provides a clear image of how diatomic and polyatomic molecules interact by looking at the Frequency, Wavelength, Wave number, Energy, and molecular process. We will also be able to see the absorption properties of molecules in various regions from the electromagnetic spectrum.
Electromagnetic Radiation
Electromagnetic radiation—light—is a form of energy whose behavior is described by the properties of both waves and particles. Some properties of electromagnetic radiation, such as its refraction when it passes from one medium to another are explained best by describing light as a wave. Other properties, such as absorption and emission, are better described by treating light as a particle. The exact nature of electromagnetic radiation remains unclear, as it has since the development of quantum mechanics in the first quarter of the 20 th century. Nevertheless, the dual models of wave and particle behavior provide a useful description for electromagnetic radiation.
Electromagnetic radiation consists of oscillating electric and magnetic fields that propagate through space along a linear path and with a constant velocity. In a vacuum electromagnetic radiation travels at the speed of light, \(c\), which is \(2.997 92 \times 10^8\, m/s\). When electromagnetic radiation moves through a medium other than a vacuum its velocity, \(v\), is less than the speed of light in a vacuum. The difference between \(v\) and \(c\) is sufficiently small (<0.1%) that the speed of light to three significant figures, \(3.00 \times 10^8\, m/s\), is accurate enough for most purposes.
The oscillations in the electric and magnetic fields are perpendicular to each other, and to the direction of the wave’s propagation. Figure 13.1.1 shows an example of plane-polarized electromagnetic radiation, consisting of a single oscillating electric field and a single oscillating magnetic field.
An electromagnetic wave is characterized by several fundamental properties, including its velocity, amplitude, frequency, phase angle, polarization, and direction of propagation. 2 For example, the amplitude of the oscillating electric field at any point along the propagating wave is
\[A_\ce{t} = A_\ce{e}\sin(2πνt + \phi) \nonumber \]
where \(A_t\) is the magnitude of the electric field at time \(t\), \(A_e\) is the electric field’s maximum amplitude , \(\nu\) is the wave’s frequency —the number of oscillations in the electric field per unit time—and \(\phi\) is a phase angle , which accounts for the fact that \(A_t\) need not have a value of zero at \(t = 0\). The identical equation for the magnetic field is
\[A_\ce{t} =A_\ce{m}\sin(2πνt + \phi) \nonumber \]
where \(A_m\) is the magnetic field’s maximum amplitude.
The frequency and wavelength of electromagnetic radiation vary over many orders of magnitude. For convenience, we divide electromagnetic radiation into different regions—the electromagnetic spectrum —based on the type of atomic or molecular transition that gives rise to the absorption or emission of photons (Figure 13.1.2 ). The boundaries between the regions of the electromagnetic spectrum are not rigid, and overlap between spectral regions is possible.
Other Units
Other properties also are useful for characterizing the wave behavior of electromagnetic radiation. The wavelength , λ , is defined as the distance between successive maxima (Figure 13.1.1 ). For ultraviolet and visible electromagnetic radiation the wavelength is usually expressed in nanometers (1 nm = 10 –9 m), and for infrared radiation it is given in microns (1 μm = 10 –6 m). The relationship between wavelength and frequency is
\[λ = \dfrac{c}{ν} \nonumber \]
Another unit useful unit is the wavenumber , \(\tilde{ν}\), which is the reciprocal of wavelength
\[\tilde{ν} = \dfrac{1}{λ} \nonumber \]
Wavenumbers are frequently used to characterize infrared radiation, with the units given in cm –1 .
Example 13.1.1
In 1817, Josef Fraunhofer studied the spectrum of solar radiation, observing a continuous spectrum with numerous dark lines. Fraunhofer labeled the most prominent of the dark lines with letters. In 1859, Gustav Kirchhoff showed that the D line in the sun’s spectrum was due to the absorption of solar radiation by sodium atoms. The wavelength of the sodium D line is 589 nm. What are the frequency and the wavenumber for this line?
Solution
The frequency and wavenumber of the sodium D line are
\[ν = \dfrac{c}{λ} = \mathrm{\dfrac{3.00×10^8\: m/s}{589×10^{−9}\: m} = 5.09×10^{14}\: s^{−1}} \nonumber \]
\[\tilde{ν} = \dfrac{1}{λ} = \mathrm{\dfrac{1}{589×10^{−9}\: m} × \dfrac{1\: m}{100\: cm} = 1.70×10^4\: cm^{−1}} \nonumber \]
Exercise 13.1.1
Another historically important series of spectral lines is the Balmer series of emission lines form hydrogen. One of the lines has a wavelength of 656.3 nm. What are the frequency and the wavenumber for this line?
Above, we defined several characteristic properties of electromagnetic radiation, including its energy, velocity, amplitude, frequency, phase angle, polarization, and direction of propagation. A spectroscopic measurement is possible only if the photon’s interaction with the sample leads to a change in one or more of these characteristic properties. We can divide spectroscopy into two broad classes of techniques. In one class of techniques there is a transfer of energy between the photon and the sample. Table 13.1.1 provides a list of several representative examples.
| Type of Energy Transfer | Region of Electromagnetic Spectrum | Spectroscopic Technique |
|---|---|---|
| absorption | γ -ray | Mossbauer spectroscopy |
| X-ray | X-ray absorption spectroscopy | |
| UV/Vis |
UV/Vis spectroscopy atomic absorption spectroscopy |
|
| IR |
infrared spectroscopy raman spectroscopy |
|
| Microwave | microwave spectroscopy | |
| Radio wave |
electron spin resonance spectroscopy nuclear magnetic resonance spectroscopy |
|
| emission (thermal excitation) | UV/Vis | atomic emission spectroscopy |
| photoluminescence | X-ray | X-ray fluorescence |
| UV/Vis |
fluorescence spectroscopy phosphorescence spectroscopy atomic fluorescence spectroscopy |
|
| chemiluminescence | UV/Vis | chemiluminescence spectroscopy |
Electromagnetic spectrum provides clearly information of molecules if they are rotational transitions, vibrational transitions, or electronic transitions. A molecule or a set of molecules can be read by the absorption of microwave radiation which provides transitions between rotational energy levels. In addition, if the molecules absorbs infrared radiation provides the transitions between vibrational levels follows by transitions between rotational energy levels. Finally, when molecules absorbs visible and ultraviolet radiation gives transitions between electronic energy levels follows by simultaneous transitions between vibrational and rotational levels.
When given the energy level of the molecules along with wavelength, we can easily figure the frequency of the molecules where they fall in the electromagnetic spectrum regions:
\[\Delta E=E_u-E_l=h \nu \nonumber \]
The above equation describes the energy change between upper state and lower state of energy.
- Frequency falls between 10 9 - 10 11 which is in the microwave range correlates to the rotation of polyatomic molecules.
- Frequency falls between 10 11 - 10 13 which is in the far infrared range correlates to the rotation of small molecules.
- Frequency falls between 10 13 - 10 14 which is in the infrared range correlates to the vibrations of flexible bonds.
- Frequency falls between 10 14 - 10 16 which is in the visible and ultraviolet range correlates to the electronic transitions.
The powerful technique of figuring out the the frequency of the molecules can help us determine the bond length, temperature, probability distribution as you will learn later on from the degree of freedoms and how the process is undergo in specific a reaction. | libretexts | 2025-03-17T22:26:41.286270 | 2023-01-07T00:44:03 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/09%3A_Molecular_Spectroscopy/9.01%3A_The_Electromagnetic_Spectrum",
"book_url": "https://commons.libretexts.org/book/chem-424908",
"title": "9.1: The Electromagnetic Spectrum",
"author": null
} |
https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/09%3A_Molecular_Spectroscopy/9.02%3A_Electronic_Spectra_Contain_Electronic_Vibrational_and_Rotational_Information | 9.2: Electronic Spectra Contain Electronic, Vibrational, and Rotational Information
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Molecules can also undergo changes in electronic transitions during microwave and infrared absorptions. The energy level differences are usually high enough that it falls into the visible to UV range; in fact, most emissions in this range can be attributed to electronic transitions.
Electron Transitions are not Purely Electronic
We have thus far studied rovibrational transitions --that is, transitions involving both the vibrational and rotational states. Similarly, electronic transitions tend to accompany both rotational and vibrational transitions. These are often portrayed as an electronic potential energy cure with the vibrational level drawn on each curve. Additionally, each vibrational level has a set of rotational levels associated with it.
Recall that in the Born-Oppenheimer approximation, nuclear kinetic energies can be ignored (e.g., fixed) to solve for electronic wavefunctions and energies, which are much faster than rotation or vibration. As such, it is important to note that unlike rovibrational transitions, electronic transitions aren't dependent on rotational or transitional terms and are assumed to be separate. Therefore, when using an anharmonic oscillator-nonrigid rotator approximation (and excluding translation energy), the total energy of a diatomic is:
\[ \tilde{E}_{total} = \tilde{\nu}_{el} + G(v) + F(J) \label{Eqa1} \]
where \( \tilde{\nu}_{el}\) is the electronic transition energy change in wavenumbers, \(G(n)\) is the vibrational energy with energy level \(v\) (assuming anharmonic oscillator), and \(F(J)\) is the rotational energy, assuming a nonrigid rotor. Equation \(\ref{Eqa1}\) can be expanded accordingly:
\[ \tilde{E}_{total} = \underbrace{\tilde{\nu}_{el}}_{\text{electronic}} + \underbrace{\tilde{\nu}_e \left (v + \dfrac{1}{2} \right) - \tilde{\chi}_e \tilde{\nu}_e \left (v + \dfrac{1}{2} \right)^2}_{\text{vibrational}} + \underbrace{\tilde{B} J(J + 1) - \tilde{D} J^2(J + 1)^2}_{\text{rotational}} \label{Eqa2} \]
Notice that both the vibration constant (\( \tilde{\nu}_e\)) and anharmonic constant ( \(\tilde{\chi}_e\)) are electronic state dependent (and hence the rotational constants would be too, but are ignored here). Since rotational energies tend to be so small compared to electronic, their effects are minimal and are typically ignored when we do calculations and are referred to as vibronic transitions.
The eigenstate-to-eigenstate transitions (e.g., \(1 \rightarrow 2\)) possible are numerous and have absorption lines at
\[ \tilde{\nu}_{obs} = \tilde{E}_{2} - \tilde{E}_{1} \label{Eqa21} \]
and for simplification, we refer to constants associated with these states as \(| ' \rangle \) and \(| '' \rangle \), respectively. So Equation \(\ref{Eqa21}\) is
\[ \tilde{\nu}_{obs} = \tilde{E''(v'')} - \tilde{E'(v')} \nonumber \]
Also important to note that typically vibronic transitions are usually the result of the vibrational \(v'=0\) vibratonal state. Within this assumption and excluding the rotational contributions (due to their low energies), Equation \(\ref{Eqa2}\) can be used with Equation \(\ref{Eqa21}\) to get
\[ \tilde{\nu}_{obs} = \tilde{T}_{el} + \left( \dfrac{1}{2} \tilde{\nu}'_e - \dfrac{1}{4} \tilde{\chi}'_e \tilde{\nu}_e' \right) - \left( \dfrac{1}{2} \tilde{\nu}''_e - \dfrac{1}{4} \tilde{\chi}''_e \tilde{\nu}_e'' \right) + \tilde{\nu}'_e v'' - \tilde{\chi}'_e \tilde{\nu}_e' v''(v''+1) \label{Eqa3} \]
A common transition of importance is the \( \tilde{\nu}_{00}\), which is the \(0 \rightarrow 0\) transition and include no vibrational change. For this case, equation \(\ref{Eqa3}\) is then
\[ \tilde{\nu}_{00} = \tilde{T}_{el} + \left( \dfrac{1}{2} \tilde{\nu}'_e - \dfrac{1}{4} \tilde{\chi}'_e \tilde{\nu}_e' \right) - \left( \dfrac{1}{2} \tilde{\nu}''_e - \dfrac{1}{4} \tilde{\chi}''_e \tilde{\nu}_e'' \right) \nonumber \]
This is the lowest energy possible to observe in an electronic transition although it may be of low intensity as discussed in the following section.
Iodine
The absorption spectrum of iodine yields information about the excited state well rather than the ground state well (notice that equation \(\ref{Eqa3}\) depends primarily on excited state parameters). In this experiment you will characterize the excited state well by extracting values for the following excited state parameters.
Recall that as \(v'\) increases, the vibrational energy spacing decreases. At the upper edge of the well, the vibrational energy spacing decreases to 0, which means that the energies form a continuum rather than being quantized. It is at this limit that bond dissociation occurs . The energy required to dissociate the bond is actually \(D_o'\) rather than \(D_e'\) because the molecule cannot have less than the zero point energy .
The vibrational-electronic spectrum of I 2 in the region from 500-650 nm displays a large number of well-defined bands which, for the most part, correspond to v'<-- 0 transitions connecting the v" = 0 vibrational level of the ground electronic state (denoted as X 1 Σ + ) to many different vibrational levels v' of the excited B 3 Π electronic state. Under the conditions of this experiment (i.e., low resolution), the rotational lines within each band are not resolved. However, the peaks may be identified as R-branch band heads (1). For a molecule as heavy as I 2 , the position of each band head is within a few tenths of one cm -1 of the band origin (2), and for the purposes of this experiment, the distinction between the two may be ignored. The general features of the absorption spectrum are shown below:
Each small bump, or peak, such as the (26,0) band labelled on the spectrum, corresponds to a transition between two vibrational levels and is called a band . Each band is comprised of several hundred lines , each of which involves different upper and lower rotational quantum numbers; as mentioned, these lines are not resolved in the present experiment. The region of maximum absorption in each band is caused by many of these lines falling together; it is called the band head . The set of all of these bands is referred to as the visible band system of I 2 .
If the sample is hot, then excited vibrational levels of the ground state may be populated, and these also will absorb light. The hot bands arising from absorption from v"=1 and v"=2 are shown very approximately on the absorption spectrum above.
At a point called the convergence limit , the spacing between bands decreases to zero. Beyond this convergence limit, the spectrum is continuous because the excited state of the I 2 molecule is not bound. One of the purposes of this experiment is to identify this convergence limit accurately. | libretexts | 2025-03-17T22:26:41.432853 | 2023-01-07T00:44:05 | {
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"url": "https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/09%3A_Molecular_Spectroscopy/9.02%3A_Electronic_Spectra_Contain_Electronic_Vibrational_and_Rotational_Information",
"book_url": "https://commons.libretexts.org/book/chem-424908",
"title": "9.2: Electronic Spectra Contain Electronic, Vibrational, and Rotational Information",
"author": null
} |
https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/09%3A_Molecular_Spectroscopy/9.05%3A_Fluorescence_and_Phosphorescence | 9.5: Fluorescence and Phosphorescence
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Fluorescence and phosphorescence are types of molecular luminescence methods. A molecule of analyte absorbs a photon and excites a species. The emission spectrum can provide qualitative and quantitative analysis. The term fluorescence and phosphorescence are usually referred as photoluminescence because both are alike in excitation brought by absorption of a photon. Fluorescence differs from phosphorescence in that the electronic energy transition that is responsible for fluorescence does not change in electron spin, which results in short-live electrons (<10 -5 s) in the excited state of fluorescence. In phosphorescence, there is a change in electron spin, which results in a longer lifetime of the excited state (second to minutes). Fluorescence and phosphorescence occurs at longer wavelength than the excitation radiation.
Introduction
Fluorescence can occur in gaseous, liquid, and solid chemical systems. The simple kind of fluorescence is by dilute atomic vapors. A fluorescence example would be if a 3s electron of a vaporized sodium atom is excited to the 3p state by absorption of a radiation at wavelength 589.6 and 589.0 nm. After 10 -8 s, the electron returns to ground state and on its return it emits radiation of the two wavelengths in all directions. This type of fluorescence in which the absorbed radiation is remitted without a change in frequency is known as resonance fluorescence. Resonance fluorescence can also occur in molecular species. Molecular fluorescence band centers at wavelengths longer than resonance lines. The shift toward longer wavelength is referred to as the Stokes Shift.
Singlet and Triplet Excited State
Understanding the difference between fluorescence and phosphorescence requires the knowledge of electron spin and the differences between singlet and triplet states. The Pauli Exclusion principle states that two electrons in an atom cannot have the same four quantum numbers (\(n\), \(l\), \(m_l\), \(m_s\)) and only two electrons can occupy each orbital where they must have opposite spin states. These opposite spin states are called spin pairing. Because of this spin pairing, most molecules do not exhibit a magnetic field and are diamagnetic. In diamagnetic molecules, electrons are not attracted or repelled by the static electric field. Free radicals are paramagnetic because they contain unpaired electrons have magnetic moments that are attracted to the magnetic field.
Singlet state is defined when all the electron spins are paired in the molecular electronic state and the electronic energy levels do not split when the molecule is exposed into a magnetic field. A doublet state occurs when there is an unpaired electron that gives two possible orientations when exposed in a magnetic field and imparts different energy to the system. A singlet or a triplet can form when one electron is excited to a higher energy level. In an excited singlet state, the electron is promoted in the same spin orientation as it was in the ground state (paired). In a triplet excited stated, the electron that is promoted has the same spin orientation (parallel) to the other unpaired electron. The difference between the spins of ground singlet, excited singlet, and excited triplet is shown in Figure \(\PageIndex{1}\). Singlet, doublet and triplet is derived using the equation for multiplicity, 2S+1, where S is the total spin angular momentum (sum of all the electron spins). Individual spins are denoted as spin up (s = +1/2) or spin down (s = -1/2). If we were to calculated the S for the excited singlet state, the equation would be 2(+1/2 + -1/2)+1 = 2(0)+1 = 1, therefore making the center orbital in the figure a singlet state. If the spin multiplicity for the excited triplet state was calculated, we obtain 2(+1/2 + +1/2)+1 = 2(1)+1 =3, which gives a triplet state as expected.
The difference between a molecule in the ground and excited state is that the electrons is diamagnetic in the ground state and paramagnetic in the triplet state.This difference in spin state makes the transition from singlet to triplet (or triplet to singlet) more improbable than the singlet-to-singlet transitions. This singlet to triplet (or reverse) transition involves a change in electronic state. For this reason, the lifetime of the triplet state is longer the singlet state by approximately 10 4 seconds fold difference.The radiation that induced the transition from ground to excited triplet state has a low probability of occurring, thus their absorption bands are less intense than singlet-singlet state absorption. The excited triplet state can be populated from the excited singlet state of certain molecules which results in phosphorescence. These spin multiplicities in ground and excited states can be used to explain transition in photoluminescence molecules by the Jablonski diagram.
Jablonski Diagrams
The Jablonski diagram that drawn below is a partial energy diagram that represents the energy of photoluminescent molecule in its different energy states. The lowest and darkest horizontal line represents the ground-state electronic energy of the molecule which is the singlet state labeled as \(S_o\). At room temperature, majority of the molecules in a solution are in this state.
The upper lines represent the energy state of the three excited electronic states: S 1 and S 2 represent the electronic singlet state (left) and T 1 represents the first electronic triplet state (right). The upper darkest line represents the ground vibrational state of the three excited electronic state.The energy of the triplet state is lower than the energy of the corresponding singlet state.
There are numerous vibrational levels that can be associated with each electronic state as denoted by the thinner lines. Absorption transitions (blues lines in Figure \(\PageIndex{2}\)) can occur from the ground singlet electronic state (S o ) to various vibrational levels in the singlet excited vibrational states. It is unlikely that a transition from the ground singlet electronic state to the triplet electronic state because the electron spin is parallel to the spin in its ground state (Figure \(\PageIndex{1}\)). This transition leads to a change in multiplicity and thus has a low probability of occurring which is a forbidden transition. Molecules also go through vibration relaxation to lose any excess vibrational energy that remains when excited to the electronic states (\(S_1\) and \(S_2\)) as demonstrated in wavy lines in Figure \(\PageIndex{2}\). The knowledge of forbidden transition is used to explain and compare the peaks of absorption and emission.
Absorption and Emission Rates
The table below compares the absorption and emission rates of fluorescence and phosphorescence.The rate of photon absorption is very rapid. Fluorescence emission occurs at a slower rate.Since the triplet to singlet (or reverse) is a forbidden transition, meaning it is less likely to occur than the singlet-to-singlet transition, the rate of triplet to singlet is typically slower. Therefore, phosphorescence emission requires more time than fluorescence.
| Process | Transition | Timescale (sec) |
|---|---|---|
| Light Absorption (Excitation) | S 0 → S n | ca. 10 -15 (instantaneous) |
| Internal Conversion | S n → S 1 | 10 -14 to 10 -11 |
| Vibrational Relaxation | S n * → S n | 10 -12 to 10 -10 |
| Intersystem Crossing | S 1 → T 1 | 10 -11 to 10 -6 |
| Fluorescence | S 1 → S 0 | 10 -9 to 10 -6 |
| Phosphorescence | T 1 → S 0 | 10 -3 to 100 |
| Non-Radiative Decay |
S
1
→ S
0
T 1 → S 0 |
10
-7
to 10
-5
10 -3 to 100 |
Deactivation Processes
A molecule that is excited can return to the ground state by several combinations of mechanical steps that will be described below and shown in Figure \(\PageIndex{2}\).The deactivation process of fluorescence and phosphorescence involve an emission of a photon radiation as shown by the straight arrow in Figure \(\PageIndex{2}\). The wiggly arrows in Figure \(\PageIndex{2}\) are deactivation processes without the use of radiation. The favored deactivation process is the route that is most rapid and spends less time in the excited state.If the rate constant for fluorescence is more favorable in the radiationless path, the fluorescence will be less intense or absent.
- Vibrational Relaxation: A molecule maybe to promoted to several vibrational levels during the electronic excitation process.Collision of molecules with the excited species and solvent leads to rapid energy transfer and a slight increase in temperature of the solvent. Vibrational relaxation is so rapid that the lifetime of a vibrational excited molecule (<10 -12 ) is less than the lifetime of the electronically excited state. For this reason, fluorescence from a solution always involves the transition of the lowest vibrational level of the excited state. Since the space of the emission lines are so close together, the transition of the vibrational relaxation can terminate in any vibrational level of the ground state.
- Internal Conversion: Internal conversion is an intermolecular process of molecule that passes to a lower electronic state without the emission of radiation.It is a crossover of two states with the same multiplicity meaning singlet-to-singlet or triplet-to-triplet states.The internal conversion is more efficient when two electronic energy levels are close enough that two vibrational energy levels can overlap as shown in between S 1 and S 2 . Internal conversion can also occur between S 0 and S 1 from a loss of energy by fluorescence from a higher excited state, but it is less probable. The mechanism of internal conversion from S 1 to S 0 is poorly understood. For some molecules, the vibrational levels of the ground state overlaps with the first excited electronic state, which leads to fast deactivation.These usually occur with aliphatic compounds (compound that do not contain ring structure), which would account for the compound is seldom fluorescing. Deactivation by energy transfer of these molecules occurs so rapidly that the molecule does not have time to fluoresce.
- External Conversion : Deactivation of the excited electronic state may also involve the interaction and energy transfer between the excited state and the solvent or solute in a process called external conversion. Low temperature and high viscosity leads to enhanced fluorescence because they reduce the number of collision between molecules, thus slowing down the deactivation process.
- Intersystem Crossing: Intersystem crossing is a process where there is a crossover between electronic states of different multiplicity as demonstrated in the singlet state to a triplet state (S 1 to T 1 ) on Figure \(\PageIndex{1}\). The probability of intersystem crossing is enhanced if the vibration levels of the two states overlap. Intersystem crossing is most commonly observed with molecules that contain heavy atom such as iodine or bromine. The spin and orbital interaction increase and the spin become more favorable.Paramagnetic species also enhances intersystem crossing, which consequently decreases fluorescence.
- Phosphorescence : Deactivation of the electronic excited state is also involved in phosphorescence. After the molecule transitions through intersystem crossing to the triplet state, further deactivation occurs through internal or external fluorescence or phosphorescence. A triplet-to-singlet transition is more probable than a singlet-to-singlet internal crossing. In phosphorescence, the excited state lifetime is inversely proportional to the probability that the molecule will transition back to the ground state. Since the lifetime of the molecule in the triplet state is large (10 -4 to 10 second or more), transition is less probable which suggest that it will persist for some time even after irradiation has stopped. Since the external and internal conversion compete so effectively with phosphorescence, the molecule has to be observed at lower temperature in highly viscous media to protect the triplet state.
Variables that affect Fluorescence
After discussing all the possible deactivation processes, variable that affect the emissions to occur. Molecular structure and its chemical environment influence whether a substance will fluoresce and the intensities of these emissions. The quantum yield or quantum efficiency is used to measure the probability that a molecule will fluoresce or phosphoresce. For fluorescence and phosphorescence is the ratio of the number of molecules that luminescent to the total number of excited molecules. For highly fluoresce molecules, the quantum efficiency approaches to one.Molecules that do not fluoresce have quantum efficiencies that approach to zero.
Fluorescence quantum yield (\( \phi \)) for a compound is determined by the relative rate constants (k) of various deactivation processes by which the lowest excited singlet state is deactivated to the ground state. The deactivation processes including fluorescence (kf), intersystem crossing (\(k_i\)), internal conversion (kic), predissociation (kpd), dissociation (kd), and external conversion (kec) allows one to qualitatively interpret the structural and environmental factors that influence the intensity of the fluorescence. They are related by the quantum yield equation given below:
\[ \dfrac{k_f}{k_f+k_i+k_{ec}+k_{ic}+k_{pd}+k_d} \]
Using this equation as an example to explain fluorescence, a high fluorescence rate (k f ) value and low values of the all the other relative rate constant terms (k f +k i +k ec +k ic +k pd +k d ) will give a large \( \phi \), which suggest that fluorescence is enhanced. The magnitudes of kf , kd, and kpd depend on the chemical structure, while the rest of the constants ki, kec, and kic are strongly influenced by the environment.
Fluorescence rarely results from absorption of ultraviolet radiation of wavelength shorter than 250 nm because radiation at this wavelength has sufficient energy to deactivate the electron in the excited state by predissociation or dissociation. The bond of some organic molecules would rupture at 140 kcal/mol, which corresponds to 200-nm of radiation. For this reason, \(\sigma \rightarrow \sigma^{*}\) transition in fluorescence are rarely observed. Instead, emissions from the less energetic transition will occur which are either \(\pi^{*} \rightarrow \pi \) or \(\pi^{*} \rightarrow n \) transition.
Molecules that are excited electronically will return to the lowest excited state by rapid vibrational relaxation and internal conversion, which produces no radiation emission. Fluorescence arises from a transition from the lowest vibrational level of the first excited electronic state to one of the vibrational levels in the electronic ground state. In most fluorescent compounds, radiation is produced by a \(\pi^{*} \rightarrow \pi \) or \(\pi^{*} \rightarrow n \) transition depending on which requires the least energy for the transition to occur.
Fluorescence is most commonly found in compounds in which the lowest energy transition is \(\pi \rightarrow \pi^{*} \) (excited singlet state) than \(n \rightarrow \pi^{*} \) which suggest that the quantum efficiency is greater for \(\pi \rightarrow \pi^{*} \) transitions. The reason for this is that the molar absorptivity, which measures the probability that a transition will occur, of the \(\pi \rightarrow \pi^{*} \) transition is 100 to 1000 fold greater than \(n \rightarrow \pi^{*} \) process. The lifetime of \(\pi \rightarrow \pi^{*} \) (10 -7 to 10 -9 s) is shorter than the lifetime of \(n \rightarrow \pi^{*} \) (10 -5 to 10 -7 ).
Phosphorescent quantum efficiency is the opposite of fluorescence in that it occurs in the \(n \rightarrow \pi^{*} \) excited state which tends to be short lived and less suceptable to deactivation than the \(\pi \rightarrow \pi^{*} \) triplet state. Intersystem crossing is also more probable for \(\pi \rightarrow \pi^{*} \) excited state than for the \(n \rightarrow \pi^{*} \)state because the energy difference between the singlet and triplet state is large and spin-orbit coupling is less likely to occur.
Fluorescence and Structure
The most intense fluorescence is found in compounds containing aromatic group with low-energy \(\pi \rightarrow \pi^{*} \) transitions. A few aliphatic, alicyclic carbonyl, and highly conjugated double-bond structures also exhibit fluorescence as well. Most unsubstituted aromatic hydrocarbons fluoresce in solution too. The quantum efficiency increases as the number of rings and the degree of condensation increases. Simple heterocycles such as the structures listed below do not exhibit fluorescence.
Pyridine Pyrrole Furan Thiophene
With nitrogen heterocyclics, the lowest energy transitions is involved in \(n \rightarrow \pi^{*} \) system that rapidly converts to the triplet state and prevents fluorescence. Although simple heterocyclics do not fluoresce, fused-ring structures do. For instance, a fusion of a benzene ring to a hetercyclic structure results in an increase in molar absorptivity of the absorption band. The lifetime of the excited state in fused structure and fluorescence is observed. Examples of fluorescent compounds is shown below.
quinoline
Benzene ring substitution causes a shift in the absorption maxima of the wavelength and changes in fluorescence emission. The table below is used to demonstrate and visually show that as benzene is substituted with increasing methyl addition, the relative intensity of fluorescence increases.
|
Compound |
Structure |
Wavelength of Fluorescence (nm) |
Relative intensity of Fluorescence |
|---|---|---|---|
|
Benzene |
|
270-310 |
10 |
|
Toluene |
|
270-320 |
17 |
|
Propyl Benzene |
|
270-320 |
17 |
The relative intensity of fluorescence increases as oxygenated species increases in substitution. The values for such increase is demonstrated in the table below.
|
Compound |
Structure |
Wavelength of Fluorescence (nm) |
Relative intensity of Fluorescence |
|---|---|---|---|
|
Phenol |
|
285-365 |
18 |
|
Phenolate ion |
|
310-400 |
10 |
|
Anisole |
|
285-345 |
20 |
Influence of a halogen substitution decreases fluorescence as the molar mass of the halogen increases. This is an example of the “heavy atom effect” which suggest that the probability of intersystem crossing increases as the size of the molecule increases. As demonstrated in the table below, as the molar mass of the substituted compound increases, the relative intensity of the fluorescence decreases.
|
Compound |
Structure |
Wavelength of Fluorescence (nm) |
Relative intensity of Fluorescence |
|---|---|---|---|
|
Fluorobenzene |
|
270-320 |
10 |
|
Chlorobenzene |
|
275-345 |
7 |
|
Bromobenzene |
|
290-380 |
5 |
In heavy atom substitution such as nitro derivatives or heavy halogen substitution such as iodobenzene, the compounds are subject to predissociation. These compounds have bonds that easily rupture that can then absorb excitation energy and go through internal conversion. Therefore, the relative intensity of fluorescence and fluorescent wavelength is not observed and this is demonstrated in the table below.
|
Compound |
Structure |
Wavelength of Fluorescence (nm) |
Relative intensity of Fluorescence |
|---|---|---|---|
|
Iodobenzene |
|
None |
0 |
|
Anilinium ion |
|
None |
0 |
|
Nitrobenzene |
|
None |
0 |
Carboxylic acid or carbonyl group on aromatic ring generally inhibits fluorescence since the energy of the \(n \rightarrow \pi^*\) transition is less than \(\pi \rightarrow \pi^*\) transition. Therefore, the fluorescence yield from \(n \rightarrow \pi^*\) transition is low.
|
Compound |
Structure |
Wavelength of Fluorescence (nm) |
Relative intensity of Fluorescence |
|---|---|---|---|
|
Benzoic Acid |
|
310-390 |
3 |
Effect of Structural Rigidity on Fluorescence
Fluorescence is particularly favored in molecules with rigid structures. The table below compares the quantum efficiencies of fluorine and biphenyl which are both similar in structure that there is a bond between the two benzene group. The difference is that fluorene is more rigid from the addition methylene bridging group. By looking at the table below, rigid fluorene has a higher quantum efficiency than unrigid biphenyl which indicates that fluorescence is favored in rigid molecules.
|
Compound |
Structure |
Quantum Efficiency |
|---|---|---|
|
Fluorene |
|
1.0 |
|
Biphenyl |
|
0.2 |
This concept of rigidity was used to explain the increase in fluorescence of organic chelating agent when the compound is complexed with a metal ion. The fluorescence intensity of 8-hydroxyquinoline is much less than its zinc complex.
vs
8-hydroxyquinoline 8-hydroxyquinoline with Zinc complexed
The explanation for lower quantum efficiency or lack of rigidity in caused by the enhanced internal conversion rate (k ic ) which increases the probability that there will be radiationless deactivation. Nonrigid molecules can also undergo low-frequency vibration which accounts for small energy loss.
Temperature and Solvent Effects
Quantum efficiency of Fluorescence decreases with increasing temperature. As the temperature increases, the frequency of the collision increases which increases the probability of deactivation by external conversion. Solvents with lower viscosity have higher possibility of deactivation by external conversion. Fluorescence of a molecule decreases when its solvent contains heavy atoms such as carbon tetrabromide and ethyl iodide, or when heavy atoms are substituted into the fluorescing compound. Orbital spin interaction result from an increase in the rate of triplet formation, which decreases the possibility of fluorescence. Heavy atoms are usually incorporated into solvent to enhance phosphorescence.
Effect of pH on Fluorescence
The fluorescence of aromatic compound with basic or acid substituent rings are usually pH dependent. The wavelength and emission intensity is different for protonated and unprotonated forms of the compound as illustrated in the table below:
|
Compound |
Structure |
Wavelength of Fluorescence (nm) |
Relative intensity of Fluorescence |
|---|---|---|---|
|
aniline |
310-405 |
20 |
|
|
Anilinium ion |
None |
0 |
The emission changes of this compound arises from different number of resonance structures associated with the acidic and basic forms of the molecule.The additional resonance forms provides a more stable first excited state, thus leading to fluorescence in the ultraviolet region.The resonance structures of basic aniline and acidic anilinium ion is shown below:
basic Aniline Fluorescence of certain compounds have been used a detection of end points in acid-base titrations.An example of this type of fluorescence seen in compound as a function of pH is the phenolic form of 1-naphthol-4-sulfonic acid.This compound is not detectable with the eye because it occurs in the ultraviolet region, but with an addition of a base, it becomes converted to a phenolate ion, the emission band shifts to the visible wavelength where it can be visually seen. Acid dissociation constant for excited molecules differs for the same species in the ground state.These changes in acid or base dissociation constant differ in four or five orders of magnitude.
Dissolved oxygen reduces the intensity of fluorescence in solution, which results from a photochemically induced oxidation of fluorescing species.Quenching takes place from the paramagnetic properties of molecular oxygen that promotes intersystem crossing and conversion of excited molecules to triplet state.Paramagnetic properties tend to quench fluorescence.
Effects of Concentration on Fluorescence Intensity
The power of fluorescence emission \(F\) is proportional to the radiant power is proportional to the radiant power of the excitation beam that is absorbed by the system. The equation below best describes this relationship.
(1)
Since \(\phi_f K”\) is constant in the system, it is represented at K’. The table below defines the variables in this equation.
|
Variable |
Definition |
|---|---|
|
F |
Power of fluorescence emission |
|
P0 |
Power of incident beam on solution |
|
P |
Power after transversing length b in medium |
|
K” |
Constant dependent on geometry and other factors |
|
f |
Quantum efficiency |
Fluorescence emission (\(F\)) can be related to concentration (\(c\)) using Beer’s Law stating:
\[ F = \epsilon b c \label{2}\]
where \( \epsilon \) is the molar absorptivity of the molecule that is fluorescing. Rewriting Equation 2 gives:
\[ P =P_o 10^{-\epsilon b c} \label{3}\]
Plugging this Equation \(\ref{3}\) into Equation \(\ref{1}\) and factoring out \(P_ 0\) gives us this equation:
\[F=K^{\prime} P_{0}\left(1-10^{-\varepsilon b c}\right)\]
The MacLaurin series could be used to solved the exponential term.
\[F=K^{\prime} P_{0}\left[2.303 \varepsilon b c-\frac{(2.303 \varepsilon b c)^{2}}{2 !}+\frac{(2.303 \varepsilon b c)^{3}}{3 !}+\frac{(2.303 \varepsilon b c)^{4}}{4 !}+\ldots \frac{(2.303 \varepsilon b c)^{n}}{n !}\right]_{1}\]
Given that \((2.303 \epsilon b c = \text{Absorbance} <0.05\), all the subsequent terms after the first can be dropped since the maximum error is 0.13%. Using only the first term, Equation \(\ref{5}\) can be rewritten as:
\[F=K^{\prime} P_{0} 2.303 \varepsilon b c\]
Equation \(\ref{6}\) can be expanded to the equation below and simplified to compare the fluorescence emission F with concentration. If the equation below were to be plotted with F versus c, a linear relation would be observed.
\[F=\phi_{f} K^{\prime \prime} P_{0} 2.303 \varepsilon b c\]
If \(c\) becomes so great that the absorbance > 0.05, the higher terms start to become taken into account and the linearity is lost. F then lies below the extrapolation of the straight-line plot. This excessive absorption is the primary absorption. Another cause of this negative downfall of linearity is the secondary absorption when the wavelength of emission overlaps the absorption band. This occurs when the emission transverse the solution and gets reabsorbed by other molecules by analyte or other species in the solution, which leads to a decrease in fluorescence.
Quenching Methods
Dynamic Quenching is a nonradiative energy transfer between the excited and the quenching agent species (Q).The requirements for a successful dynamic quenching are that the two collision species the concentration must be high so that there is a higher possibility of collision between the two species.Temperature and quenching agent viscosity play a role on the rate of dynamic quenching.Dynamic quenching reduces fluorescence quantum yield and the fluorescence lifetime.
Dissolved oxygen in a solution increases the intensity of the fluorescence by photochemically inducing oxidation of the fluorescing species.Quenching results from the paramagnetic properties of molecular oxygen that promotes intersystem crossing and converts the excited molecules to triplet state.Paramagnetic species and dissolved oxygen tend to quench fluorescence and quench the triplet state.
Static quenching occurs when the quencher and ground state fluorophore forms a dark complex.Fluorescence is usually observed from unbound fluorophore.Static quenching can be differentiated from dynamic quenching in that the lifetime is not affected in static quenching.In long range (Förster) quenching, energy transfer occurs without collision between molecules, but dipole-dipole coupling occurs between excited fluorophore and quencher.
Emission and Excitation Spectra
One of the ways to visually distinguish the difference between each photoluminescence is to compare the relative intensities of emission/excitation at each wavelength. An example of the three types of photoluminescence (absorption, fluorescence and phosphorescence) is shown for phenanthrene in the spectrum below.In the spectrum, the luminescent intensity is measure in a wavelength is fixed while the excitation wavelength is varied. The spectrum in red represents the excitation spectrum, which is identical to the absorption spectrum because in order for fluorescence emission to occur, radiation needs to be absorbed to create an excited state.The spectrum in blue represent fluorescence and green spectrum represents the phosphorescence.
Fluorescence and Phosphorescence occur at wavelengths that are longer than their absorption wavelengths.Phosphorescence bands are found at a longer wavelength than fluorescence band because the excited triplet state is lower in energy than the singlet state.The difference in wavelength could also be used to measure the energy difference between the singlet and triplet state of the molecule. The wavelength (\(\lambda\)) of a molecule is inversely related to the energy (\(E\)) by the equation below:
\[E= \dfrac{hc}{\lambda}\]
As the wavelength increases, the energy of the molecule decrease and vice versa.
References
- D. A. Skoog, et al. "Principles of Instrumental Analysis" 6th Edition, Thomson Brooks/Cole. 2007
- D. C. Harris and M.D. Bertolucci "Symmetry and Spectroscopy, An Introduction to Vibrational and Electronic Spectroscopy" Dover Publications, Inc., New York. 1989.
Problems
- Draw and label the Jablonski Diagram.
- How do spin states differ in ground singlet state versus excite singlet state and triplet excited state?
- Describe the rates of deactivation process.
- What is quantum yield and how is it used to compare the fluorescence of different types of molecule?
- What roles do solvent play in fluorescence? | libretexts | 2025-03-17T22:26:41.650855 | 2023-01-07T00:44:09 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/09%3A_Molecular_Spectroscopy/9.05%3A_Fluorescence_and_Phosphorescence",
"book_url": "https://commons.libretexts.org/book/chem-424908",
"title": "9.5: Fluorescence and Phosphorescence",
"author": null
} |
https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/09%3A_Molecular_Spectroscopy/9.06%3A_Jablonski_diagram | 9.6: Jablonski diagram
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Aleksander Jablonski was a Polish academic who devoted his life to the study of molecular absorbance and emission of light. He developed a written representation that generally shows a portion of the possible consequences of applying photons from the visible spectrum of light to a particular molecule. These schematics are referred to as Jablonski diagrams.
Introduction
A Jablonski diagram is basically an energy diagram, arranged with energy on a vertical axis. The energy levels can be quantitatively denoted, but most of these diagrams use energy levels schematically. The rest of the diagram is arranged into columns. Every column usually represents a specific spin multiplicity for a particular species. However, some diagrams divide energy levels within the same spin multiplicity into different columns. Within each column, horizontal lines represent eigenstates for that particular molecule. Bold horizontal lines are representations of the limits of electronic energy states. Within each electronic energy state are multiple vibronic energy states that may be coupled with the electronic state. Usually only a portion of these vibrational eigenstates are represented due to the massive number of possible vibrations in a molecule. Each of these vibrational energy states can be subdivided even further into rotational energy levels; however, typical Jablonski diagrams omit such intense levels of detail. As electronic energy states increase, the difference in energy becomes continually less, eventually becoming a continuum that can be approach with classical mechanics. Additionally, as the electronic energy levels get closer together, the overlap of vibronic energy levels increases.
Through the use of straight and curved lines, these figures show transitions between eigenstates that occur from the exposure of a molecule to a particular wavelength of light. Straight lines show the conversion between a photon of light and the energy of an electron. Curved lines show transitions of electrons without any interaction with light. Within a Jablonski diagram several different pathways show how an electron may accept and then dissipate the energy from a photon of a particular wavelength. Thus, most diagrams start with arrows going from the ground electronic state and finish with arrows going to the ground electronic state.
Absorbance
The first transition in most Jablonski diagrams is the absorbance of a photon of a particular energy by the molecule of interest. This is indicated by a straight arrow pointing up. Absorbance is the method by which an electron is excited from a lower energy level to a higher energy level. The energy of the photon is transferred to the particular electron. That electron then transitions to a different eigenstate corresponding to the amount of energy transferred. Only certain wavelengths of light are possible for absorbance, that is, wavelengths that have energies that correspond to the energy difference between two different eigenstates of the particular molecule. Absorbance is a very fast transition, on the order of 10 -15 seconds. Most Jablonski diagrams, however, do not indicate a time scale for the phenomenon being indicated. This transition will usually occur from the lowest (ground) electronic state due to the statistical mechanical issue of most electrons occupying a low lying state at reasonable temperatures. There is a Boltzmann distribution of electrons within this low lying levels, based on the the energy available to the molecules. This energy available is a function of the Boltzmann's constant and the temperature of the system. These low lying electrons will transition to an excited electronic state as well as some excited vibrational state.
Vibrational Relaxation and Internal Conversion
Once an electron is excited, there are a multitude of ways that energy may be dissipated. The first is through vibrational relaxation, a non-radiative process. This is indicated on the Jablonski diagram as a curved arrow between vibrational levels. Vibrational relaxation is where the energy deposited by the photon into the electron is given away to other vibrational modes as kinetic energy. This kinetic energy may stay within the same molecule, or it may be transferred to other molecules around the excited molecule, largely depending on the phase of the probed sample. This process is also very fast, between 10 -14 and 10 -11 seconds. Since this is a very fast transition, it is extremely likely to occur immediately following absorbance. This relaxation occurs between vibrational levels, so generally electrons will not change from one electronic level to another through this method.
However, if vibrational energy levels strongly overlap electronic energy levels, a possibility exists that the excited electron can transition from a vibration level in one electronic state to another vibration level in a lower electronic state. This process is called internal conversion and mechanistically is identical to vibrational relaxation. It is also indicated as a curved line on a Jablonski diagram, between two vibrational levels in different electronic states. Internal Conversion occurs because of the overlap of vibrational and electronic energy states. As energies increase, the manifold of vibrational and electronic eigenstates becomes ever closer distributed. At energy levels greater than the first excited state, the manifold of vibrational energy levels strongly overlap with the electronic levels. This overlap gives a higher degree of probability that the electron can transition between vibrational levels that will lower the electronic state. Internal conversion occurs in the same time frame as vibrational relaxation, therefore, is a very likely way for molecules to dissipate energy from light perturbation. However, due to a lack of vibrational and electronic energy state overlap and a large energy difference between the ground state and first excited state, internal conversion is very slow for an electron to return to the ground state. This slow return to the ground state lets other transitive processes compete with internal conversion at the first electronically excited state. Both vibrational relaxation and internal conversion occur in most perturbations, yet are seldom the final transition.
Fluorescence
Another pathway for molecules to deal with energy received from photons is to emit a photon. This is termed fluorescence. It is indicated on a Jablonski diagram as a straight line going down on the energy axis between electronic states. Fluorescence is a slow process on the order of 10 -9 to 10 -7 seconds; therefore, it is not a very likely path for an electron to dissipate energy especially at electronic energy states higher than the first excited state. While this transition is slow, it is an allowed transition with the electron staying in the same multiplicity manifold. Fluorescence is most often observed between the first excited electron state and the ground state for any particular molecule because at higher energies it is more likely that energy will be dissipated through internal conversion and vibrational relaxation. At the first excited state, fluorescence can compete in regard to timescales with other non-radiative processes. The energy of the photon emitted in fluorescence is the same energy as the difference between the eigenstates of the transition; however, the energy of fluorescent photons is always less than that of the exciting photons. This difference is because energy is lost in internal conversion and vibrational relaxation, where it is transferred away from the electron. Due to the large number of vibrational levels that can be coupled into the transition between electronic states, measured emission is usually distributed over a range of wavelengths.
Intersystem Crossing
Yet another path a molecule may take in the dissipation of energy is called intersystem crossing. This where the electron changes spin multiplicity from an excited singlet state to an excited triplet state. It is indicated by a horizontal, curved arrow from one column to another. This is the slowest process in the Jablonski diagram, several orders of magnitude slower than fluorescence. This slow transition is a forbidden transition, that is, a transition that based strictly on electronic selection rules should not happen. However, by coupling vibrational factors into the selection rules, the transition become weakly allowed and able to compete with the time scale of fluorescence. Intersystem crossing leads to several interesting routes back to the ground electronic state. One direct transition is phosphorescence, where a radiative transition from an excited triplet state to a singlet ground state occurs.This is also a very slow, forbidden transition. Another possibility is delayed fluorescence, the transition back to the first excited singlet level, leading to the emitting transition to the ground electronic state.
Other non-emitting transitions from excited state to ground state exist and account for the majority of molecules not exhibiting fluorescence or phosphorescent behavior. One process is the energy transfer between molecules through molecular collisions (e.g., external conversion). Another path is through quenching, energy transfer between molecules through overlap in absorption and fluorescence spectra. These are non-emitting processes that will compete with fluorescence as the molecule relaxes back down to the ground electronic state. In a Jablonski diagram, each of these processes are indicated with a curved line going down to on the energy scale.
Time Scales
It is important to note that a Jablonski diagram shows what sorts of transitions that can possibly happen in a particular molecule. Each of these possibilities is dependent on the time scales of each transition. The faster the transition, the more likely it is to happen as determined by selection rules. Therefore, understanding the time scales each process can happen is imperative to understanding if the process may happen. Below is a table of average time scales for basic radiative and non-radiative processes.
| Transition | Time Scale | Radiative Process? |
|---|---|---|
| Internal Conversion | 10 -14 - 10 -11 s | no |
| Vibrational Relaxation | 10 -14 - 10 -11 s | no |
| Absorption | 10 -15 s | yes |
| Phosphorescence | 10 -4 - 10 -1 s | yes |
| Intersystem Crossing | 10 -8 - 10 -3 s | no |
| Fluorescence | 10 -9 - 10 -7 s | yes |
Each process outlined above can be combined into a single Jablonski diagram for a particular molecule to give a overall picture of possible results of perturbation of a molecule by light energy. Jablonski diagrams are used to easily visualize the complex inner workings of how electrons change eigenstates in different conditions. Through this simple model, specific quantum mechanical phenomena are easily communicated.
References
- H. H. Jaffe and Albert L. Miller "The fates of electronic excitation energy" J. Chem. Educ. , 1966, 43 (9), p 469 DOI: 10.1021/ed043p469
- E. B. Priestley and A. Haug "Phosphorescence Spectrum of Pure Crystalline Naphthalene" J. Chem. Phys. 49, 622 (1968), DOI:10.1063/1.1670118 | libretexts | 2025-03-17T22:26:41.727893 | 2023-01-07T00:44:10 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/09%3A_Molecular_Spectroscopy/9.06%3A_Jablonski_diagram",
"book_url": "https://commons.libretexts.org/book/chem-424908",
"title": "9.6: Jablonski diagram",
"author": null
} |
https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/13%3A_Chemical_Equilibrium/13.04%3A_Gibbs_Energy_of_a_Reaction_vs._Extent_of_Reaction_is_a_Minimum_at_Equilibrium | 13.4: Gibbs Energy of a Reaction vs. Extent of Reaction is a Minimum at Equilibrium
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As discussed in Section 26.4 , we defined the extent of the reaction (\(\xi\)) as a quantitative measure of how far along the reaction has evolved. For simple reactions like \(\ce{A <=> B}\), the extent of reaction is easy to define and is simply the number of moles of \(\ce{A}\) that has been converted to \(\ce{B}\). Since \(d\xi\) has the same value as \(dn\), we can write that
\[\begin{align*} dG &= \mu_A dn_A + \mu_B dn_B \\[4pt] &= -\mu_A d\xi + \mu_B d\xi. \end{align*} \]
The minus sign comes from the fact that when the reaction goes in the left to right direction, the amount of \(\ce{A}\) is decreasing, while the amount of \(\ce{B}\) is increasing. Looking at these equations, it is reasonable to suggest that:
\[\left(\frac{\partial G}{\partial \xi}\right)_{P T}=\mu_{B}-\mu_{A} \nonumber \]
This is the slope of the free energy with respect to the extent of the reaction. This relationship will have a region where the sign is negative, one point where the value is zero and a region where the value is positive (Figure \(\PageIndex{1}\)). If we look at a plot of \(G\) as a function of \(\xi\), we can see that the point where is the minimum of the curve:
\[\left(\frac{\partial G}{\partial \xi}\right)_{PT}=0 \nonumber \]
This is the point where \(\mu_A\) and \(\mu_B\) are the same. On one side of the minimum, the slope is negative and on the other side, the slope is positive. In both cases, the reaction is spontaneous (i.e, \(dG < 0\)) as long as the reaction is evolving towards the minimum, which will be called the equilibrium position . Hence, all reactions spontaneously move towards equilibrium; the immediate question is what value of \(\xi\) corresponds to the equilibrium position of a reaction?
Terminology
- A reaction for which \(Δ_rG < 0\) is called exergonic .
- A reaction for which \(Δ_r G > 0\) is called endergonic .
- \(Δ_r G < 0\), the forward reaction is spontaneous.
- \(Δ_r G > 0\), the reverse reaction is spontaneous.
- \(Δ_r G = 0\), the reaction is at equilibrium.
Ideal Gas Equilibrium
To understand how we can find the minimum and what the Gibbs free energy of a reaction depends on, let's first start with a reaction that converts one ideal gas into another.
\[\ce{A<=>B} \nonumber \]
Let's assume the reaction enthalpy is zero and hence the only thing that determines what the ratio of product to reactant should be is the entropy term (the mixing term), which is the most favorable when the mixture is half and half.
\[\Delta G_{\text {mix }}=n R T\left(\chi_{A} \ln \chi_{A}+ \chi_{B} \ln \chi_{B}\right) \label{Smixing} \]
Remember that the Gibbs free energy of mixing is not a molar quantity and depends on \(n\) (unlike the reaction Gibbs free energy). Also, the Gibbs free energy of mixing is defined relative to pure \(\ce{A}\) and \(\ce{B}\). The free energy of mixing in Equation \ref{Smixing} is at a minimum when the amounts of \(\ce{A}\) equal \(\ce{B}\).
Thus, if we have a reaction \(\ce{A<=>B}\) and there is no enthalpy term (and no change in the inherent entropy of A vs. B), we would expect the system to have the minimum Gibbs free energy when the mole fraction of \(\ce{A}\) and \(\ce{B}\) are each 0.5.
Think about it
If \(\ce{A}\) can form \(\ce{B}\) and \(\ce{B}\) can form \(\ce{A}\) and there is no other forces involved (no heat/enthalpy) that would favor one product over the other, probability would just state that eventually you will have statistically the same amount of both A and B present -- this is the lowest free energy state. (This, of course, will not be true if the enthalpy for the reaction is not zero or if \(\ce{A}\) and \(\ce{B}\) have different inherent entropies). Usually there are additional terms for the reaction.
Lets work through this for an ideal gas reaction:
\[\begin{align*}
\Delta_{r} G &=\mu_{B}-\mu_{A}
\\[4pt] &=\mu_{B}^{o}+R T \ln \dfrac{P_{B}}{P^{\theta}}-\mu_{A}^{o}-R T \ln \dfrac{P_{A}}{P^{\theta}} \\[4pt]
&=\mu_{B}^{o}-\mu_{A}^{o}+R T \ln \dfrac{P_{B}}{P_{A}}
\end{align*} \]
What we normally do at this point is to give the first two terms a special name. Since it is the difference between the chemical potentials at standard conditions, we refer to it as the Gibbs free energy of reaction at standard conditions or the standard Gibbs free energy of reaction .
\[\begin{align} \Delta_{r} G^{o} &= \mu_B^{o}-\mu_A^{o} \nonumber \\[4pt] \Delta_r G &= \Delta_{r} G^{o} + R T \ln \dfrac{P_{B}}{P_{A}} \label{eq30} \end{align} \]
There are two parts to Equation \ref{eq30}. The first term (\(\Delta_{r} G^{o}\)) is the Gibbs free energy for converting one mole of \(\ce{A}\) to \(\ce{B}\) under standard conditions (1 bar of both \(\ce{A}\) and \(\ce{B}\)). The second term is the mixing term that is minimized when the amounts of A and B are equal to one another (Figure \(\PageIndex{2}\)). The first term you look up in a reference book and is specific for a specific reaction. The second term is calculated knowing the partial pressures of \(\ce{A}\) and \(\ce{B}\) in the gas.
Now the minimum absolute Gibbs free energy will occur at the bottom of the curve where the slope is zero. Thus, the lowest free energy will occur when the reaction free energy (i.e., the slope of that curve in Figure \(\PageIndex{1}\)) is equal to zero. The chemical potentials of \(\ce{A}\) and \(\ce{B}\) are equal.
\[\mu_A = \mu_B \nonumber \]
At this point, the reaction will neither go forwards or backwards and we call this equilibrium. Hence, at equilibrium:
\[ \begin{align*} \Delta_{r} G &=\Delta_{r} G^{o}+R T \ln \dfrac{P_{B}}{P_{A}} \\[4pt] &=0 \end{align*} \]
and the specific ratio of \(P_A\) and \(P_B\) necessary to ensure \(\Delta_{r} G=0\) is characteristic of the reaction and is called the equilibrium constant for that reaction.
\[K_{e q}=\dfrac{P_{B}}{P_{A}} \nonumber \]
We can now relate thermodynamic quantities to concentrations of molecules at equilibrium. We can also see that at equilibrium:
\[\begin{aligned}
&0=\Delta_{r}G^{o}+R T \ln K_{e q} \\[4pt]
&\Delta_{r} G^{o}=-R T \ln K_{e q} \\[4pt]
&K_{eq}=e^{-\dfrac{\Delta_{r} G^{o}}{R T}}
\end{aligned} \nonumber \] | libretexts | 2025-03-17T22:26:43.142062 | 2023-01-07T00:45:29 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/13%3A_Chemical_Equilibrium/13.04%3A_Gibbs_Energy_of_a_Reaction_vs._Extent_of_Reaction_is_a_Minimum_at_Equilibrium",
"book_url": "https://commons.libretexts.org/book/chem-424908",
"title": "13.4: Gibbs Energy of a Reaction vs. Extent of Reaction is a Minimum at Equilibrium",
"author": null
} |
https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/13%3A_Chemical_Equilibrium/13.05%3A_Reaction_Quotient_and_Equilibrium_Constant_Ratio_Determines_Reaction_Direction | 13.5: Reaction Quotient and Equilibrium Constant Ratio Determines Reaction Direction
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The Gibbs free energy function was constructed to be able to predict which changes could occur spontaneously. If we start with a set of initial concentrations we can write them in a reaction quotient
\[Δ_rG = Δ_rG^o + RT \ln Q \nonumber \]
if we subtract the equilibrium version of this expression:
\[0= Δ_rG^o + RT\ln K \nonumber \]
we get
\[Δ_rG = RT \ln \left(\dfrac{Q}{K} \right) \nonumber \]
That gives us the sign of \(Δ_rG\). If this is negative the reaction will spontaneously proceed from left to right as written, if positive it will run in reverse. In both case the value of \(Q\) will change until \(Q=K\) and equilibrium has been reached.
The main difference between \(K\) and \(Q\) is that \(K\) describes a reaction that is at equilibrium, whereas \(Q\) describes a reaction that is not at equilibrium. To determine \(Q\), the concentrations of the reactants and products must be known. For a given general chemical equation:
\[aA + bB \rightleftharpoons cC + dD \label{1} \]
the \(Q\) equation is written by multiplying the activities for the species of the products and dividing by the activities of the reactants. If any component in the reaction has a coefficient, indicated above with lower case letters, the concentration is raised to the power of the coefficient. \(Q\) for the above equation is therefore:
\[Q = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{2} \]
Note
A comparison of \(Q\) with \(K\) indicates which way the reaction shifts and which side of the reaction is favored:
- If \(Q>K\), then the reaction favors the reactants. This means that in the \(Q\) equation, the ratio of the numerator (the concentration or pressure of the products) to the denominator (the concentration or pressure of the reactants) is larger than that for \(K\), indicating that more products are present than there would be at equilibrium. Because reactions always tend toward equilibrium ( Le Ch â telier's Principle ), the reaction produces more reactants from the excess products, therefore causing the system to shift to the LEFT. This allows the system to reach equilibrium.
- If \(Q<K\), then the reaction favors the products. The ratio of products to reactants is less than that for the system at equilibrium—the concentration or the pressure of the reactants is greater than the concentration or pressure of the products. Because the reaction tends toward reach equilibrium, the system shifts to the RIGHT to make more products.
- If \(Q=K\), then the reaction is already at equilibrium. There is no tendency to form more reactants or more products at this point. No side is favored and no shift occurs. | libretexts | 2025-03-17T22:26:43.200799 | 2023-01-07T00:45:30 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/13%3A_Chemical_Equilibrium/13.05%3A_Reaction_Quotient_and_Equilibrium_Constant_Ratio_Determines_Reaction_Direction",
"book_url": "https://commons.libretexts.org/book/chem-424908",
"title": "13.5: Reaction Quotient and Equilibrium Constant Ratio Determines Reaction Direction",
"author": null
} |
https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/15%3A_Solutions_II_-_Nonvolatile_Solutes/15.07%3A_Extending_Debye-Huckel_Theory_to_Higher_Concentrations | 15.7: Extending Debye-Hückel Theory to Higher Concentrations
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The equation for \(\log \gamma _{\pm }\) predicted from Debye–Hückel limiting law is:
\[\log _{10}\gamma _{\pm }=-Az_{j}^{2}{\frac {\sqrt {I}}{1+Ba_{0}{\sqrt {I}}}} \label{DH} \]
It gives satisfactory agreement with experimental measurements for low electrolyte concentrations, typically less than \(10^{−3} mol/L\). Deviations from the theory occur at higher concentrations and with electrolytes that produce ions of higher charges, particularly asymmetrical electrolytes. These deviations occur because the model is oversimplified, so there is little to be gained by making small adjustments to the model. Instead, we must challenge the individual assumptions of the model:
- Ions do not interact with each other . Ion association may take place, particularly with ions of higher charge. This was followed up in detail by Niels Bjerrum. The Bjerrum length is the separation at which the electrostatic interaction between two ions is comparable in magnitude to \(kT\).
- Complete dissociation . A weak electrolyte is one that is not fully dissociated. As such it has a dissociation constant. The dissociation constant can be used to calculate the extent of dissociation and hence, make the necessary correction needed to calculate activity coefficients.
- Ions are spherical point charges that cannot be polarizable . Ions, as all other atoms and molecules, have a finite size. Many ions, such as the nitrate ion \(\ce{NO3^{−}}\), are not spherical. Polyatomic ions are polarizable.
- The solvent composition does not matter . The solvent is not a structureless medium but is made up of molecules. The water molecules in aqueous solution are both dipolar and polarizable. Both cations and anions have a strong primary solvation shell and a weaker secondary solvation shell. Ion–solvent interactions are ignored in Debye–Hückel theory.
- Ionic radius is negligible . At higher concentrations, the ionic radius becomes comparable to the radius of the ionic atmosphere.
Most extensions to the Debye–Hückel theory are empirical in nature. They usually allow the Debye–Hückel equation to be followed at low concentration and add further terms in some power of the ionic strength to fit experimental observations. Several approaches have been proposed to extend the validity of the Debye–Hückel theory.
Extended Debye-Hückel Equation
One such approach is the Extended Debye-Hückel Equation:
\[- \log(\gamma) = \dfrac{A|z_+z_-|\sqrt{I}}{1 + Ba\sqrt{I}} \nonumber \]
where \(\gamma\) is the activity coefficient, \(z\) is the integer charge of the ion \(\mu\) is the ionic strength of the aqueous solution, and \(a\), is the size or effective diameter of the ion in angstrom. The effective hydrated radius of the ion, \(a\) is the radius of the ion and its closely bound water molecules. Large ions and less highly charged ions bind water less tightly and have smaller hydrated radii than smaller, more highly charged ions. Typical values are 3 Å for ions such as H+,Cl-,CN-, and HCOO-. The effective diameter for the hydronium ion is 9 Å. \ (A\) and \(B\) are constants with values of respectively 0.5085 and 0.3281 at 25°C in water.
Other approaches include the Davies equation, Pitzer equations and specific ion interaction theory.
Davis Equation
The Davies equation is an empirical extension of Debye–Hückel theory which can be used to calculate activity coefficients of electrolyte solutions at relatively high concentrations at 25 °C. The equation, originally published in 1938, was refined by fitting to experimental data. The final form of the equation gives the mean molal activity coefficient f± of an electrolyte that dissociates into ions having charges z1 and z2 as a function of ionic strength I:
\[ -\log f_{\pm }=0.5z_{1}z_{2}\left({\frac {\sqrt {I}}{1+{\sqrt {I}}}}-0.30I\right). \nonumber \]
The second term, 0.30 \(I\), goes to zero as the ionic strength goes to zero, so the equation reduces to the Debye–Hückel equation at low concentration. However, as concentration increases, the second term becomes increasingly important, so the Davies equation can be used for solutions too concentrated to allow the use of the Debye–Hückel equation. For 1:1 electrolytes the difference between measured values and those calculated with this equation is about 2% of the value for 0.1 M solutions. The calculations become less precise for electrolytes that dissociate into ions with higher charges. Further discrepancies will arise if there is association between the ions, with the formation of ion pairs, such as \(\ce{Mg^{2+}SO^{2-}4}\).
Plot of activity coefficients calculated using the Davies equation.
Pitzer Equations
Pitzer equations are important for the understanding of the behaviour of ions dissolved in natural waters such as rivers, lakes and sea-water. They were first described by physical chemist Kenneth Pitzer. The parameters of the Pitzer equations are linear combinations of parameters, of a virial expansion of the excess Gibbs free energy, which characterize interactions amongst ions and solvent. The derivation is thermodynamically rigorous at a given level of expansion. The parameters may be derived from various experimental data such as the osmotic coefficient, mixed ion activity coefficients, and salt solubility. They can be used to calculate mixed ion activity coefficients and water activities in solutions of high ionic strength for which the Debye–Hückel theory is no longer adequate.
An expression is obtained for the mean activity coefficient.
\[\ln \gamma _{\pm }= \dfrac {p\ln \gamma _{M}+q\ln \gamma _{X}}{p+q} \nonumber \]
\[\ln \gamma _{\pm }=|z^{+}z^{-}|f^{\gamma }+m\left({\frac {2pq}{p+q}}\right)B_{MX}^{\gamma }+m^{2}\left[2{\frac {(pq)^{3/2}}{p+q}}\right]C_{MX}^{\gamma } \nonumber \]
These equations were applied to an extensive range of experimental data at 25 °C with excellent agreement to about 6 mol kg−1 for various types of electrolyte. The treatment can be extended to mixed electrolytes and to include association equilibria. Values for the parameters β(0), β(1) and C for inorganic and organic acids, bases and salts have been tabulated. Temperature and pressure variation is also discussed.
Specific ion interaction theory
Specific ion Interaction Theory (SIT theory) is a theory used to estimate single-ion activity coefficients in electrolyte solutions at relatively high concentrations. It does so by taking into consideration interaction coefficients between the various ions present in solution. Interaction coefficients are determined from equilibrium constant values obtained with solutions at various ionic strengths. The determination of SIT interaction coefficients also yields the value of the equilibrium constant at infinite dilution.
The activity coefficient of the jth ion in solution is written as \(γ_j\) when concentrations are on the molal concentration scale and as yj when concentrations are on the molar concentration scale. (The molality scale is preferred in thermodynamics because molal concentrations are independent of temperature). The basic idea of SIT theory is that the activity coefficient can be expressed as
\[\log \gamma _{j}=-z_{j}^{2}{\frac {0.51{\sqrt {I}}}{1+1.5{\sqrt {I}}}}+\sum _{k}\epsilon _{jk}m_{k} \nonumber \]
where z is the electrical charge on the ion, \(I\) is the ionic strength, \(ε\) and \(b\) are interaction coefficients and \(m\) are concentrations. The summation extends over the other ions present in solution, which includes the ions produced by the background electrolyte. The first term in these expressions comes from Debye-Hückel theory. The second term shows how the contributions from "interaction" are dependent on concentration. Thus, the interaction coefficients are used as corrections to Debye-Hückel theory when concentrations are higher than the region of validity of that theory. | libretexts | 2025-03-17T22:26:44.119254 | 2023-01-07T00:46:17 | {
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"url": "https://chem.libretexts.org/Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/15%3A_Solutions_II_-_Nonvolatile_Solutes/15.07%3A_Extending_Debye-Huckel_Theory_to_Higher_Concentrations",
"book_url": "https://commons.libretexts.org/book/chem-424908",
"title": "15.7: Extending Debye-Hückel Theory to Higher Concentrations",
"author": "Wikipedia"
} |
https://human.libretexts.org/Bookshelves/Languages/Spanish/Instructor_Guide%3A_Tarea_Libre_2-Second-Year_Spanish_(Moon_Lee_Harmon_and_Diaz_Rodil)/01%3A_Introducing_Tarea_Libre | 1: Introducing Tarea Libre 2
Introduction to Tarea Libre 2
This is an instructor guide to Tarea Libre 2 . The Tarea Libre 2 project is a comprehensive and accessible Open Educational Resources question bank of 800 interactive activities created in ADAPT for second-year Spanish courses for California Community Colleges (CCC) that can be used via ADAPT or linked to a campus Learning Management System via LTI (Learning Tools Interoperability). This LTI integration seamlessly connects the assigned activities to the grade book streamlining the assigning and grading of homework.
The project follows the success of Tarea Libre for first-year Spanish and aims to create the foundation for a cohesive OER Spanish language homework set equivalent to what is available through paid-subscriptions licenses provided by commercial publishers. This homework ancillary fills an existing gap in OER and increases the likelihood of OER adoption in the first two years of Spanish courses.
Tarea Libre 2 aligns with California’s C-ID SPAN 200 and SPAN 210 , the curriculum standard for higher education disciplines. Additionally, the project aligns with existing Spanish OER text materials already identified on the ASCCC OERI Spanish Discipline page . It includes a variety of formative and summative assessments that cover a range of skills. Each module has multiple levels of engagement, allowing instructors to create homework assignments that address different skill levels. Additionally, it will enable students to practice speaking, listening, reading, and writing skills.
The Academic Senate for California Community Colleges (ASCCC) IDEA framework of inclusion, diversity, equity, and anti-racism is at the foundation of this project. All 8 modules have been designed to address them through various readings that underscore the cultural practices of minoritized groups in the Spanish-speaking world, including the United States.
About the Project
Tarea Libre covers 32 grammar topics typically taught in second-year Spanish courses. To increase adaptability and flexibility for instructors, the project has 4 modules for SPAN 200 and 4 modules for SPAN 210 , based on California’s C-ID curriculum model.
Each module is comprised of 4 to 5 grammar topics and listening, speaking, and reading comprehension activities. As found on most homework (workbook and lab activities), each grammar topic has a four levels of mastering steps:
-
Paso 1 "Empecemos"
-
Paso 2 "Continuemos"
-
Paso 3 "Practiquemos"
-
Paso 4 "Avancemos”
-
De todo un poco
Each grammar module has an average of 20 auto-graded activities.
In addition to these auto-graded activities, Tarea Libre 2 includes four sections of instructor-graded and auto-graded activities that target oral, speaking, and reading skills:
-
Hablemos: Using the ADAPT recorder, students answer the prompt questions.
-
Escuchemos: These short thematic videos were created in Powtoon. Using the ADAPT recorder, students provide their answers after watching these videos.
-
Leamos: These cultural readings with narration and auto-graded comprehension questions were created on various topics.
-
Conversemos: Through these interactive discussion assessments, students are able to post audio, video, or text submissions. Topics are directly connected to topics in Acceso .
About the Integration of IDEA Principles and Culturally Responsive Pedagogy
IDEA principles and culturally responsive pedagogy (CRP) are integral to our project. The project’s eight modules contain a variety of cultural themes across the Spanish-speaking world. We, as authors, have curated modules that celebrate the rich cultural tapestry of the Spanish-speaking world. Through the inclusion of voices, stories, and traditions that are often overlooked in standard textbooks, we aim to provide a more inclusive and engaging learning experience.
From personal narratives like those of Chef José Andrés and Albert Espinosa, to exploring the multicultural influences shaping Latin American societies, such as the Levantine roots of tacos al pastor and Korean immigration in Mexico, our content highlights the complexity of these communities. Moreover, our focus on Indigenous groups—through articles on the Garifuna, Ngäbe-Buglé, Mapuche, Rapa Nui, and sustainable initiatives in the Moskitia—underscores the importance of preserving cultural diversity and ancestral knowledge.
By embracing these diverse perspectives, we hope to inspire a deeper appreciation for the varied and dynamic cultures that make up the Spanish-speaking world.
Thumbnail image by Wikimedia under CC BY-SA 4.0 license | libretexts | 2025-03-17T22:26:45.765533 | 2024-09-28T15:49:54 | {
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"url": "https://human.libretexts.org/Bookshelves/Languages/Spanish/Instructor_Guide%3A_Tarea_Libre_2-Second-Year_Spanish_(Moon_Lee_Harmon_and_Diaz_Rodil)/01%3A_Introducing_Tarea_Libre",
"book_url": "https://commons.libretexts.org/book/human-285761",
"title": "1: Introducing Tarea Libre 2",
"author": "Cristina Moon, Alejandro Lee, Sarah Harmon, and Hugo Díaz-Rodil"
} |
https://human.libretexts.org/Bookshelves/Languages/Spanish/Instructor_Guide%3A_Tarea_Libre_2-Second-Year_Spanish_(Moon_Lee_Harmon_and_Diaz_Rodil)/02%3A_Overview_of_the_Grammar_and_Vocabulary_Content | 2: Overview of the Grammar and Thematic Modules
Grammar and Thematic Modules
Tarea Libre 2 covers more than 32 grammar topics typically taught in second-year Spanish courses in the California Community Colleges. The thematic modules follow closely the Acceso project from the University of Kansas.
Tarea Libre 2 aims to increase adaptability and flexibility for Spanish instructors. Based on California’s C-ID curriculum model, the project has 4 modules for SPAN 200 and 4 modules for SPAN 210 .
Each of the modules is composed of the following:
-
Four to five grammar topics. As found on most homework (workbook and lab activities), each grammar topic includes activities that connect to the
Acceso
the thematic modules and is follows the 4 levels of mastering steps:
- Paso 1 Empecemos : Focus on recognizing grammatical concepts, mostly multiple choice and select choice questions.
- Paso 2 Continuemos : Focus on the initial application of grammatical topics, mostly fill-in-the-blank in shorter contexts.
- Paso 3 Practiquemos : Focus on the expanded application of grammatical topics, mostly fill-in-the-blank and select choices in longer paragraphs.
- Paso 4 Avancemos : Focus on the expanded application of grammatical topics in dialogues, mostly fill-in-the-blanks.
- De todo un poco : Focus on a variety of concepts that confuses students such as punctuation, false cognates, spelling, cultural and linguistic differences.
- Listening, speaking, and reading comprehension activities.
Section |
Grammar Topic |
Thematic Alignment with Acceso Module |
Number of Activities |
|---|---|---|---|
|
1.1 |
Números: ordinales, cardinales, romanos |
Module 1: Estados Unidos |
10 |
|
1.2 |
Comparativos y superlativos |
Module 1: Estados Unidos |
16 |
|
1.3 |
Presente del indicativo |
Module 1: Estados Unidos |
15 |
|
1.4 |
Ser y estar |
Module 1: Estados Unidos |
16 |
|
1.5 |
Usos del infinitivo y gerundio |
Module 1: Estados Unidos |
16 |
|
1.6 |
Énfasis silábico, división silábica, clasificación, diacríticos |
Module 1: Estados Unidos |
27 |
|
2.1 |
El imperfecto |
Module 2: España |
16 |
|
2.2 |
El pretérito |
Module 2: España |
16 |
|
2.3 |
Pretérito vs. imperfecto |
Module 2: España |
16 |
|
2.4 |
Verbos como gustar |
Module 2: España |
16 |
|
2.5 |
Reglas de puntuación, verbos como gustar en el pretérito y en el imperfecto |
Module 2: España |
19 |
|
3.1 |
El presente del subjuntivo y sus usos |
Module 3: El Caribe |
16 |
|
3.2 |
El subjuntivo enn cláusulas adjetivales |
Module 3: El Caribe |
17 |
|
3.3 |
El subjuntivo en cláusulas adverbiales |
Module 3: El Caribe |
16 |
|
3.4 |
El subjuntivo en combinación |
Module 3: El Caribe |
16 |
|
3.5 |
Mayúsculas y minúsculas, ortografía (c, s, z), ortografía (b, v), ortografía hache |
Module 3: El Caribe |
17 |
|
4.1 |
El objeto directo |
Module 4: México |
23 |
|
4.2 |
El objeto indirecto |
Module 4: México |
16 |
|
4.3 |
El pretérito perfecto del indicativo y el pretérito perfecto del subjuntivo |
Module 4: México |
16 |
|
4.4 |
Cognados falsos, participios adjetivales |
Module 4: México |
16 |
|
5.1 |
El futuro simple |
Module 5: Centroaméricoa |
18 |
|
5.2 |
El condicional |
Module 5: Centroamérica |
16 |
|
5.3 |
Pronombres de objeto directo e indirecto |
Module 5: Centroamérica |
16 |
|
5.4 |
Por y para |
Module 5: Centroamérica |
16 |
|
5.5 |
Transiciones, verbos con preposición |
Module 5: Centroamérica |
24 |
|
6.1 |
Cláusulas con "si" en situaciones reales |
Module 6: Caribe Continental |
16 |
|
6.2 |
Imperfecto del subjuntivo |
Module 6: Caribe continental |
16 |
|
6.3 |
Cláusulas con "si" en situaciones hipotéticas |
Module 6: Caribe continental |
16 |
|
6.4 |
Mandatos formales e informales |
Module 6: Caribe continental |
16 |
|
6.5 |
Más ortografía (j,g), (c, k), (ll, y, i) |
Module 6: Caribe continental |
16 |
|
7.1 |
Ser y estar |
Module 7: Andes y Amazonas |
16 |
|
7.2 |
El pretérito perfecto del indicativo y el pretérito perfecto del subjuntivo |
Module 7: Andes y Amazonas |
16 |
|
7.3 |
El pluscuamperfecto del indicativo y el pluscuamperfecto del subjuntivo |
Module 7: Andes y Amazonas |
18 |
|
7.4 |
Los tiempos perfectos en construcciones hipotéticas |
Module 7: Andes y Amazonas |
16 |
|
7.5 |
Diferencias culturales, lingüísticas, léxicas y dialectales |
Module 7: Andes y Amazonas |
18 |
|
8.1 |
Pretérito vs. imperfecto |
Module 8: Cono Sur |
16 |
|
8.2 |
Usos impersonales de hacer |
Module 8: Cono Sur |
16 |
|
8.3 |
La voz pasiva |
Module 8: Cono Sur |
20 |
|
8.4 |
Usos comunes de se |
Module 8: Cono Sur |
16 |
|
8.5 |
Cognados falsos |
Module 8: Cono Sur |
16 |
Thumbnail image by Amador Loureiro on Unsplash . | libretexts | 2025-03-17T22:26:45.879737 | 2024-09-28T15:50:08 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://human.libretexts.org/Bookshelves/Languages/Spanish/Instructor_Guide%3A_Tarea_Libre_2-Second-Year_Spanish_(Moon_Lee_Harmon_and_Diaz_Rodil)/02%3A_Overview_of_the_Grammar_and_Vocabulary_Content",
"book_url": "https://commons.libretexts.org/book/human-285761",
"title": "2: Overview of the Grammar and Thematic Modules",
"author": "Cristina Moon, Alejandro Lee, Sarah Harmon, and Hugo Díaz-Rodil"
} |
https://human.libretexts.org/Bookshelves/Languages/Spanish/Instructor_Guide%3A_Tarea_Libre_2-Second-Year_Spanish_(Moon_Lee_Harmon_and_Diaz_Rodil)/03%3A_Overview_of_Comprehensive_Assessments | 3: Overview of Comprehensive Assessments
Comprehensive Assessments in ADAPT
The comprehensive assessments for Tarea Libre 2 build on the success of similar assignments in Tarea Libre . With the addition of incorporated media and a new question type, the Tarea Libre 2 team members were able to expand the assessment types to include interactive questions that allow for more opportunities for learners to showcase their comprehension and production skills, doing so with greater depth. Just as in Tarea Libre , all media used in the assignments is either openly licensed or was created by the team.
Types of Comprehensive Assessments
Leamos
Building on the success of this assessment type in Tarea Libre , the team wrote original pieces and adapted from known open-access news sites for Leamos. These readings were chosen to mirror the topics of Acceso , particularly the readings showcased in the "Almanaque" and "Un paso más" sections. Of note is that the topics represent a wide array of peoples and cultures from the Spanish-speaking world, in compliance with the IDEA Framework of the Academic Senate of California Community College Open Educational Resource Initiative (ASCCC OERI), thereby incorporating culturally responsive educational best practices. The readings are accompanied by audio narrations by members of our team, and they are followed by comprehension questions. There is more information on these readings in the next chapter, Overview of Reading Content . Examples of the readings types can be seen in the images below.
Escuchemos
Building on the success of this assessment type in Tarea Libre , the team created original videos using Powtoon that focused on topics related to various Acceso topics, particularly those showcased in the "Almanaque" and "Un paso más" sections. These original videos include either monologues on the topics or dialogues between people discussing the topics. The videos have then been incorporated into ADAPT and are captioned and transcribed into Spanish. An example of the Escuchemos content can be seen below.
After watching the video, students are then asked to record answers to comprehension questions using the recording widget embedded within ADAPT. Alternatively, students can submit an audio file with that information. Regardless of the submission, ADAPT will automatically caption and transcribe the submission for instructors. An example of the recording widget can be seen below.
Hablemos
Building on the success of this assessment type in Tarea Libre, the team created new comprehension assessments that tie directly to various readings in Acceso , particularly to "Almanaque" and "Un paso más" sections. Students will submit audio recordings that answer the questions in the prompt using the same audio recording widget that is used with Escuchemos assessments. An example of the Hablemos content is shown below.
Conversemos
New for this project, Tarea Libre 2 incorporates the question type Discuss-It to create interactive discussion assessments. These discussions are directly connected to various topics in Acceso , particularly readings in "Almanaque" and "Un paso más" sections. Students will be required to post an initial audio, video, or text recording, followed by posting reactions to the recordings of their classmates. This assessment type can be used in online courses to satisfy regular and substantive interaction (RSI), and can be used in all courses to encourage oral production. An example of the Conversemos content is shown below. | libretexts | 2025-03-17T22:26:46.021083 | 2024-10-12T18:17:00 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://human.libretexts.org/Bookshelves/Languages/Spanish/Instructor_Guide%3A_Tarea_Libre_2-Second-Year_Spanish_(Moon_Lee_Harmon_and_Diaz_Rodil)/03%3A_Overview_of_Comprehensive_Assessments",
"book_url": "https://commons.libretexts.org/book/human-285761",
"title": "3: Overview of Comprehensive Assessments",
"author": "Cristina Moon, Alejandro Lee, Sarah Harmon, and Hugo Díaz-Rodil"
} |
https://human.libretexts.org/Bookshelves/Languages/Spanish/Instructor_Guide%3A_Tarea_Libre_2-Second-Year_Spanish_(Moon_Lee_Harmon_and_Diaz_Rodil)/04%3A_Overview_of_Reading_Content | 4: Overview of Reading Content
More on Leamos
To accompany the content in Acceso , 16 cultural readings with recorded narration and auto-graded comprehension questions have been created. In ADAPT, we have titled these readings "Leamos". There are at least 2 readings per module on various topics. We have included people, traditions, history, geography, music, and other elements from the Spanish-speaking world that are not traditionally found on commercial textbooks for second-year Spanish courses. Our project aims to highlight the cultural diversity of the Spanish-speaking world and its people by applying the Academic Senate of California Community College Open Educational Initiative's IDEA Framework on all the readings.
Below is a reading on the Albert Espinosa, a Spanish industrial engineer, writer, director, and actor with disabilities.
Below is a reading on the history of tacos al pastor and the Levantine influence in México.
Below you will find a list of titles by modules and sections.
List of Titles
-
1A Censo 2020
-
1B Las aventuras del chef José Andrés y sus hijas
-
2A Albert Espinosa
-
2B Ventajas e inconvenientes de vivir en España
-
3A Ponce de León
-
3B El son cubano y el Buena Vista Social Club
-
4A Tacos al pastor y la influencia levantina en México
-
4B Una nueva ola de inmigrantes: los coreanos en México
-
5A Los garífunas de América Central
-
5B Proyecto de desarrollo sostenible en la Moskitia
-
6A La lucha por la preservación: Los ngäbe-buglé de Panamá
-
6B Descubrimientos arqueológicos en el Caribe continental
-
7A El artista peruano Félix Espinosa
-
7B Nuevos descubrimientos en la selva amazónica en Ecuador
-
8A ¿Qué es el lunfardo?
-
8B El rongo rongo de Rapa Nui
-
8C El mapudungun, la lengua del pueblo mapuche
Thumbnail image by Dip Devices on Unsplash . | libretexts | 2025-03-17T22:26:46.084213 | 2024-09-28T15:50:05 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://human.libretexts.org/Bookshelves/Languages/Spanish/Instructor_Guide%3A_Tarea_Libre_2-Second-Year_Spanish_(Moon_Lee_Harmon_and_Diaz_Rodil)/04%3A_Overview_of_Reading_Content",
"book_url": "https://commons.libretexts.org/book/human-285761",
"title": "4: Overview of Reading Content",
"author": "Cristina Moon, Alejandro Lee, Sarah Harmon, and Hugo Díaz-Rodil"
} |
https://human.libretexts.org/Bookshelves/Languages/Spanish/Instructor_Guide%3A_Tarea_Libre_2-Second-Year_Spanish_(Moon_Lee_Harmon_and_Diaz_Rodil)/05%3A_Suggestions_for_using_Tarea_Libre | 5: Suggestions for using Tarea Libre 2
How to Use Tarea Libre 2
This guide provides instructors with suggestions on how to integrate Tarea Libre 2 into their intermediate and advanced Spanish courses. Designed as an accessible OER, this question bank includes 800 interactive activities that align with the standards established for intermediate and advanced Spanish courses in California higher education, SPAN 200 and SPAN 210 California’s C-ID curriculum model. Tarea Libre 2 can be used directly in ADAPT or seamlessly integrated with your Learning Management System (LMS) via LTI. This integration ensures automatic gradebook synchronization with most LMSs.
Getting Started
-
For Instructors Already Using Entrada Libre and Acceso :
The modules and topics in Tarea Libre 2 mirror the structure and progression of these popular OERs. This continuity makes it easy to adapt your course materials and align homework assignments with these OERs. -
Replacing Commercial Platforms :
Tarea Libre 2 serves as an OER alternative to publisher-based homework platforms. With its extensive question bank, you can select activities that best address your students' specific learning needs.
Activity Structure
Each grammar module includes four levels of progressively challenging activities:
- Paso 1: Empecemos – Introduction to foundational grammar concepts.
- Paso 2: Continuemos – Intermediate practice with guided activities.
- Paso 3: Practiquemos – Advanced practice focusing on mastery.
- Paso 4: Avancemos – Synthesis of learned material in a more advanced format.
Each module also features a De todo un poco section, offering a variety of auto-graded activities to reinforce learning.
Skills-Based Assessments
Beyond grammar, Tarea Libre 2 provides skill-building activities in:
- Speaking (Hablemos) – Prompts using the ADAPT recorder widget.
- Listening (Escuchemos) – Short, thematic videos paired with comprehension exercises using the ADAPT recorder widget.
- Reading (Leamos) – Culturally enriched readings with comprehension checks.
- Interactive Discussions (Conversemos) – Open-ended prompts linked to Acceso topics, supporting threaded audio, video, or text submissions. This activity uses the ADAPT Discuss-It question type.
Cultural Themes and IDEA Integration
All eight modules in Tarea Libre 2 embody the principles of inclusion, diversity, equity, and anti-racism (IDEA), alongside culturally responsive pedagogy (CRP). Through these features, students engage with:
- Underrepresented voices and narratives , such as Indigenous groups and diverse cultural traditions.
- Explorations of multicultural influences , like the Levantine roots of tacos al pastor or Korean-Mexican connections.
- Inspirational stories from figures like Chef José Andrés and Albert Espinosa.
By emphasizing these diverse perspectives, Tarea Libre 2 enriches the language-learning experience and fosters deeper cultural appreciation.
OER Textbook Remixes for Second-year Spanish
To help with the adoption of Tarea Libre 2 , we have created two possible remixes of Spanish OERs for second-year Spanish courses. You can take any of the already created remixes and use them as-is or re-remix them by adding, deleting, or moving content as needed. Remember, this is your opportunity to create your free customized Spanish OER.
We will be working on a second edition of Entrada Libre with updated modules and to address the additional content added on Tarea Libre 2 . We expect to release it by the end of summer 2025.
Below we show examples of third-semester and fourth-semester textbook remixes with the sources and a list of content and activities by chapter.
Spanish 2A: Intermediate Spanish (Chabot College). Below you can find the breakdown of grammar topics covered on this remix:
-
1.1 Los números -
1.2 Los comparativos
-
1.3 El presente del indicativo
-
1.4 Ser y estar. Ir + a + infinitivo
-
2.1 El imperfecto
-
2.2 El pretérito
-
2.3 El pretérito en contraste con el imperfecto
-
2.4 Verbos como gustar
-
3.1 Parte 1: Las formas del subjuntivo
-
3.2 Parte 2: El subjuntivo en cláusulas nominales
-
3.3 Parte 3: El subjuntivo en cláusulas adjetivales
-
3.4 Parte 4: El subjuntivo en cláusulas adverbiales
-
4.1 Los pronombres de objeto directo
-
4.2 Los pronombres de objeto indirecto
-
4.3 Usos comunes de "se"
-
4.4 El pretérito perfecto
Spanish 2B: Advanced Spanish (Chabot College). Below you can find the breakdown of the grammar topics on this remix:
-
5.1 El futuro
-
5.2 El condicional
-
5.3 Los pronombres de objeto directo y de objeto indirecto
-
5.4 Las preposiciones por y para
-
6.1 Cláusulas con “si”
-
6.2 El imperfecto del subjuntivo
-
6.3 Cláusulas con “si” en situaciones hipotéticas
- 6.4 Los mandatos formales e informales
-
7.1 Ser y estar—Repaso y más funciones
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7.2 El pretérito perfecto
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7.3 El pluscuamperfecto
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7.4 Los tiempos perfectos en construcciones hipotéticas—más cláusulas con “si”
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8.1 Pretérito vs. imperfecto
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8.2 Más usos comunes de "se"
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8.3 Las reglas de acentuación
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"book_url": "https://commons.libretexts.org/book/human-285761",
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/1%3A_Introduction_to_Physiology | Skip to main content
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1.1: What is Physiology?
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Physiology is a study of how animal form or structure and function sustain life and shape responses to environmental conditions. The multicellular bodies of animals consist of tissues that make up more complex organs and organ systems. The organ systems of an animal maintain homeostasis within the multicellular body.
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1.2: Limits of Life
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Life on Earth exists in nearly every niche on our planet. However, life forms are not equally successful in all conditions on Earth, and one challenge is to find under which conditions life can be sustained. Certainly, life on Earth is very adaptable, which has led to an immense biomass and an incredible biodiversity. Therefore, it is a challenge to find where the limits to this adaptability are.
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1.3: Homeostasis
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Homeostasis refers to the relatively stable state inside the body of an animal. Animal organs and organ systems constantly adjust to internal and external changes in order to maintain this steady state. The goal of homeostasis is the maintenance of equilibrium around a specific value of some aspect of the body or its cells called a set point. While there are normal fluctuations from the set point, the body’s systems will usually attempt to go back to this point. | libretexts | 2025-03-17T22:26:51.534879 | 2018-10-18T16:30:14 | {
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"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "1: Introduction to Physiology",
"author": "Sanja Hinić-Frlog"
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/1%3A_Introduction_to_Physiology/1.1%3A_What_is_Physiology | 1.1: What is Physiology?
- Define physiology and explain its importance and connections to other fields of biological studies
Physiology is a study of how animal form or structure and function sustain life and shape responses to environmental conditions. The arctic fox (Figure 1.1), a complex animal that has adapted to its environment, illustrates the relationships between an animal’s form and function. The multicellular bodies of animals consist of tissues that make up more complex organs and organ systems. The organ systems of an animal maintain homeostasis within the multicellular body. These systems are adapted to obtain the necessary nutrients and other resources needed by the cells of the body, to remove the wastes those cells produce, to coordinate the activities of the cells, tissues, and organs throughout the body, and to coordinate the many responses of the individual organism to its environment.
Some examples of arctic fox physiological adaptions to its environment include:
- Fur with insulative properties to keep the arctic fox warm;
- Reduced heat loss through adaptations to body components such as shorter ears, legs, muzzle;
- Ability to stand on cold surfaces due to capillary rete in pads;
- Can reduce metabolic rate to conserve energy, especially helpful in times of food scarcity.
SOURCE: PESTRUD, P. (1991). Adaptations by the Arctic Fox (Alopex lagopus) to the Polar Winter. Arctic, 44 (2), 132-138. Retrieved 2018 February 5 from www.jstor.org/stable/40511073.
The study of physiology is interrelated with other fields of biological studies, such as
- Anatomy (how organisms are structured),
- Biogeography (spatial and temporal distribution of organisms),
- Biomechanics (the study of how organisms move),
- Conservation Biology (the study of the natural environment and ecosystems),
- Ecology (the study of how organisms interact with other living organisms and their environment),
- Ethology (the study of animal behaviour), and
- Other fields of Biology.
Define physiology in your own words.
Watch this video to see how Arctic foxes find prey in the winter. | libretexts | 2025-03-17T22:26:51.594385 | 2018-10-18T16:30:14 | {
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"title": "1.1: What is Physiology?",
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/1%3A_Introduction_to_Physiology/1.2%3A_Limits_of_Life | 1.2: Limits of Life
- Describe physical, chemical and evolutionary limitations to sustaining life.
Life on Earth exists in nearly every niche on our planet. However, life forms are not equally successful in all conditions on Earth, and one challenge is to find under which conditions life can be sustained. Certainly, life on Earth is very adaptable, which has led to an immense biomass and an incredible biodiversity. Therefore, it is a challenge to find where the limits to this adaptability are.
Try to come up with the best answer for the following question:
Which of the following may be the most limiting physical factor for animals living in aquatic environments? Why?
- locomotion
- being wet all the time
- access to nutrients
- access to oxygen
On Earth, life is based on carbon as a major building block, water as a solvent and chemical bonds and light as life-sustaining energy sources, which seems to be an ideal combination on a terrestrial planet like Earth, which has an average surface temperature of about 15 °C under a pressure of 1 bar. In principle, other building blocks than carbon are feasible, such as silicon, but these would require very different planetary conditions to be used in a comparable fashion to carbon on Earth. Water as a solvent has great benefits and also some challenges (reviewed by Schulze-Makuch and Irwin [Schulze-Makuch & Irwin., 2008]), but being the most abundant molecule on our planet that exists in liquid form, life had simply to adapt to some of its drawbacks. Light is plentiful on our planet, and organic compounds with covalent bonds are versatile at average Earth temperatures, thus providing a powerful combination that resulted on Earth in a biosphere with a large biomass and an incredible biodiversity, including complex life.
There are limits to the conditions under which life can exist on our planet. Most pronounced is the temperature envelope under which active life can exist. Organismic growth can usually occur at temperatures from at least −15 °C to about 113 °C. There are also reports in the literature that the temperature range may even be broader. For example, metabolic activity was inferred down to temperatures of −40 °C due to anomalous concentrations of gases [Campen et al ., 2003], and the upper temperature limit may be as high as about 122 °C. At that temperature, methanogenic archaea could be cultured under a pressure of 20 MPa [Takai et al ., 2008], only limited by the solubility of lipids in water and protein stability [McKay, 2014]. In principle, if the biochemistry of organisms could be adapted to these extremes, perhaps even higher temperatures may be tolerated, but the practical limit due to energetic and biochemical constraints under which life can still metabolize and reproduce is surely much lower [Schulze-Makuch, 2015]. Hyperthermophilic microorganisms require specialized cell components, like proteins and membranes, to be stable and function at high temperatures. Particularly, at temperatures of 100 °C and beyond, some low molecular weight compounds, such as ATP and NAD, hydrolyze quite rapidly, and thermolabile amino acids, like cysteine and glutamic acid, are decomposing [Stetter, 1999]. The pressure tolerance of life, though, is high and extends to at least 1100 bar [ 1 ] [Stan-Lotter, 2007].
Organisms, particularly microbes and other microscopic organisms, are quite tolerant to extreme pH values, from just below 0 to about 13. Ferroplasma sp. and Cephalosporium are examples of organisms that live at low pH-values, Natronobacterium, and several species of protists and rotifers are examples of organisms that live at very high pH-values [Schulze-Makuch & Irwin, 2008, Baross et al ., 2007].
However, life on Earth is relatively sensitive to a lack of water: bacteria, archaea, and fungi can only metabolize at water activities down to about 0.6 [Stevenson, et al., 2014]. Adaptation to water with high salt content, however, is quite common, as some Halobacteria and archaea can grow in 35% NaCl solution [Schulze-Makuch & Irwin, 2008].
Another physical limit to life is radiation, both UV, and ionizing radiation. Tolerances to radiation vary widely. Tardigrades, microscopic animals that usually live in mosses and lichen, can withstand ionizing radiation doses up to 5000 Gy when in the dormant state [Schulze-Makuchh & Seckbach, 2013] and display additional special adaptation traits, such as anhydrobiosis and cryptobiosis [Watanabe, 2006] (elaborated on below). Deinococcus radiodurans can still tolerate higher radiation doses and grow at doses upward of 10,000 Gy.
One non-chemical limitation of life on earth is related to scaling of body size.
Some examples of physical and chemical limitations of life were listed above. One aspect that was not discussed is size. How do you think size is a physical limitation for sustaining life for animals? Find one specific example with one of your peers. Once you have found your example, you should think about why the size is the limit in your example.
Add exercises text here. For the automatic numbDescribe five different limitations to life on Earth.
- Stan-Lotter, H. Extremophiles, the physicochemical limits of life (growth and survival). In Complete Course in Astrobiology; Horneck, G., Rettberg, P., Eds.; Wiley-VCH: Weinheim, Germany, 2007; pp. 121–150. ↵ | libretexts | 2025-03-17T22:26:51.656340 | 2018-10-18T16:30:14 | {
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/1%3A_Introduction_to_Physiology/1.3%3A_Homeostasis | 1.3: Homeostasis
- Provide a general description of and some examples of homeostasis
To function properly, cells require appropriate conditions such as proper temperature, pH, and appropriate concentration of diverse chemicals. These conditions may, however, change from one moment to the next. Organisms are able to maintain internal conditions within a narrow range almost constantly, despite environmental changes, through homeostasis (literally, “steady state”). For example, an organism needs to regulate body temperature through the thermoregulation process. Organisms that live in cold climates, such as the polar bear, have body structures that help them withstand low temperatures and conserve body heat. Structures that aid in this type of insulation include fur, feathers, blubber, and fat. In hot climates, organisms have methods (such as perspiration in humans or panting in dogs) that help them to shed excess body heat.
Homeostasis refers to the relatively stable state inside the body of an animal. Animal organs and organ systems constantly adjust to internal and external changes in order to maintain this steady state. Examples of internal conditions maintained homeostatically are the level of blood glucose, body temperature, blood calcium level. These conditions remain stable because of physiologic processes that result in negative feedback relationships. If the blood glucose or calcium rises, this sends a signal to organs responsible for lowering blood glucose or calcium. The signals that restore the normal levels are examples of negative feedback. When homeostatic mechanisms fail, the results can be unfavorable for the animal. Homeostatic mechanisms keep the body in dynamic equilibrium by constantly adjusting to the changes that the body’s systems encounter. Even an animal that is apparently inactive is maintaining this homeostatic equilibrium. Two examples of factors that are regulated homeostatically are temperature and water content. The processes that maintain homeostasis of these two factors are called thermoregulation and osmoregulation .
Homeostasis
The goal of homeostasis is the maintenance of equilibrium around a specific value of some aspect of the body or its cells called a set point. While there are normal fluctuations from the set point, the body’s systems will usually attempt to go back to this point. A change in the internal or external environment is called a stimulus and is detected by a receptor; the response of the system is to adjust the activities of the system so the value moves back toward the set point. For instance, if the body becomes too warm, adjustments are made to cool the animal. If glucose levels in the blood rise after a meal, adjustments are made to lower them and to get the nutrient into tissues that need it or to store it for later use.
When a change occurs in an animal’s environment, an adjustment must be made so that the internal environment of the body and cells remains stable. The receptor that senses the change in the environment is part of a feedback mechanism. The stimulus—temperature, glucose, or calcium levels—is detected by the receptor. The receptor sends information to a control center, often the brain, which relays appropriate signals to an effector organ that is able to cause an appropriate change, either up or down, depending on the information the sensor was sending.
Thermoregulation
Animals can be divided into two groups: those that maintain a constant body temperature in the face of differing environmental temperatures, and those that have a body temperature that is the same as their environment and thus varies with the environmental temperature. Animals that do not have internal control of their body temperature are called ectotherms. The body temperature of these organisms is generally similar to the temperature of the environment, although the individual organisms may do things that keep their bodies slightly below or above the environmental temperature. This can include burrowing underground on a hot day or resting in the sunlight on a cold day. The ectotherms have been called cold-blooded, a term that may not apply to an animal in the desert with a very warm body temperature.
An animal that maintains a constant body temperature in the face of environmental changes is called an endotherm. These animals are able to maintain a level of activity that an ectothermic animal cannot because they generate internal heat that keeps their cellular processes operating optimally even when the environment is cold.
Watch this Discovery Channel video on thermoregulation to see illustrations of the process in a variety of animals
Animals conserve or dissipate heat in a variety of ways. Endothermic animals have some form of insulation. They have fur, fat, or feathers. Animals with thick fur or feathers create an insulating layer of air between their skin and internal organs. Polar bears and seals live and swim in a subfreezing environment and yet maintain a constant, warm, body temperature. The arctic fox, for example, uses its fluffy tail as extra insulation when it curls up to sleep in cold weather. Mammals can increase body heat production by shivering, which is an involuntary increase in muscle activity. In addition, arrector pili muscles can contract causing individual hairs to stand up when the individual is cold. This increases the insulating effect of the hair. Humans retain this reaction, which does not have the intended effect on our relatively hairless bodies; it causes “goose bumps” instead. Mammals use layers of fat as insulation also. Loss of significant amounts of body fat will compromise an individual’s ability to conserve heat.
Ectotherms and endotherms use their circulatory systems to help maintain body temperature. Vasodilation, the opening up of arteries to the skin by relaxation of their smooth muscles, brings more blood and heat to the body surface, facilitating radiation and evaporative heat loss, cooling the body. Vasoconstriction, the narrowing of blood vessels to the skin by contraction of their smooth muscles, reduces blood flow in peripheral blood vessels, forcing blood toward the core and vital organs, conserving heat. Some animals have adaptions to their circulatory system that enable them to transfer heat from arteries to veins that are flowing next to each other, warming blood returning to the heart. This is called a countercurrent heat exchange; it prevents the cold venous blood from cooling the heart and other internal organs. The countercurrent adaptation is found in dolphins, sharks, bony fish, bees, and hummingbirds.
Some ectothermic animals use changes in their behavior to help regulate body temperature. They simply seek cooler areas during the hottest part of the day in the desert to keep from getting too warm. The same animals may climb onto rocks in the evening to capture heat on a cold desert night before entering their burrows.
Thermoregulation is coordinated by the nervous system (Figure 1.2). The processes of temperature control are centered in the hypothalamus of the advanced animal brain. The hypothalamus maintains the set point for body temperature through reflexes that cause vasodilation or vasoconstriction and shivering or sweating. The sympathetic nervous system under control of the hypothalamus directs the responses that effect the changes in temperature loss or gain that return the body to the set point. The set point may be adjusted in some instances. During an infection, compounds called pyrogens are produced and circulate to the hypothalamus resetting the thermostat to a higher value. This allows the body’s temperature to increase to a new homeostatic equilibrium point in what is commonly called a fever. The increase in body heat makes the body less optimal for bacterial growth and increases the activities of cells so they are better able to fight the infection.
When bacteria are destroyed by leukocytes, pyrogens are released into the blood. Pyrogens reset the body’s thermostat to a higher temperature, resulting in fever. How might pyrogens cause the body temperature to rise? Note: Pyrogens increase body temperature by causing the blood vessels to constrict, inducing shivering, and stopping sweat glands from secreting fluid
What is homeostasis?
Describe a thermoregulatory homeostatic loop.
Describe an osmoregulatory homeostatic loop.
Examples of maintenance of homeostasis through negative feedback
Negative feedback is a mechanism that reverses a deviation from the set point. Therefore, negative feedback maintains body parameters within their normal range. The maintenance of homeostasis by negative feedback goes on throughout the body at all times, and an understanding of negative feedback is thus fundamental to an understanding of human physiology. A negative feedback system has three basic components (Figure 1.3a). A sensor , also referred to a receptor, is a component of a feedback system that monitors a physiological value. This value is reported to the control center. The control center is the component in a feedback system that compares the value to the normal range. If the value deviates too much from the set point, then the control center activates an effector. An effector is the component in a feedback system that causes a change to reverse the situation and return the value to the normal range.
In order to set the system in motion, a stimulus must drive a physiological parameter beyond its normal range (that is, beyond homeostasis). This stimulus is “heard” by a specific sensor. For example, in the control of blood glucose, specific endocrine cells in the pancreas detect excess glucose (the stimulus) in the bloodstream. These pancreatic beta cells respond to the increased level of blood glucose by releasing the hormone insulin into the bloodstream. The insulin signals skeletal muscle fibers, fat cells (adipocytes), and liver cells to take up the excess glucose, removing it from the bloodstream. As glucose concentration in the bloodstream drops, the decrease in concentration—the actual negative feedback—is detected by pancreatic alpha cells, and insulin release stops. This prevents blood sugar levels from continuing to drop below the normal range.
Humans have a similar temperature regulation feedback system that works by promoting either heat loss or heat gain (Figure 1.3b). When the brain’s temperature regulation center receives data from the sensors indicating that the body’s temperature exceeds its normal range, it stimulates a cluster of brain cells referred to as the “heat-loss center.” This stimulation has three major effects:
- Blood vessels in the skin begin to dilate allowing more blood from the body core to flow to the surface of the skin allowing the heat to radiate into the environment.
- As blood flow to the skin increases, sweat glands are activated to increase their output. As the sweat evaporates from the skin surface into the surrounding air, it takes heat with it.
- The depth of respiration increases, and a person may breathe through an open mouth instead of through the nasal passageways. This further increases heat loss from the lungs.
In contrast, activation of the brain’s heat-gain center by exposure to cold reduces blood flow to the skin, and blood returning from the limbs is diverted into a network of deep veins. This arrangement traps heat closer to the body core and restricts heat loss. If heat loss is severe, the brain triggers an increase in random signals to skeletal muscles, causing them to contract and producing shivering. The muscle contractions of shivering release heat while using up ATP. The brain triggers the thyroid gland in the endocrine system to release thyroid hormone, which increases metabolic activity and heat production in cells throughout the body. The brain also signals the adrenal glands to release epinephrine (adrenaline), a hormone that causes the breakdown of glycogen into glucose, which can be used as an energy source. The breakdown of glycogen into glucose also results in increased metabolism and heat production.
Watch this video to learn more about water concentration in the body.
Water concentration in the body is critical for proper functioning. A person’s body retains very tight control on water levels without conscious control by the person. Which organ has primary control over the amount of water in the body?
Examples of maintenance of homeostasis through negative feedback
Positive feedback intensifies a change in the body’s physiological condition rather than reversing it. A deviation from the normal range results in more change, and the system moves farther away from the normal range. Positive feedback in the body is normal only when there is a definite end point. Childbirth and the body’s response to blood loss are two examples of positive feedback loops that are normal but are activated only when needed.
Childbirth at full term is an example of a situation in which the maintenance of the existing body state is not desired. Enormous changes in the mother’s body are required to expel the baby at the end of pregnancy. And the events of childbirth, once begun, must progress rapidly to a conclusion or the life of the mother and the baby are at risk. The extreme muscular work of labor and delivery are the result of a positive feedback system (Figure 1.4).
The first contractions of labor (the stimulus) push the baby toward the cervix (the lowest part of the uterus). The cervix contains stretch-sensitive nerve cells that monitor the degree of stretching (the sensors). These nerve cells send messages to the brain, which in turn causes the pituitary gland at the base of the brain to release the hormone oxytocin into the bloodstream. Oxytocin causes stronger contractions of the smooth muscles in the uterus (the effectors), pushing the baby further down the birth canal. This causes even greater stretching of the cervix. The cycle of stretching, oxytocin release, and increasingly more forceful contractions stops only when the baby is born. At this point, the stretching of the cervix halts, stopping the release of oxytocin.
A second example of positive feedback centers on reversing extreme damage to the body. Following a penetrating wound, the most immediate threat is excessive blood loss. Less blood circulating means reduced blood pressure and reduced perfusion (penetration of blood) to the brain and other vital organs. If perfusion is severely reduced, vital organs will shut down and the person will die. The body responds to this potential catastrophe by releasing substances in the injured blood vessel wall that begin the process of blood clotting. As each step of clotting occurs, it stimulates the release of more clotting substances. This accelerates the processes of clotting and sealing off the damaged area. Clotting is contained in a local area based on the tightly controlled availability of clotting proteins. This is an adaptive, life-saving cascade of events.
After you eat lunch, nerve cells in your stomach respond to the distension (the stimulus) resulting from the food. They relay this information to ________.
a. a control center
b. a set point
c. effectors
d. sensory
Stimulation of the heat-loss center causes ________.
a. blood vessels in the skin to constrict
b. breathing to become slow and shallow
c. sweat glands to increase their output
d. All of the above
Which of the following is an example of a normal physiologic process that uses a positive feedback loop?
a. blood pressure regulation
b. childbirth
c. regulation of fluid balance
d. temperature regulation
Identify the four components of a negative feedback loop and explain what would happen if secretion of a body chemical controlled by a negative feedback system became too great.
What regulatory processes would your body use if you were trapped by a blizzard in an unheated, uninsulated cabin in the woods? | libretexts | 2025-03-17T22:26:51.738237 | 2018-10-18T16:30:14 | {
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/2%3A_Gaseous_Exchange | 2: Gaseous Exchange Last updated Save as PDF Page ID 9547 Sanja Hinić-Frlog University of Toronto Mississauga 2.1: Introduction to Respiratory Systems 2.2: Respiratory Structures and Their Function 2.3: Gaseous Exchange Mechanism | libretexts | 2025-03-17T22:26:51.817309 | 2018-10-18T16:30:14 | {
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"title": "2: Gaseous Exchange",
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/2%3A_Gaseous_Exchange/2.1%3A_Introduction_to_Respiratory_Systems | 2.1: Introduction to Respiratory Systems
Discuss general respiration advantages and disadvantages with respect to aquatic and terrestrial environments.
Breathing is an involuntary event. How often a breath is taken and how much air is inhaled or exhaled are tightly regulated by the respiratory center in the brain. Humans, when they aren’t exerting themselves, breathe approximately 15 times per minute on average. Canines, like the dog in Figure 2.1, have a respiratory rate of about 15–30 breaths per minute. With every inhalation, air fills the lungs, and with every exhalation, air rushes back out. That air is doing more than just inflating and deflating the lungs in the chest cavity. The air contains oxygen that crosses the lung tissue, enters the bloodstream, and travels to organs and tissues. Oxygen (O 2 ) enters the cells where it is used for metabolic reactions that produce ATP, a high-energy compound. At the same time, these reactions release carbon dioxide (CO 2 ) as a by-product. CO 2 is toxic and must be eliminated. Carbon dioxide exits the cells, enters the bloodstream, travels back to the lungs, and is expired out of the body during exhalation.
The primary function of the respiratory system is to deliver oxygen to the cells of the body’s tissues and remove carbon dioxide, a cell waste product. The main structures of the human respiratory system are the nasal cavity, the trachea, and the lungs.
All aerobic organisms require oxygen to carry out their metabolic functions. Along the evolutionary tree, different organisms have devised different means of obtaining oxygen from the surrounding atmosphere. The environment in which the animal lives greatly determines how an animal respires. The complexity of the respiratory system is correlated with the size of the organism. As animal size increases, diffusion distances increase and the ratio of surface area to volume drops. In unicellular organisms, diffusion across the cell membrane is sufficient for supplying oxygen to the cell (Figure 2.2). Diffusion is a slow, passive transport process. In order for diffusion to be a feasible means of providing oxygen to the cell, the rate of oxygen uptake must match the rate of diffusion across the membrane. In other words, if the cell were very large or thick, diffusion would not be able to provide oxygen quickly enough to the inside of the cell. Therefore, dependence on diffusion as a means of obtaining oxygen and removing carbon dioxide remains feasible only for small organisms or those with highly-flattened bodies, such as many flatworms (Platyhelminthes). Larger organisms had to evolve specialized respiratory tissues, such as gills, lungs, and respiratory passages accompanied by a complex circulatory system, to transport oxygen throughout their entire body.
Direct diffusion
For small multicellular organisms, diffusion across the outer membrane is sufficient to meet their oxygen needs. Gas exchange by direct diffusion across surface membranes is efficient for organisms less than 1 mm in diameter. In simple organisms, such as cnidarians and flatworms, every cell in the body is close to the external environment. Their cells are kept moist and gases diffuse quickly via direct diffusion. Flatworms are small, literally flatworms, which ‘breathe’ through diffusion across the outer membrane (Figure 2.3). The flat shape of these organisms increases the surface area for diffusion, ensuring that each cell within the body is close to the outer membrane surface and has access to oxygen. If the flatworm had a cylindrical body, then the cells in the center would not be able to get oxygen.
Which physical and chemical factors affect the rate of diffusion of a gas and how? (You will be able to find some answers to this question in the next section of this chapter).
How are some fish able to breathe on land? | libretexts | 2025-03-17T22:26:51.877289 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/2%3A_Gaseous_Exchange/2.1%3A_Introduction_to_Respiratory_Systems",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "2.1: Introduction to Respiratory Systems",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/2%3A_Gaseous_Exchange/2.2%3A_Respiratory_Structures_and_Their_Function | 2.2: Respiratory Structures and Their Function
- Describe select respiratory structures in aquatic and terrestrial animals
- Explain relationships between respiratory structure and function in generalized aquatic and terrestrial animal systems.
Skin and Gills
Earthworms and amphibians use their skin (integument) as a respiratory organ. A dense network of capillaries lies just below the skin and facilitates gas exchange between the external environment and the circulatory system. The respiratory surface must be kept moist in order for the gases to dissolve and diffuse across cell membranes.
Organisms that live in water need to obtain oxygen from the water. Oxygen dissolves in water but at a lower concentration than in the atmosphere. The atmosphere has roughly 21 percent oxygen. In water, the oxygen concentration is much smaller than that. Fish and many other aquatic organisms have evolved gills to take up the dissolved oxygen from water (Figure 2.4). Gills are thin tissue filaments that are highly branched and folded. When water passes over the gills, the dissolved oxygen in water rapidly diffuses across the gills into the bloodstream. The circulatory system can then carry the oxygenated blood to the other parts of the body. In animals that contain coelomic fluid instead of blood, oxygen diffuses across the gill surfaces into the coelomic fluid. Gills are found in mollusks, annelids, and crustaceans.
The folded surfaces of the gills provide a large surface area to ensure that the fish gets sufficient oxygen. Diffusion is a process in which material travels from regions of high concentration to low concentration until equilibrium is reached. In this case, blood with a low concentration of oxygen molecules circulates through the gills. The concentration of oxygen molecules in water is higher than the concentration of oxygen molecules in gills. As a result, oxygen molecules diffuse from water (high concentration) to blood (low concentration), as shown in Figure 2.5 . Similarly, carbon dioxide molecules in the blood diffuse from the blood (high concentration) to water (low concentration).
Tracheal systems
Insect respiration is independent of its circulatory system; therefore, the blood does not play a direct role in oxygen transport. Insects have a highly specialized type of respiratory system called the tracheal system, which consists of a network of small tubes that carries oxygen to the entire body. The tracheal system is the most direct and efficient respiratory system in active animals. The tubes in the tracheal system are made of a polymeric material called chitin.
Insect bodies have openings, called spiracles along the thorax and abdomen. These openings connect to the tubular network, allowing oxygen to pass into the body (Figure 2.6) and regulating the diffusion of CO 2 and water vapor. Air enters and leaves the tracheal system through the spiracles. Some insects can ventilate the tracheal system with body movements.
Mammalian systems
In mammals, pulmonary ventilation occurs via inhalation (breathing). During inhalation, air enters the body through the nasal cavity located just inside the nose (Figure 2.7). As air passes through the nasal cavity, the air is warmed to body temperature and humidified. The respiratory tract is coated with mucus to seal the tissues from direct contact with air. Mucus is high in water. As air crosses these surfaces of the mucous membranes, it picks up water. These processes help equilibrate the air to the body conditions, reducing any damage that cold, dry air can cause. Particulate matter that is floating in the air is removed in the nasal passages via mucus and cilia. The processes of warming, humidifying, and removing particles are important protective mechanisms that prevent damage to the trachea and lungs. Thus, inhalation serves several purposes in addition to bringing oxygen into the respiratory system.
Which of the following statements about the mammalian respiratory system is false?
- When we breathe in, air travels from the pharynx to the trachea.
- The bronchioles branch into bronchi.
- Alveolar ducts connect to alveolar sacs.
- Gas exchange between the lung and blood takes place in the alveolus
From the nasal cavity, air passes through the pharynx (throat) and the larynx (voice box), as it makes its way to the trachea (Figure 2.8). The main function of the trachea is to funnel the inhaled air to the lungs and the exhaled air back out of the body. The human trachea is a cylinder about 10 to 12 cm long and 2 cm in diameter that sits in front of the esophagus and extends from the larynx into the chest cavity where it divides into the two primary bronchi at the mid-thorax. It is made of incomplete rings of hyaline cartilage and smooth muscle (Figure 2.8). The trachea is lined with mucus-producing goblet cells and ciliated epithelia. The cilia propel foreign particles trapped in the mucus toward the pharynx. The cartilage provides strength and support to the trachea to keep the passage open. The smooth muscle can contract, decreasing the trachea’s diameter, which causes expired air to rush upwards from the lungs at a great force. The forced exhalation helps to expel mucus when we cough. Smooth muscle can contract or relax, depending on stimuli from the external environment or the body’s nervous system.
Lungs: bronchi and alveoli
The end of the trachea bifurcates (divides) to the right and left lungs. The lungs are not identical. The right lung is larger and contains three lobes, whereas the smaller left lung contains two lobes (Figure 2.9). The muscular diaphragm, which facilitates breathing, is inferior (below) to the lungs and marks the end of the thoracic cavity.
In the lungs, air is diverted into smaller and smaller passages or bronchi. Air enters the lungs through the two primary (main) bronchi (singular: bronchus). Each bronchus divides into secondary bronchi, then into tertiary bronchi, which in turn divide, creating smaller and smaller diameter bronchioles as they split and spread through the lung. Like the trachea, the bronchi are made of cartilage and smooth muscle. At the bronchioles, the cartilage is replaced with elastic fibers. Bronchi are innervated by nerves of both the parasympathetic and sympathetic nervous systems that control muscle contraction (parasympathetic) or relaxation (sympathetic) in the bronchi and bronchioles, depending on the nervous system’s cues. In humans, bronchioles with a diameter smaller than 0.5 mm are the respiratory bronchioles . They lack cartilage and therefore rely on inhaled air to support their shape. As the passageways decrease in diameter, the relative amount of smooth muscle increases.
The terminal bronchioles subdivide into microscopic branches called respiratory bronchioles. The respiratory bronchioles subdivide into several alveolar ducts. Numerous alveoli and alveolar sacs surround the alveolar ducts. The alveolar sacs resemble bunches of grapes tethered to the end of the bronchioles (Figure 2.10). In the acinar region, the alveolar ducts are attached to the end of each bronchiole. At the end of each duct are approximately 100 alveolar sacs, each containing 20 to 30 alveoli that are 200 to 300 microns in diameter. Gas exchange occurs only in alveoli. Alveoli are made of thin-walled parenchymal cells, typically one-cell thick, that look like tiny bubbles within the sacs. Alveoli are in direct contact with capillaries (one-cell thick) of the circulatory system. Such intimate contact ensures that oxygen will diffuse from alveoli into the blood and be distributed to the cells of the body. In addition, the carbon dioxide that was produced by cells as a waste product will diffuse from the blood into alveoli to be exhaled. The anatomical arrangement of capillaries and alveoli emphasizes the structural and functional relationship of the respiratory and circulatory systems. Because there are so many alveoli (~300 million per lung) within each alveolar sac and so many sacs at the end of each alveolar duct, the lungs have a sponge-like consistency. This organization produces a very large surface area that is available for gas exchange. The surface area of alveoli in the lungs is approximately 75 m2. This large surface area, combined with the thin-walled nature of the alveolar parenchymal cells, allows gases to easily diffuse across the cells.
Which of the following statements about the human respiratory system is false?
a. When we breathe in, air travels from the pharynx to the trachea.
b. The bronchioles branch into bronchi.
c. Alveolar ducts connect to alveolar sacs.
d. Gas exchange between the lungs and blood takes place in the alveolus | libretexts | 2025-03-17T22:26:51.943350 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/2%3A_Gaseous_Exchange/2.2%3A_Respiratory_Structures_and_Their_Function",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "2.2: Respiratory Structures and Their Function",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/2%3A_Gaseous_Exchange/2.3%3A_Gaseous_Exchange_Mechanism | 2.3: Gaseous Exchange Mechanism
- Compare counter- and crosscurrent exchange using specific respiratory system examples.
Use the following gif animation as you are reviewing counter-current gas exchange.
This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License (Eleanor Lutz).
How would you describe gas exchange and breathing cycle in each of the examples you see in the gif animation? What are their similarities? What are their differences
Which one do you think is an example of counter-current and which one is an example of crosscurrent exchange
Pulmonary ventilation is the act of breathing, which can be described as the movement of air into and out of the lungs. The major mechanisms that drive pulmonary ventilation are atmospheric pressure ( P atm ); the air pressure within the alveoli, called intra-alveolar pressure ( P alv ); and the pressure within the pleural cavity, called intrapleural pressure ( P ip ).
Mechanisms of breathing
The intra-alveolar and intrapleural pressures are dependent on certain physical features of the lung. However, the ability to breathe—to have air enter the lungs during inspiration and air leave the lungs during expiration—is dependent on the air pressure of the atmosphere and the air pressure within the lungs.
- Pressure relationships
Inspiration (or inhalation) and expiration (or exhalation) are dependent on the differences in pressure between the atmosphere and the lungs. In a gas, the pressure is a force created by the movement of gas molecules that are confined. For example, a certain number of gas molecules in a two-liter container has more room than the same number of gas molecules in a one-liter container (Figure 2.11). In this case, the force exerted by the movement of the gas molecules against the walls of the two-liter container is lower than the force exerted by the gas molecules in the one-liter container. Therefore, the pressure is lower in the two-liter container and higher in the one-liter container. At a constant temperature, changing the volume occupied by the gas changes the pressure, as does changing the number of gas molecules. Boyle’s law describes the relationship between volume and pressure in a gas at a constant temperature. Boyle discovered that the pressure of a gas is inversely proportional to its volume: If volume increases, pressure decreases. Likewise, if volume decreases, pressure increases. Pressure and volume are inversely related (P = k/V). Therefore, the pressure in the one-liter container (one-half the volume of the two-liter container) would be twice the pressure in the two-liter container. Boyle’s law is expressed by the following formula:
[P 1 V 1 =P 2 V 2 ]
In this formula, P 1 represents the initial pressure and V 1 represents the initial volume, whereas the final pressure and volume are represented by P 2 and V 2 , respectively. If the two- and one-liter containers were connected by a tube and the volume of one of the containers were changed, then the gases would move from higher pressure (lower volume) to lower pressure (higher volume).
Pulmonary ventilation is dependent on three types of pressure: atmospheric, intra-alveolar, and intrapleural. Atmospheric pressure is the amount of force that is exerted by gases in the air surrounding any given surface, such as the body. Atmospheric pressure can be expressed in terms of the unit atmosphere, abbreviated atm, or in millimeters of mercury (mm Hg). One atm is equal to 760 mm Hg, which is the atmospheric pressure at sea level. Typically, for respiration, other pressure values are discussed in relation to atmospheric pressure. Therefore, negative pressure is pressure lower than the atmospheric pressure, whereas positive pressure is the pressure that it is greater than the atmospheric pressure. A pressure that is equal to the atmospheric pressure is expressed as zero.
Intra-alveolar pressure (intrapulmonary pressure) is the pressure of the air within the alveoli, which changes during the different phases of breathing (Figure 2.12). Because the alveoli are connected to the atmosphere via the tubing of the airways (similar to the two- and one-liter containers in the example above), the intrapulmonary pressure of the alveoli always equalizes with the atmospheric pressure.
Intrapleural pressure is the pressure of the air within the pleural cavity, between the visceral and parietal pleurae. Similar to intra-alveolar pressure, intrapleural pressure also changes during the different phases of breathing. However, due to certain characteristics of the lungs, the intrapleural pressure is always lower than, or negative to, the intra-alveolar pressure (and therefore also to atmospheric pressure). Although it fluctuates during inspiration and expiration, intrapleural pressure remains approximately –4 mm Hg throughout the breathing cycle.
Competing forces within the thorax cause the formation of the negative intrapleural pressure. One of these forces relates to the elasticity of the lungs themselves—elastic tissue pulls the lungs inward, away from the thoracic wall. Surface tension of the alveolar fluid, which is mostly water, also creates an inward pull of the lung tissue. This inward tension from the lungs is countered by opposing forces from the pleural fluid and thoracic wall. Surface tension within the pleural cavity pulls the lungs outward. Too much or too little pleural fluid would hinder the creation of the negative intrapleural pressure; therefore, the level must be closely monitored by the mesothelial cells and drained by the lymphatic system. Since the parietal pleura is attached to the thoracic wall, the natural elasticity of the chest wall opposes the inward pull of the lungs. Ultimately, the outward pull is slightly greater than the inward pull, creating the –4 mm Hg intrapleural pressure relative to the intra-alveolar pressure. Transpulmonary pressure is the difference between the intrapleural and intra-alveolar pressures, and it determines the size of the lungs. A higher transpulmonary pressure corresponds to a larger lung.
Physical factors affecting ventilation
In addition to the differences in pressures, breathing is also dependent upon the contraction and relaxation of muscle fibers of both the diaphragm and thorax. The lungs themselves are passive during breathing, meaning they are not involved in creating the movement that helps inspiration and expiration. This is because of the adhesive nature of the pleural fluid, which allows the lungs to be pulled outward when the thoracic wall moves during inspiration. The recoil of the thoracic wall during expiration causes compression of the lungs. Contraction and relaxation of the diaphragm and intercostals muscles (found between the ribs) cause most of the pressure changes that result in inspiration and expiration. These muscle movements and subsequent pressure changes cause air to either rush in or be forced out of the lungs.
Other characteristics of the lungs influence the effort that must be expended to ventilate. Resistance is a force that slows motion, in this case, the flow of gases. The size of the airway is the primary factor affecting resistance. A small tubular diameter forces air through a smaller space, causing more collisions of air molecules with the walls of the airways. The following formula helps to describe the relationship between airway resistance and pressure changes:
[F=∆P/RF=∆P/R]
As noted earlier, there is surface tension within the alveoli caused by water present in the lining of the alveoli. This surface tension tends to inhibit expansion of the alveoli. However, pulmonary surfactant secreted by type II alveolar cells mixes with that water and helps reduce this surface tension. Without pulmonary surfactant, the alveoli would collapse during expiration.
Thoracic wall compliance is the ability of the thoracic wall to stretch while under pressure. This can also affect the effort expended in the process of breathing. In order for inspiration to occur, the thoracic cavity must expand. The expansion of the thoracic cavity directly influences the capacity of the lungs to expand. If the tissues of the thoracic wall are not very compliant, it will be difficult to expand the thorax to increase the size of the lungs.
Gas laws and air composition
Gas molecules exert a force on the surfaces with which they are in contact; this force is called pressure. In natural systems, gases are normally present as a mixture of different types of molecules. For example, the atmosphere consists of oxygen, nitrogen, carbon dioxide, and other gaseous molecules, and this gaseous mixture exerts a certain pressure referred to as atmospheric pressure (Table 2.1). Partial pressure ( P x ) is the pressure of a single type of gas in a mixture of gases. For example, in the atmosphere, oxygen exerts a partial pressure, and nitrogen exerts another partial pressure, independent of the partial pressure of oxygen (Figure 2.13). Total pressure is the sum of all the partial pressures of a gaseous mixture. Dalton’s law describes the behavior of nonreactive gases in a gaseous mixture and states that a specific gas type in a mixture exerts its own pressure; thus, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the gases in the mixture.
| Gas | Percent of total composition | Partial pressure (mm Hg) |
|---|---|---|
| Nitrogen (N 2 ) | 78.6 | 597.4 |
| Oxygen (O 2 ) | 20.9 | 158.8 |
| Water (H 2 O) | 0.4 | 3.0 |
| Carbon dioxide (CO 2 ) | 0.04 | 0.3 |
| Others | 0.06 | 0.5 |
| Total composition/total atmospheric pressure | 100% | 760.0 |
Partial pressure is extremely important in predicting the movement of gases. Recall that gases tend to equalize their pressure in two regions that are connected. A gas will move from an area where its partial pressure is higher to an area where its partial pressure is lower. In addition, the greater the partial pressure difference between the two areas, the more rapid is the movement of gases.
Solubility of gases in liquids
Henry’s law describes the behavior of gases when they come into contact with a liquid, such as blood. Henry’s law states that the concentration of gas in a liquid is directly proportional to the solubility and partial pressure of that gas. The greater the partial pressure of the gas, the greater the number of gas molecules that will dissolve in the liquid. The concentration of the gas in a liquid is also dependent on the solubility of the gas in the liquid. For example, although nitrogen is present in the atmosphere, very little nitrogen dissolves into the blood because the solubility of nitrogen in blood is very low. The exception to this occurs in scuba divers; the composition of the compressed air that divers breathe causes nitrogen to have a higher partial pressure than normal, causing it to dissolve in the blood in greater amounts than normal. Too much nitrogen in the bloodstream results in a serious condition that can be fatal if not corrected. Gas molecules establish an equilibrium between those molecules dissolved in liquid and those in the air.
The composition of air in the atmosphere and in the alveoli differs. In both cases, the relative concentration of gases is nitrogen > oxygen > water vapor > carbon dioxide. The amount of water vapor present in the alveolar air is greater than that in atmospheric air (Table 2.2). Recall that the respiratory system works to humidify incoming air, thereby causing the air present in the alveoli to have a greater amount of water vapor than atmospheric air. In addition, alveolar air contains a greater amount of carbon dioxide and less oxygen than atmospheric air. This is no surprise, as gas exchange removes oxygen from and adds carbon dioxide to the alveolar air. Both deep and forced breathing cause the alveolar air composition to be changed more rapidly than during quiet breathing. As a result, the partial pressures of oxygen and carbon dioxide change, affecting the diffusion process that moves these materials across the membrane. This will cause oxygen to enter and carbon dioxide to leave the blood more quickly.
| Gas | Percent of total composition | Partial pressure (mm Hg) |
|---|---|---|
| Nitrogen (N 2 ) | 74.9 | 569 |
| Oxygen (O 2 ) | 13.7 | 104 |
| Water (H 2 O) | 6.2 | 40 |
| Carbon dioxide (CO 2 ) | 5.2 | 47 |
| Total composition/total alveolar pressure | 100% | 760.0 |
Ventilation and perfusion
Two important aspects of gas exchange in the lung are ventilation and perfusion. Ventilation is the movement of air into and out of the lungs, and perfusion is the flow of blood in the pulmonary capillaries. For gas exchange to be efficient, the volumes involved in ventilation and perfusion should be compatible. However, factors such as regional gravity effects on the blood, blocked alveolar ducts, or disease can cause ventilation and perfusion to be unbalanced.
The partial pressure of oxygen in the alveolar air is about 104 mm Hg, whereas the partial pressure of oxygenated blood in pulmonary veins is about 100 mm Hg. When ventilation is sufficient, oxygen enters the alveoli at a high rate, and the partial pressure of oxygen in the alveoli remains high. In contrast, when ventilation is insufficient, the partial pressure of oxygen in the alveoli drops. Without the large difference in partial pressure between the alveoli and the blood, oxygen does not diffuse efficiently across the respiratory membrane. The body has mechanisms that counteract this problem. In cases when ventilation is not sufficient for an alveolus, the body redirects blood flow to alveoli that are receiving sufficient ventilation. This is achieved by constricting the pulmonary arterioles that serve the dysfunctional alveolus, which redirects blood to other alveoli that have sufficient ventilation. At the same time, the pulmonary arterioles that serve alveoli receiving sufficient ventilation vasodilate, which brings in greater blood flow. Factors such as carbon dioxide, oxygen, and pH levels can all serve as stimuli for adjusting blood flow in the capillary networks associated with the alveoli.
Ventilation is regulated by the diameter of the airways, whereas perfusion is regulated by the diameter of the blood vessels. The diameter of the bronchioles is sensitive to the partial pressure of carbon dioxide in the alveoli. A greater partial pressure of carbon dioxide in the alveoli causes the bronchioles to increase their diameter as will a decreased level of oxygen in the blood supply, allowing carbon dioxide to be exhaled from the body at a greater rate. As mentioned above, a greater partial pressure of oxygen in the alveoli causes the pulmonary arterioles to dilate, increasing blood flow.
Gas exchange
Gas exchange occurs at two sites in the body: in the lungs, where oxygen is picked up and carbon dioxide is released at the respiratory membrane, and at the tissues, where oxygen is released and carbon dioxide is picked up. External respiration is the exchange of gases with the external environment and occurs in the alveoli of the lungs. Internal respiration is the exchange of gases with the internal environment and occurs in the tissues. The actual exchange of gases occurs due to simple diffusion. Energy is not required to move oxygen or carbon dioxide across membranes. Instead, these gases follow pressure gradients that allow them to diffuse. The anatomy of the lung maximizes the diffusion of gases: The respiratory membrane is highly permeable to gases; the respiratory and blood capillary membranes are very thin, and there is a large surface area throughout the lungs.
External respiration
The pulmonary artery carries deoxygenated blood into the lungs from the heart, where it branches and eventually becomes the capillary network composed of pulmonary capillaries. These pulmonary capillaries create the respiratory membrane with the alveoli (Figure 2.14). As the blood is pumped through this capillary network, gas exchange occurs. Although a small amount of the oxygen is able to dissolve directly into the plasma from the alveoli, most of the oxygen is picked up by erythrocytes (red blood cells) and binds to a protein called hemoglobin, a process described later in this chapter. Oxygenated hemoglobin is red, causing the overall appearance of bright red oxygenated blood, which returns to the heart through the pulmonary veins. Carbon dioxide is released in the opposite direction of oxygen, from the blood to the alveoli. Some of the carbon dioxide is returned on hemoglobin, but can also be dissolved in plasma or is present as a converted form, also explained in greater detail later in this chapter.
External respiration occurs as a function of partial pressure differences in oxygen and carbon dioxide between the alveoli and the blood in the pulmonary capillaries.
External Respiration
Figure 2.14. In external respiration, oxygen diffuses across the respiratory membrane from the alveolus to the capillary, whereas carbon dioxide diffuses out of the capillary into the alveolus.
Although the solubility of oxygen in the blood is not high, there is a drastic difference in the partial pressure of oxygen in the alveoli versus in the blood of the pulmonary capillaries. This difference is about 64 mm Hg: The partial pressure of oxygen in the alveoli is about 104 mm Hg, whereas its partial pressure in the blood of the capillary is about 40 mm Hg. This large difference in partial pressure creates a very strong pressure gradient that causes oxygen to rapidly cross the respiratory membrane from the alveoli into the blood.
The partial pressure of carbon dioxide is also different between the alveolar air and the blood of the capillary. However, the partial pressure difference is less than that of oxygen, about 5 mm Hg. The partial pressure of carbon dioxide in the blood of the capillary is about 45 mm Hg, whereas its partial pressure in the alveoli is about 40 mm Hg. However, the solubility of carbon dioxide is much greater than that of oxygen—by a factor of about 20—in both blood and alveolar fluids. As a result, the relative concentrations of oxygen and carbon dioxide that diffuse across the respiratory membrane are similar.
Internal respiration
Internal respiration is the gas exchange that occurs at the level of body tissues (Figure 2.15). Similar to external respiration, internal respiration also occurs as simple diffusion due to a partial pressure gradient. However, the partial pressure gradients are the opposite of those present at the respiratory membrane. The partial pressure of oxygen in tissues is low, about 40 mm Hg, because oxygen is continuously used for cellular respiration. In contrast, the partial pressure of oxygen in the blood is about 100 mm Hg. This creates a pressure gradient that causes oxygen to dissociate from hemoglobin, diffuse out of the blood, cross the interstitial space, and enter the tissue. Hemoglobin that has little oxygen bound to it loses much of its brightness, so that blood returning to the heart is more burgundy in color.
Considering that cellular respiration continuously produces carbon dioxide, the partial pressure of carbon dioxide is lower in the blood than it is in the tissue, causing carbon dioxide to diffuse out of the tissue, cross the interstitial fluid, and enter the blood. It is then carried back to the lungs either bound to hemoglobin, dissolved in plasma, or in a converted form. By the time blood returns to the heart, the partial pressure of oxygen has returned to about 40 mm Hg, and the partial pressure of carbon dioxide has returned to about 45 mm Hg. The blood is then pumped back to the lungs to be oxygenated once again during external respiration.
A hyperbaric chamber is a unit that can be sealed and expose a patient to either 100 percent oxygen with increased pressure or a mixture of gases that includes a higher concentration of oxygen than normal atmospheric air, also at a higher partial pressure than the atmosphere (Figure 2.15). Hyperbaric chamber treatment is based on the behaviour of gases. As you recall, gases move from a region of higher partial pressure to a region of lower partial pressure. In a hyperbaric chamber, the atmospheric pressure is increased, causing a greater amount of oxygen than normal to diffuse into the bloodstream of the patient. Hyperbaric chamber therapy is used to treat a variety of medical problems, such as wound and graft healing, anaerobic bacterial infections, and carbon monoxide poisoning. Exposure to and poisoning by carbon monoxide is difficult to reverse, because hemoglobin’s affinity for carbon monoxide is much stronger than its affinity for oxygen, causing carbon monoxide to replace oxygen in the blood. Hyperbaric chamber therapy can treat carbon monoxide poisoning because the increased atmospheric pressure causes more oxygen to diffuse into the bloodstream. At this increased pressure and increased concentration of oxygen, carbon monoxide is displaced from hemoglobin. Another example is the treatment of anaerobic bacterial infections, which are created by bacteria that cannot or prefer not to live in the presence of oxygen. An increase in blood and tissue levels of oxygen helps to kill the anaerobic bacteria that are responsible for the infection, as oxygen is toxic to anaerobic bacteria. For wounds and grafts, the chamber stimulates the healing process by increasing energy production needed for repair. Increasing oxygen transport allows cells to ramp up cellular respiration and thus ATP production, the energy needed to build new structures.
Gas moves from an area of ________ partial pressure to an area of ________ partial pressure.
a. low; high
b. low; low
c. high; high
d. high; low
When ventilation is not sufficient, which of the following occurs?
a. The capillary constricts.
b. The capillary dilates.
c. The partial pressure of oxygen in the affected alveolus increases.
d. The bronchioles dilate
Gas exchange that occurs at the level of the tissues is called ________.
a. external respiration
b. interpulmonary respiration
c. internal respiration
d. pulmonary ventilation
The partial pressure of carbon dioxide is 45 mm Hg in the blood and 40 mm Hg in the alveoli. What happens to the carbon dioxide?
a. It diffuses into the blood.
b. It diffuses into the alveoli.
c. The gradient is too small for carbon dioxide to diffuse.
d. It decomposes into carbon and oxygen
Compare and contrast Dalton’s law and Henry’s law
A smoker develops damage to several alveoli that then can no longer function. How does this affect gas exchange? | libretexts | 2025-03-17T22:26:52.040108 | 2018-10-18T16:30:14 | {
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"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/2%3A_Gaseous_Exchange/2.3%3A_Gaseous_Exchange_Mechanism",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "2.3: Gaseous Exchange Mechanism",
"author": "Sanja Hinić-Frlog"
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/3%3A_Transport_Gases | 3: Transport Gases Last updated Save as PDF Page ID 9554 Sanja Hinić-Frlog University of Toronto Mississauga 3.1: Gas Transport 3.2: Circulatory Systems | libretexts | 2025-03-17T22:26:52.114866 | 2018-10-18T16:30:14 | {
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"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "3: Transport Gases",
"author": "Sanja Hinić-Frlog"
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/3%3A_Transport_Gases/3.1%3A_Gas_Transport | 3.1: Gas Transport
- Explain how oxygen and carbon dioxide are transported in blood or equivalent fluid medium.
The other major activity in the lungs is the process of respiration, the process of gas exchange. The function of respiration is to provide oxygen for use by body cells during cellular respiration and to eliminate carbon dioxide, a waste product of cellular respiration, from the body. In order for the exchange of oxygen and carbon dioxide to occur, both gases must be transported between the external and internal respiration sites. Although carbon dioxide is more soluble than oxygen in blood, both gases require a specialized transport system for the majority of the gas molecules to be moved between the lungs and other tissues.
Oxygen Transport in the Blood
Even though oxygen is transported via the blood, you may recall that oxygen is not very soluble in liquids. A small amount of oxygen does dissolve in the blood and is transported in the bloodstream, but it is only about 1.5% of the total amount. The majority of oxygen molecules are carried from the lungs to the body’s tissues by a specialized transport system, which relies on the erythrocyte—the red blood cell. Erythrocytes contain a metalloprotein, hemoglobin, which serves to bind oxygen molecules to the erythrocyte (Figure 3.1). Heme is the portion of hemoglobin that contains iron, and it is heme that binds oxygen. One hemoglobin molecule contains iron-containing Heme molecules, and because of this, each hemoglobin molecule is capable of carrying up to four molecules of oxygen. As oxygen diffuses across the respiratory membrane from the alveolus to the capillary, it also diffuses into the red blood cell and is bound by hemoglobin. The following reversible chemical reaction describes the production of the final product, oxyhemoglobin (\(\ce{Hb–O2}\)), which is formed when oxygen binds to hemoglobin. Oxyhemoglobin is a bright red-colored molecule that contributes to the bright red color of oxygenated blood.
\[\ce{Hb + O2 <=> Hb − O2Hb + O2 <=> Hb − O2}\]
In this formula, Hb represents reduced hemoglobin, that is, hemoglobin that does not have oxygen bound to it. There are multiple factors involved in how readily heme binds to and dissociates from oxygen, which will be discussed in the subsequent sections.
Function of hemoglobin
Hemoglobin is composed of subunits, a protein structure that is referred to as a quaternary structure. Each of the four subunits that make up hemoglobin is arranged in a ring-like fashion, with an iron atom covalently bound to the heme in the center of each subunit. Binding of the first oxygen molecule causes a conformational change in hemoglobin that allows the second molecule of oxygen to bind more readily. As each molecule of oxygen is bound, it further facilitates the binding of the next molecule, until all four heme sites are occupied by oxygen. The opposite occurs as well: After the first oxygen molecule dissociates and is “dropped off” at the tissues, the next oxygen molecule dissociates more readily. When all four heme sites are occupied, the hemoglobin is said to be saturated. When one to three heme sites are occupied, the hemoglobin is said to be partially saturated. Therefore, when considering the blood as a whole, the percent of the available heme units that are bound to oxygen at a given time is called hemoglobin saturation. Hemoglobin saturation of 100 percent means that every heme unit in all of the erythrocytes of the body is bound to oxygen. In a healthy individual with normal hemoglobin levels, hemoglobin saturation generally ranges from 95 percent to 99 percent.
Oxygen dissociation from hemoglobin
Partial pressure is an important aspect of the binding of oxygen to and disassociation from heme. An oxygen-hemoglobin dissociation curve is a graph that describes the relationship of partial pressure to the binding of oxygen to heme and its subsequent dissociation from heme (Figure 3.2). Remember that gases travel from an area of higher partial pressure to an area of lower partial pressure. In addition, the affinity of an oxygen molecule for heme increases as more oxygen molecules are bound. Therefore, in the oxygen-hemoglobin saturation curve, as the partial pressure of oxygen increases, a proportionately greater number of oxygen molecules are bound by heme. Not surprisingly, the oxygen-hemoglobin saturation/dissociation curve also shows that the lower the partial pressure of oxygen, the fewer oxygen molecules are bound to heme. As a result, the partial pressure of oxygen plays a major role in determining the degree of binding of oxygen to heme at the site of the respiratory membrane, as well as the degree of dissociation of oxygen from heme at the site of body tissues.
The mechanisms behind the oxygen-hemoglobin saturation/dissociation curve also serve as automatic control mechanisms that regulate how much oxygen is delivered to different tissues throughout the body. This is important because some tissues have a higher metabolic rate than others. Highly active tissues, such as muscle, rapidly use oxygen to produce ATP, lowering the partial pressure of oxygen in the tissue to about 20 mm Hg. The partial pressure of oxygen inside capillaries is about 100 mm Hg, so the difference between the two becomes quite high, about 80 mm Hg. As a result, a greater number of oxygen molecules dissociate from hemoglobin and enter the tissues. The reverse is true of tissues, such as adipose (body fat), which have lower metabolic rates. Because less oxygen is used by these cells, the partial pressure of oxygen within such tissues remains relatively high, resulting in fewer oxygen molecules dissociating from hemoglobin and entering the tissue interstitial fluid. Although venous blood is said to be deoxygenated, some oxygen is still bound to hemoglobin in its red blood cells. This provides an oxygen reserve that can be used when tissues suddenly demand more oxygen.
Factors other than partial pressure also affect the oxygen-hemoglobin saturation/dissociation curve. For example, a higher temperature promotes hemoglobin and oxygen to dissociate faster, whereas a lower temperature inhibits dissociation (see Figure 3.2b). However, the human body tightly regulates temperature, so this factor may not affect gas exchange throughout the body. The exception to this is in highly active tissues, which may release a larger amount of energy than is given off as heat. As a result, oxygen readily dissociates from hemoglobin, which is a mechanism that helps to provide active tissues with more oxygen.
Certain hormones, such as androgens, epinephrine, thyroid hormones, and growth hormone, can affect the oxygen-hemoglobin saturation/disassociation curve by stimulating the production of a compound called 2,3-bisphosphoglycerate (BPG) by erythrocytes. BPG is a byproduct of glycolysis. Because erythrocytes do not contain mitochondria, glycolysis is the sole method by which these cells produce ATP. BPG promotes the disassociation of oxygen from hemoglobin. Therefore, the greater the concentration of BPG, the more readily oxygen dissociates from hemoglobin, despite its partial pressure.
The pH of the blood is another factor that influences the oxygen-hemoglobin saturation/dissociation curve (see Figure 3.2b). The Bohr effect is a phenomenon that arises from the relationship between pH and oxygen’s affinity for hemoglobin: A lower, more acidic pH promotes oxygen dissociation from hemoglobin. In contrast, a higher, or more basic, pH inhibits oxygen dissociation from hemoglobin. The greater the amount of carbon dioxide in the blood, the more molecules that must be converted, which in turn generates hydrogen ions and thus lowers blood pH. Furthermore, blood pH may become more acidic when certain byproducts of cell metabolism, such as lactic acid, carbonic acid, and carbon dioxide, are released into the bloodstream.
Carbon dioxide transport in the blood
Carbon dioxide is transported by three major mechanisms. The first mechanism of carbon dioxide transport is by blood plasma, as some carbon dioxide molecules dissolve in the blood. The second mechanism is transport in the form of bicarbonate (HCO 3 – ), which also dissolves in plasma. The third mechanism of carbon dioxide transport is similar to the transport of oxygen by erythrocytes (Figure 3.3).
Dissolved carbon dioxide
Although carbon dioxide is not considered to be highly soluble in blood, a small fraction—about 7 to 10 percent—of the carbon dioxide that diffuses into the blood from the tissues dissolves in plasma. The dissolved carbon dioxide then travels in the bloodstream and when the blood reaches the pulmonary capillaries, the dissolved carbon dioxide diffuses across the respiratory membrane into the alveoli, where it is then exhaled during pulmonary ventilation.
Bicarbonate buffer
A large fraction—about 70 percent—of the carbon dioxide molecules that diffuse into the blood is transported to the lungs as bicarbonate. Most bicarbonate is produced in erythrocytes after carbon dioxide diffuses into the capillaries, and subsequently into red blood cells. Carbonic anhydrase (CA) causes carbon dioxide and water to form carbonic acid (H 2 CO 3 ), which dissociates into two ions: bicarbonate (HCO 3 – ) and hydrogen (H + ). The following formula depicts this reaction:
\[\ce{CO2 + H2O CA↔ H2CO3↔H+ + HCO3−CO2 + H2O CA↔ H2CO3↔H+ + HCO3−}\]
Bicarbonate tends to build up in the erythrocytes so that there is a greater concentration of bicarbonate in the erythrocytes than in the surrounding blood plasma. As a result, some of the bicarbonate will leave the erythrocytes and move down its concentration gradient into the plasma in exchange for chloride (Cl – ) ions. This phenomenon is referred to as the chloride shift and occurs because by exchanging one negative ion for another negative ion, neither the electrical charge of the erythrocytes nor that of the blood is altered.
At the pulmonary capillaries, the chemical reaction that produced bicarbonate (shown above) is reversed, and carbon dioxide and water are the products. Much of the bicarbonate in the plasma re-enters the erythrocytes in exchange for chloride ions. Hydrogen ions and bicarbonate ions join to form carbonic acid, which is converted into carbon dioxide and water by carbonic anhydrase. Carbon dioxide diffuses out of the erythrocytes and into the plasma, where it can further diffuse across the respiratory membrane into the alveoli to be exhaled during pulmonary ventilation.
Carbaminohemoglobin
About 20 percent of carbon dioxide is bound by hemoglobin and is transported to the lungs. Carbon dioxide does not bind to iron as oxygen does; instead, carbon dioxide binds amino acid moieties on the globin portions of hemoglobin to form carbaminohemoglobin, which forms when hemoglobin and carbon dioxide bind. When hemoglobin is not transporting oxygen, it tends to have a bluish-purple tone to it, creating the darker maroon color typical of deoxygenated blood. The following formula depicts this reversible reaction:
\[\ce{CO2 + Hb↔HbCO2CO2 + Hb↔HbCO2}\]
Similar to the transport of oxygen by heme, the binding and dissociation of carbon dioxide to and from hemoglobin is dependent on the partial pressure of carbon dioxide. Because carbon dioxide is released from the lungs, blood that leaves the lungs and reaches body tissues has a lower partial pressure of carbon dioxide than is found in the tissues. As a result, carbon dioxide leaves the tissues because of its higher partial pressure, enters the blood, and then moves into red blood cells, binding to hemoglobin. In contrast, in the pulmonary capillaries, the partial pressure of carbon dioxide is high compared to within the alveoli. As a result, carbon dioxide dissociates readily from hemoglobin and diffuses across the respiratory membrane into the air.
In addition to the partial pressure of carbon dioxide, the oxygen saturation of hemoglobin and the partial pressure of oxygen in the blood also influence the affinity of hemoglobin for carbon dioxide. The Haldane effect is a phenomenon that arises from the relationship between the partial pressure of oxygen and the affinity of hemoglobin for carbon dioxide. Hemoglobin that is saturated with oxygen does not readily bind carbon dioxide. However, when oxygen is not bound to heme and the partial pressure of oxygen is low, hemoglobin readily binds to carbon dioxide.
- Watch this to see the transport of oxygen from the lungs to the tissues.
Oxyhemoglobin forms by a chemical reaction between which of the following?
a. hemoglobin and carbon dioxide
b. carbonic anhydrase and carbon dioxide
c. hemoglobin and oxygen
d. carbonic anhydrase and oxygen
Which of the following factors play a role in the oxygen-hemoglobin saturation/dissociation curve?
a. temperature
b. pH
c. BPG
d. All of the above
Which of the following occurs during the chloride shift?
a. Chloride is removed from the erythrocyte.
b. Chloride is exchanged for bicarbonate.
c. Bicarbonate is removed from the erythrocyte.
d. Bicarbonate is removed from the blood
A low partial pressure of oxygen promotes hemoglobin binding to carbon dioxide. This is an example of the ________.
a. Haldane effect
b. Bohr effect
c. Dalton’s law
d. Henry’s law
Why is oxygenated blood bright red, whereas deoxygenated blood tends to be more of a purple color? (think back to the video about oxygen transport)
Describe the relationship between the partial pressure of oxygen and the binding of oxygen to hemoglobin. | libretexts | 2025-03-17T22:26:52.195105 | 2018-10-18T16:30:14 | {
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"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "3.1: Gas Transport",
"author": "Sanja Hinić-Frlog"
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/3%3A_Transport_Gases/3.2%3A_Circulatory_Systems | 3.2: Circulatory Systems
- Compare and contrast different circulatory systems using specific animal examples and evolution.
The circulatory system
The circulatory system is a network of vessels—the arteries, veins, and capillaries—and a pump, the heart. In all vertebrate organisms this is a closed-loop system, in which the blood is largely separated from the body’s other extracellular fluid compartment, the interstitial fluid, which is the fluid bathing the cells. Blood circulates inside blood vessels and circulates unidirectionally from the heart around one of two circulatory routes, then returns to the heart again; this is a closed circulatory system. Open circulatory systems are found in invertebrate animals in which the circulatory fluid bathes the internal organs directly even though it may be moved about with a pumping heart.
The heart
The heart is a complex muscle that consists of two pumps: one that pumps blood through pulmonary circulation to the lungs, and the other that pumps blood through systemic circulation to the rest of the body’s tissues (and the heart itself).
The heart is asymmetrical, with the left side being larger than the right side, correlating with the different sizes of the pulmonary and systemic circuits (Figure 3.4). In humans, the heart is about the size of a clenched fist; it is divided into four chambers: two atria and two ventricles. There is one atrium and one ventricle on the right side and one atrium and one ventricle on the left side. The right atrium receives deoxygenated blood from the systemic circulation through the major veins: the superior vena cava, which drains blood from the head and from the veins that come from the arms, as well as the inferior vena cava, which drains blood from the veins that come from the lower organs and the legs. This deoxygenated blood then passes to the right ventricle through the tricuspid valve, which prevents the backflow of blood. After it is filled, the right ventricle contracts, pumping the blood to the lungs for reoxygenation. The left atrium receives the oxygen-rich blood from the lungs. This blood passes through the bicuspid valve to the left ventricle where the blood is pumped into the aorta. The aorta is the major artery of the body, taking oxygenated blood to the organs and muscles of the body. This pattern of pumping is referred to as double circulation and is found in all mammals. (Figure 3.4).
Which of the following statements about the circulatory system is false?
a. Blood in the pulmonary vein is deoxygenated.
b. Blood in the inferior vena cava is deoxygenated.
c. Blood in the pulmonary artery is deoxygenated.
d. Blood in the aorta is oxygenated.
The blood from the heart is carried through the body by a complex network of blood vessels (Figure 3.5). Arteries take blood away from the heart. The main artery of the systemic circulation is the aorta; it branches into major arteries that take blood to different limbs and organs. The aorta and arteries near the heart have heavy but elastic walls that respond to and smooth out the pressure differences caused by the beating heart. Arteries farther away from the heart have more muscle tissue in their walls that can constrict to affect flow rates of blood. The major arteries diverge into minor arteries, and then smaller vessels called arterioles, to reach more deeply into the muscles and organs of the body.
Arterioles diverge into capillary beds. Capillary beds contain a large number, 10’s to 100’s of capillaries that branch among the cells of the body. Capillaries are narrow-diameter tubes that can fit single red blood cells and are the sites for the exchange of nutrients, waste, and oxygen with tissues at the cellular level. Fluid also leaks from the blood into the interstitial space from the capillaries. The capillaries converge again into venules that connect to minor veins that finally connect to major veins. Veins are blood vessels that bring blood high in carbon dioxide back to the heart. Veins are not as thick-walled as arteries, since pressure is lower, and they have valves along their length that prevent backflow of blood away from the heart. The major veins drain blood from the same organs and limbs that the major arteries supply.
Where would flow be the fastest, narrow or wide tube? Why?
How is this relevant for blood transport through a vertebrate body?
How does the giraffe pump blood to its head? | libretexts | 2025-03-17T22:26:52.267489 | 2018-10-18T16:30:14 | {
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"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "3.2: Circulatory Systems",
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/4%3A_Ion_and_Water_Balance | 4: Ion and Water Balance
-
- 4.1: Osmoregulation in Animals Living in Aquatic Environment
- Osmoregulation is the process of maintaining salt and water balance (osmotic balance) across membranes within the body. The fluids inside and surrounding cells are composed of water, electrolytes, and nonelectrolytes. An electrolyte is a compound that dissociates into ions when dissolved in water. A nonelectrolyte, in contrast, does not dissociate into ions in water. | libretexts | 2025-03-17T22:26:52.322550 | 2018-10-18T16:30:14 | {
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"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/4%3A_Ion_and_Water_Balance",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "4: Ion and Water Balance",
"author": "Sanja Hinić-Frlog"
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/4%3A_Ion_and_Water_Balance/4.1%3A_Osmoregulation_in_Animals_Living_in_Aquatic_Environment | 4.1: Osmoregulation in Animals Living in Aquatic Environment
- Discuss osmoregulatory function challenges of animals living in terrestrial versus aquatic environments;
- Explain how ion and water balance function in sample animal saltwater systems; and,
- Explain how ion and water balance function in sample animal freshwater systems.
Osmoregulation
Osmoregulation is the process of maintaining salt and water balance (osmotic balance) across membranes within the body. The fluids inside and surrounding cells are composed of water, electrolytes, and nonelectrolytes. An electrolyte is a compound that dissociates into ions when dissolved in water. A nonelectrolyte, in contrast, does not dissociate into ions in water. The body’s fluids include blood plasma, the fluid that exists within cells, and the interstitial fluid that exists in the spaces between cells and tissues of the body. The membranes of the body (both the membranes around cells and the “membranes” made of cells lining body cavities) are semipermeable membranes. Semipermeable membranes are permeable to certain types of solutes and to water, but typically cell membranes are impermeable to solutes.
The body does not exist in isolation. There is a constant input of water and electrolytes into the system. Excess water, electrolytes, and wastes are transported to the kidneys and excreted, helping to maintain osmotic balance. Insufficient fluid intake results in fluid conservation by the kidneys. Biological systems constantly interact and exchange water and nutrients with the environment by way of consumption of food and water and through excretion in the form of sweat, urine, and feces. Without a mechanism to regulate osmotic pressure, or when a disease damages this mechanism, there is a tendency to accumulate toxic waste and water, which can have dire consequences.
Mammalian systems have evolved to regulate not only the overall osmotic pressure across membranes, but also specific concentrations of important electrolytes in the three major fluid compartments: blood plasma, interstitial fluid, and intracellular fluid. Since osmotic pressure is regulated by the movement of water across membranes, the volume of the fluid compartments can also change temporarily. Since blood plasma is one of the fluid components, osmotic pressures have a direct bearing on blood pressure.
Excretory system
The human excretory system functions to remove waste from the body through the skin as sweat, the lungs in the form of exhaled carbon dioxide, and through the urinary system in the form of urine. All three of these systems participate in osmoregulation and waste removal. Here we focus on the urinary system, which is comprised of the paired kidneys, the ureter, urinary bladder and urethra (Figure 4.1). The kidneys are a pair of bean-shaped structures that are located just below the liver in the body cavity. Each of the kidneys contains more than a million tiny units called nephrons that filter blood containing the metabolic wastes from cells. All the blood in the human body is filtered about 60 times a day by the kidneys. The nephrons remove wastes, concentrate them, and form urine that is collected in the bladder.
Internally, the kidney has three regions—an outer cortex, a medulla in the middle, and the renal pelvis, which is the expanded end of the ureter. The renal cortex contains the nephrons—the functional unit of the kidney. The renal pelvis collects the urine and leads to the ureter on the outside of the kidney. The ureters are urine-bearing tubes that exit the kidney and empty into the urinary bladder.
Blood enters each kidney from the aorta, the main artery supplying the body below the heart, through a renal artery. It is distributed in smaller vessels until it reaches each nephron in capillaries. Within the nephron, the blood comes in intimate contact with the waste-collecting tubules in a structure called the glomerulus. Water and many solutes present in the blood, including ions of sodium, calcium, magnesium, and others; as well as wastes and valuable substances such as amino acids, glucose, and vitamins, leave the blood and enter the tubule system of the nephron. As materials pass through the tubule much of the water, required ions, and useful compounds are reabsorbed back into the capillaries that surround the tubules leaving the wastes behind. Some of this reabsorption requires active transport and consumes ATP. Some wastes, including ions and some drugs remaining in the blood, diffuse out of the capillaries into the interstitial fluid and are taken up by the tubule cells. These wastes are then actively secreted into the tubules. The blood then collects in larger and larger vessels and leaves the kidney in the renal vein. The renal vein joins the inferior vena cava, the main vein that returns blood to the heart from the lower body. The amounts of water and ions reabsorbed into the circulatory system are carefully regulated and this is an important way the body regulates its water content and ion levels. The waste is collected in larger tubules and then leaves the kidney in the ureter, which leads to the bladder where urine, the combination of waste materials and water, is stored.
The bladder contains sensory nerves, stretch receptors that signal when it needs to be emptied. These signals create the urge to urinate, which can be voluntarily suppressed up to a limit. The conscious decision to urinate sets in play signals that open the sphincters, rings of smooth muscle that close off the opening, to the urethra that allows urine to flow out of the bladder and the body.
Osmosis is the diffusion of water across a membrane in response to osmotic pressure caused by an imbalance of molecules on either side of the membrane. Osmoregulation is the process of maintenance of salt and water balance ( osmotic balance ) across membranes within the body’s fluids, which are composed of water, plus electrolytes and non-electrolytes. An electrolyte is a solute that dissociates into ions when dissolved in water. A non-electrolyte , in contrast, doesn’t dissociate into ions during water dissolution. Both electrolytes and non-electrolytes contribute to the osmotic balance. The body’s fluids include blood plasma, the cytosol within cells, and interstitial fluid, the fluid that exists in the spaces between cells and tissues of the body. The membranes of the body (such as the pleural, serous, and cell membranes) are semi-permeable membranes . Semi-permeable membranes are permeable (or permissive) to certain types of solutes and water. Solutions on two sides of a semi-permeable membrane tend to equalize in solute concentration by movement of solutes and/or water across the membrane. As seen in Figure 4.2, a cell placed in water tends to swell due to gain of water from the hypotonic or “low salt” environment. A cell placed in a solution with higher salt concentration, on the other hand, tends to make the membrane shrivel up due to loss of water into the hypertonic or “high salt” environment. Isotonic cells have an equal concentration of solutes inside and outside the cell; this equalizes the osmotic pressure on either side of the cell membrane which is a semi-permeable membrane.
The body does not exist in isolation. There is a constant input of water and electrolytes into the system. While osmoregulation is achieved across membranes within the body, excess electrolytes and wastes are transported to the kidneys and excreted, helping to maintain osmotic balance.
Need for osmoregulation
Biological systems constantly interact and exchange water and nutrients with the environment by way of consumption of food and water and through excretion in the form of sweat, urine, and feces. Without a mechanism to regulate osmotic pressure, or when a disease damages this mechanism, there is a tendency to accumulate toxic waste and water, which can have dire consequences.
Mammalian systems have evolved to regulate not only the overall osmotic pressure across membranes, but also specific concentrations of important electrolytes in the three major fluid compartments: blood plasma, extracellular fluid, and intracellular fluid. Since osmotic pressure is regulated by the movement of water across membranes, the volume of the fluid compartments can also change temporarily. Because blood plasma is one of the fluid components, osmotic pressures have a direct bearing on blood pressure.
Transport of electrolytes across cell membranes
Electrolytes, such as sodium chloride, ionize in water, meaning that they dissociate into their component ions. In water, sodium chloride (NaCl), dissociates into the sodium ion (Na+) and the chloride ion (Cl–). The most important ions, whose concentrations are very closely regulated in body fluids, are the cations sodium (Na+), potassium (K+), calcium (Ca+2),
magnesium (Mg+2), and the anions chloride (Cl–), carbonate (CO3-2), bicarbonate (HCO3–), and phosphate(PO3–). Electrolytes are lost from the body during urination and perspiration. For this reason, athletes are encouraged to replace electrolytes and fluids during periods of increased activity and perspiration.
Osmotic pressure is influenced by the concentration of solutes in a solution. It is directly proportional to
the number of solute atoms or molecules and not dependent on the size of the solute molecules. Because electrolytes dissociate into their component ions, they, in essence, add more solute particles into the solution and have a greater effect on osmotic pressure, per mass than compounds that do not dissociate in water, such as glucose.
Water can pass through membranes by passive diffusion. If electrolyte ions could passively diffuse across membranes, it would be impossible to maintain specific concentrations of ions in each fluid compartment, therefore, they require special mechanisms to cross the semi-permeable membranes in the body. This movement can be accomplished by facilitated diffusion and active transport. Facilitated diffusion requires protein-based channels for moving the solute. Active transport requires energy in the form of ATP conversion, carrier proteins, or pumps in order to move ions against the concentration gradient.
Concept of osmolality and milliequivalent
In order to calculate osmotic pressure, it is necessary to understand how solute concentrations are measured. The unit for measuring solutes is the mole . One mole is defined as the gram molecular weight of the solute. For example, the molecular weight of sodium chloride is 58.44. Thus, one mole of sodium chloride weighs 58.44 grams. The molarity of a solution is the number of moles of solute per liter of solution. The molality of a solution is the number of moles of solute per kilogram of solvent. If the solvent is water, one kilogram of water is equal to one liter of water. While molarity and molality are used to express the concentration of solutions, electrolyte concentrations are usually expressed in terms of milliequivalents per liter (mEq/L): the mEq/L is equal to the ion concentration (in millimoles) multiplied by the number of electrical charges on the ion. The unit of milliequivalent takes into consideration the ions present in the solution (since electrolytes form ions in aqueous solutions) and the charge on the ions.
Thus, for ions that have a charge of one, one milliequivalent is equal to one millimole. For ions that have a charge of two (like calcium), one milliequivalent is equal to 0.5 millimoles. Another unit for the expression of electrolyte concentration is the milliosmole (mOsm), which is the number of milliequivalents of solute per kilogram of solvent. Body fluids are usually maintained within the range of 280 to 300 mOsm.
Osmoregulators and osmoconformers
Persons lost at sea without any fresh water to drink, are at risk of severe dehydration because the human body cannot adapt to drinking seawater, which is hypertonic in comparison to body fluids. Organisms such as goldfish that can tolerate only a relatively narrow range of salinity are referred to as stenohaline. About 90 percent of all bony fish are restricted to either freshwater or seawater. They are incapable of osmotic regulation in the opposite environment. It is possible, however, for a few fishes like salmon to spend part of their life in freshwater and part in sea water. Organisms like the salmon and molly that can tolerate a relatively wide range of salinity are referred to as euryhaline organisms. This is possible because some fish have evolved osmoregulatory mechanisms to survive in all kinds of aquatic environments. When they live in fresh water, their bodies tend to take up water because the environment is relatively hypotonic, as illustrated in Figure 4.3a. In such hypotonic environments, these fish do not drink much water. Instead, they pass a lot of very dilute urine, and they achieve electrolyte balance by active transport of salts through the gills. When they move to a hypertonic marine environment, these fish start drinking sea water; they excrete the excess salts through their gills and their urine, as illustrated in Figure 4.3b. Most marine invertebrates, on the other hand, maybe isotonic with sea water ( osmoconformers ). Their body fluid concentrations conform to changes in seawater concentration. Cartilaginous fishes’ salt composition of the blood is similar to bony fishes; however, the blood of sharks contains the organic compounds urea and trimethylamine oxide (TMAO). This does not mean that their electrolyte composition is similar to that of seawater. They achieve isotonicity with the sea by storing large concentrations of urea. These animals that secrete urea are called ureotelic animals. TMAO stabilizes proteins in the presence of high urea levels, preventing the disruption of peptide bonds that would occur in other animals exposed to similar levels of urea. Sharks are cartilaginous fish with a rectal gland to secrete salt and assist in osmoregulation.
What is the difference between osmoregulator and osmoconformer?
What are the biggest osmoregulatory challenges for fish in saltwater?
Summarize how salmon survive in both salt and freshwater. | libretexts | 2025-03-17T22:26:52.393000 | 2018-10-18T16:30:14 | {
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"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "4.1: Osmoregulation in Animals Living in Aquatic Environment",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/4%3A_Ion_and_Water_Balance/4.2%3A_Ion_and_Water_Balance_on_Land | 4.2: Ion and Water Balance on Land
- Explain how ion and water balance function in sample animal terrestrial systems.
Although the kidneys are the major osmoregulatory organ, the skin and lungs also play a role in the process. Water and electrolytes are lost through sweat glands in the skin, which helps moisturize and cool the skin surface, while the lungs expel a small amount of water in the form of mucous secretions and via evaporation of water vapor.
Kidneys: the main osmoregulatory organ
The kidneys, illustrated in Figure 4.4, are a pair of bean-shaped structures that are located just below and posterior to the liver in the peritoneal cavity. The adrenal glands sit on top of each kidney and are also called the suprarenal glands. Kidneys filter blood and purify it. All the blood in the human body is filtered many times a day by the kidneys; these organs use up almost 25 percent of the oxygen absorbed through the lungs to perform this function. Oxygen allows the kidney cells to efficiently manufacture chemical energy in the form of ATP through aerobic respiration. The filtrate coming out of the kidneys is called urine.
Kidney structure
Externally, the kidneys are surrounded by three layers, illustrated in Figure 4.5. The outermost layer is a tough connective tissue layer called the renal fascia. The second layer is called the perirenal fat capsule, which helps anchor the kidneys in place. The third and innermost layer is the renal capsule. Internally, the kidney has three regions—an outer cortex, a medulla in the middle, and the renal pelvis in the region called the hilum of the kidney. The hilum is the concave part of the bean-shape where blood vessels and nerves enter and exit the kidney; it is also the point of exit for the ureters. The renal cortex is granular due to the presence of nephrons—the functional unit of the kidney. The medulla consists of multiple pyramidal tissue masses, called the renal pyramids. In between the pyramids are spaces called renal columns through which the blood vessels pass. The tips of the pyramids, called renal papillae, point toward the renal pelvis. There are, on average, eight renal pyramids in each kidney. The renal pyramids along with the adjoining cortical region are called the lobes of the kidney. The renal pelvis leads to the ureter on the outside of the kidney. On the inside of the kidney, the renal pelvis branches out into two or three extensions called the major calyces, which further branch into the minor calyces. The ureters are urine-bearing tubes that exit the kidney and empty into the urinary bladder.
Which of the following statements about the kidney is false?
a. The renal pelvis drains into the ureter.
b. The renal pyramids are in the medulla.
c. The cortex covers the capsule.
d. Nephrons are in the renal cortex
Because the kidney filters blood, its network of blood vessels is an important component of its structure and function. The arteries, veins, and nerves that supply the kidney enter and exit at the renal hilum. Renal blood supply starts with the branching of the aorta into the renal arteries(which are each named based on the region of the kidney they pass through) and ends with the exiting of the renal veins to join the inferior vena cava. The renal arteries split into several segmental arteries upon entering the kidneys. Each segmental artery splits further into several interlobar arteries and enters the renal columns, which supply the renal lobes. The interlobar arteries split at the junction of the renal cortex and medulla to form the arcuate arteries. The arcuate “bow-shaped” arteries form arcs along the base of the medullary pyramids. Corticalradiate arteries, as the name suggests, radiate out from the arcuate arteries. The cortical radiate arteries branch into numerous afferent arterioles and then enter the capillaries supplying the nephrons. Veins trace the path of the arteries and have similar names, except there are no segmental veins.
As mentioned previously, the functional unit of the kidney is the nephron, illustrated in Figure 4.6. Each kidney is made up of over one million nephrons that dot the renal cortex, giving it a granular appearance when sectioned sagittally. There are two types of nephrons— cortical nephrons (85 percent), which are deep in the renal cortex, and juxtamedullary nephrons, which lie in the renal cortex close to the renal medulla. A nephron consists of three parts—a renal corpuscle, a renal tubule, and the associated capillary network, which originates from the cortical radiate arteries.
Nephrons: the functional unit
Nephrons take a simple filtrate of the blood and modify it into the urine. Many changes take place in the different parts of the nephron before the urine is created for disposal. The term forming urine will be used hereafter to describe the filtrate as it is modified into the true urine. The principal task of the nephron population is to balance the plasma to homeostatic set points and excrete potential toxins in the urine. They do this by accomplishing three principle functions—filtration, reabsorption, and secretion. They also have additional secondary functions that exert control in three areas: blood pressure (via production of renin), red blood cell production (via the hormone EPO), and calcium absorption (via conversion of calcidiol into calcitriol, the active form of vitamin D).
Renal Corpuscle
The renal corpuscle, located in the renal cortex, is made up of a network of capillaries known as the glomerulus and the capsule, a cup-shaped chamber that surrounds it, called the glomerular or Bowman’s capsule. The renal corpuscle consists of a tuft of capillaries called the glomerulus that is largely surrounded by Bowman’s (glomerular) capsule. The glomerulus is a high-pressure capillary bed between afferent and efferent arterioles. Bowman’s capsule surrounds the glomerulus to form a lumen and captures and directs this filtrate to the PCT. The outermost part of Bowman’s capsule, the parietal layer, is a simple squamous epithelium. It transitions onto the glomerular capillaries in an intimate embrace to form the visceral layer of the capsule. Here, the cells are not squamous, but uniquely shaped cells (podocytes) extending finger-like arms (pedicels) to cover the glomerular capillaries (Figure 4.7). These projections interdigitate to form filtration slits, leaving small gaps between the digits to form a sieve. As blood passes through the glomerulus, 10 to 20 percent of the plasma filters between these sieve-like fingers to be captured by Bowman’s capsule and funneled to the PCT. Where the fenestrae (windows) in the glomerular capillaries match the spaces between the podocyte “fingers,” the only thing separating the capillary lumen and the lumen of Bowman’s capsule is their shared basement membrane (Figure 4.8). These three features comprise what is known as the filtration membrane. This membrane permits very rapid movement of filtrate from capillary to capsule through pores that are only 70 nm in diameter.
The fenestrations prevent filtration of blood cells or large proteins but allow most other constituents through. These substances cross readily if they are less than 4 nm in size and most pass freely up to 8 nm in size. An additional factor affecting the ability of substances to cross this barrier is their electric charge. The proteins associated with these pores are negatively charged, so they tend to repel negatively charged substances and allow positively charged substances to pass more readily. The basement membrane prevents filtration of medium-to-large proteins such as globulins. There are also mesangial cells in the filtration membrane that can contract to help regulate the rate of filtration of the glomerulus. Overall, filtration is regulated by fenestrations in capillary endothelial cells, podocytes with filtration slits, membrane charge, and the basement membrane between capillary cells. The result is the creation of a filtrate that does not contain cells or large proteins and has a slight predominance of positively charged substances.
Lying just outside Bowman’s capsule and the glomerulus is the juxtaglomerular apparatus (JGA) (Figure 4.9). At the juncture where the afferent and efferent arterioles enter and leave Bowman’s capsule, the initial part of the distal convoluted tubule (DCT) comes into direct contact with the arterioles. The wall of the DCT at that point forms a part of the JGA known as the macula densa. This cluster of cuboidal epithelial cells monitors the fluid composition of fluid flowing through the DCT. In response to the concentration of Na + in the fluid flowing past them, these cells release paracrine signals. They also have a single, nonmotile cilium that responds to the rate of fluid movement in the tubule. The paracrine signals released in response to changes in flow rate and Na + concentration are adenosine triphosphate (ATP) and adenosine.
The renal tubule is a long and convoluted structure that emerges from the glomerulus and can be divided into three parts based on function. The first part is called the proximal convoluted tubule (PCT) due to its proximity to the glomerulus; it stays in the renal cortex. The second part is called the loop of Henle, or nephritic loop because it forms a loop (with descending and ascending limbs) that goes through the renal medulla. The third part of the renal tubule is called the distal convoluted tubule (DCT) and this part is also restricted to the renal cortex. The DCT, which is the last part of the nephron, connects and empties its contents into collecting ducts that line the medullary pyramids. The collecting ducts amass contents from multiple nephrons and fuse together as they enter the papillae of the renal medulla.
Capillary network within the nephron
The capillary network that originates from the renal arteries supplies the nephron with blood that needs to be filtered. The branch that enters the glomerulus is called the afferent arteriole. The branch that exits the glomerulus is called the efferent arteriole. Within the glomerulus, the network of capillaries is called the glomerular capillary bed. Once the efferent arteriole exits the glomerulus, it forms the peritubular capillary network, which surrounds and interacts with parts of the renal tubule. In cortical nephrons, the peritubular capillary network surrounds the PCT and DCT. In juxtamedullary nephrons, the peritubular capillary network forms a network around the loop of Henle and is called the vasa recta.
Kidney function and physiology
Kidneys filter blood in a three-step process. First, the nephrons filter blood that runs through the capillary network in the glomerulus. Almost all solutes, except for proteins, are filtered out into the glomerulus by a process called glomerular filtration. Second, the filtrate is collected in the renal tubules. Most of the solutes get reabsorbed in the PCT by a process called tubular reabsorption. In the loop of Henle, the filtrate continues to exchange solutes and water with the renal medulla and the peritubular capillary network. Water is also reabsorbed during this step. Then, additional solutes and wastes are secreted into the kidney tubules during tubular secretion, which is, in essence, the opposite process of tubular reabsorption. The collecting ducts collect filtrate coming from the nephrons and fuse in the medullary papillae. From here, the papillae deliver the filtrate, now called urine, into the minor calyces that eventually connect to the ureters through the renal pelvis. This entire process is illustrated in Figure 4.10.
Glomerular filtration
Glomerular filtration filters out most of the solutes due to high blood pressure and specialized membranes in the afferent arteriole. The blood pressure in the glomerulus is maintained independent of factors that affect systemic blood pressure. The “leaky” connections between the endothelial cells of the glomerular capillary network allow solutes to pass through easily. All solutes in the glomerular capillaries, except for macromolecules like proteins, pass through by passive diffusion. There is no energy requirement at this stage of the filtration process. Glomerular filtration rate (GFR) is the volume of glomerular filtrate formed per minute by the kidneys. GFR is regulated by multiple mechanisms and is an important indicator of kidney function.
Tubular reabsorption and secretion
Tubular reabsorption occurs in the PCT part of the renal tubule. Almost all nutrients are reabsorbed, and this occurs either by passive or active transport. Reabsorption of water and some key electrolytes are regulated and can be influenced by hormones. Sodium (Na+) is the most abundant ion and most of it is reabsorbed by active transport and then transported to the peritubular capillaries. Because Na+ is actively transported out of the tubule, water follows it to even out the osmotic pressure. Water is also independently reabsorbed into the peritubular capillaries due to the presence of aquaporins, or water channels, in the PCT. This occurs due to the low blood pressure and high osmotic pressure in the peritubular capillaries. However, every solute has a transport maximum and the excess is not reabsorbed.
In the loop of Henle, the permeability of the membrane changes. The descending limb is permeable to water, not solutes; the opposite is true for the ascending limb. Additionally, the loop of Henle invades the renal medulla, which is naturally high in salt concentration and tends to absorb water from the renal tubule and concentrate the filtrate. The osmotic gradient increases as it moves deeper into the medulla. Because two sides of the loop of Henle perform opposing functions, as illustrated in Figure 4.11, it acts as a countercurrent multiplier. The vasa recta around it acts as the countercurrent exchanger.
Loop diuretics are drugs sometimes used to treat hypertension. These drugs inhibit the reabsorption of Na+ and Cl– ions by the ascending limb of the loop of Henle. A side effect is that they increase urination. Why do you think this is the case?
By the time the filtrate reaches the DCT, most of the urine and solutes have been reabsorbed. If the body requires additional water, all of it can be reabsorbed at this point. Further reabsorption is controlled by hormones, which will be discussed in a later section. Excretion of wastes occurs due to lack of reabsorption combined with tubular secretion. Undesirable products like metabolic wastes, urea, uric acid, and certain drugs, are excreted by tubular secretion. Most of the tubular secretion happens in the DCT, but some occurs in the early part of the collecting duct. Kidneys also maintain an acid-base balance by secreting excess H+ ions.
Although parts of the renal tubules are named proximal and distal, in a cross-section of the kidney, the tubules are placed close together and in contact with each other and the glomerulus. This allows for the exchange of chemical messengers between the different cell types. For example, the DCT ascending limb of the loop of Henle has masses of cells called macula densa, which are in contact with cells of the afferent arterioles called juxtaglomerular cells. Together, the macula densa and juxtaglomerular cells form the juxtaglomerular complex (JGC). The JGC is an endocrine structure that secretes the enzyme renin and the hormone erythropoietin. When hormones trigger the macula densa cells in the DCT due to variations in blood volume, blood pressure, or electrolyte balance, these cells can immediately communicate the problem to the capillaries in the afferent and efferent arterioles, which can constrict or relax to change the glomerular filtration rate of the kidneys. | libretexts | 2025-03-17T22:26:52.466943 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/4%3A_Ion_and_Water_Balance/4.2%3A_Ion_and_Water_Balance_on_Land",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "4.2: Ion and Water Balance on Land",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/5%3A_Digestion_and_Energy | Skip to main content
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5.1: Biological Energy
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Scientists use the term bioenergetics to describe the concept of energy flow through living systems, such as cells. Cellular processes such as the building and breaking down of complex molecules occur through stepwise chemical reactions. Some of these chemical reactions are spontaneous and release energy, whereas others require energy to proceed.
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5.2: Components of the Digestive System
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The nutrients and macromolecules present in food are not immediately accessible to the cells. There are a number of processes that modify food within the animal body in order to make the nutrients and organic molecules accessible for cellular function. As animals evolved in the complexity of form and function, their digestive systems have also evolved to accommodate their various dietary needs.
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5.3: Nutrient Transport and Energy Metabolism
- | libretexts | 2025-03-17T22:26:52.524381 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/5%3A_Digestion_and_Energy",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "5: Digestion and Energy",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/5%3A_Digestion_and_Energy/5.1%3A_Biological_Energy | 5.1: Biological Energy
- List and briefly explain different types of energy relevant for biology.
Cell’s metabolism and energy
Scientists use the term bioenergetics to describe the concept of energy flow ( Figure 5.1 ) through living systems, such as cells. Cellular processes such as the building and breaking down of complex molecules occur through stepwise chemical reactions . Some of these chemical reactions are spontaneous and release energy, whereas others require energy to proceed. Just as living things must continually consume food to replenish their energy supplies, cells must continually produce more energy to replenish that used by the many energy-requiring chemical reactions that constantly take place. Together, all of the chemical reactions that take place inside cells, including those that consume or generate energy, are referred to as the cell’s metabolism .
Metabolic pathways
Consider the metabolism of sugar. This is a classic example of one of the many cellular processes that use and produce energy. Living things consume sugars as a major energy source because sugar molecules have a great deal of energy stored within their bonds. For the most part, photosynthesizing organisms like plants produce these sugars. During photosynthesis, plants use energy (originally from sunlight) to convert carbon dioxide gas (CO 2 ) into sugar molecules (like glucose: C 6 H 12 O 6 ). They consume carbon dioxide and produce oxygen as a waste product. This reaction is summarized as:
\[\ce{6CO2 + 6H2O + energy -> C6H12O6 + 6O2}\]
Because this process involves synthesizing an energy-storing molecule, it requires energy input to proceed. During the light reactions of photosynthesis, energy is provided by a molecule called adenosine triphosphate (ATP) , which is the primary energy currency of all cells. Just as the dollar is used as currency to buy goods, cells use molecules of ATP as energy currency to perform immediate work. In contrast, energy-storage molecules such as glucose are consumed only to be broken down to use their energy. The reaction that harvests the energy of a sugar molecule in cells requiring oxygen to survive can be summarized by the reverse reaction to photosynthesis. In this reaction, oxygen is consumed and carbon dioxide is released as a waste product. The reaction is summarized as:
\[\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O + energy}\]
Both of these reactions involve many steps.
The processes of making and breaking down sugar molecules illustrate two examples of metabolic pathways. A metabolic pathway is a series of chemical reactions that takes a starting molecule and modifies it, step-by-step, through a series of metabolic intermediates, eventually yielding a final product. In the example of sugar metabolism, the first metabolic pathway synthesized sugar from smaller molecules, and the other pathway broke sugar down into smaller molecules. These two opposite processes—the first requiring energy and the second producing energy—are referred to as anabolic pathways (building polymers) and catabolic pathways (breaking down polymers into their monomers) , respectively. Consequently, metabolism is composed of synthesis (anabolism) and degradation (catabolism) ( Figure 5.2 ).
It is important to know that the chemical reactions of metabolic pathways do not take place on their own. Each reaction step is facilitated, or catalyzed, by a protein called an enzyme. Enzymes are important for catalyzing all types of biological reactions —those that require energy as well as those that release energy.
Energy
Thermodynamics refers to the study of energy and energy transfer involving physical matter. The matter relevant to a particular case of energy transfer is called a system, and everything outside of that matter is called the surroundings. For instance, when heating a pot of water on the stove, the system includes the stove, the pot, and the water. Energy is transferred within the system (between the stove, pot, and water). There are two types of systems: open and closed. In an open system, energy can be exchanged with its surroundings. The stovetop system is open because heat can be lost to the air. A closed system cannot exchange energy with its surroundings.
Biological organisms are open systems. Energy is exchanged between them and their surroundings as they use energy from the sun to perform photosynthesis or consume energy-storing molecules and release energy to the environment by doing work and releasing heat. Like all things in the physical world, energy is subject to physical laws. The laws of thermodynamics govern the transfer of energy in and among all systems in the universe.
In general, energy is defined as the ability to do work or to create some kind of change. Energy exists in different forms. For example, electrical energy, light energy, and heat energy are all different types of energy. To appreciate the way energy flows into and out of biological systems, it is important to understand two of the physical laws that govern energy.
Thermodynamics
The first law of thermodynamics states that the total amount of energy in the universe is constant and conserved. In other words, there has always been, and always will be, exactly the same amount of energy in the universe. Energy exists in many different forms . According to the first law of thermodynamics, energy may be transferred from place to place or transformed into different forms, but it cannot be created or destroyed . The transfers and transformations of energy take place around us all the time. Light bulbs transform electrical energy into light and heat energy. Gas stoves transform chemical energy from natural gas into heat energy. Plants perform one of the most biologically useful energy transformations on earth: that of converting the energy of sunlight to chemical energy stored within organic molecules (Figure 5.2). Some examples of energy transformations are shown in Figure 5.3 .
The challenge for all living organisms is to obtain energy from their surroundings in forms that they can transfer or transform into usable energy to do work. Living cells have evolved to meet this challenge. Chemical energy stored within organic molecules such as sugars and fats is transferred and transformed through a series of cellular chemical reactions into energy within molecules of ATP. Energy in ATP molecules is easily accessible to do work. Examples of the types of work that cells need to do include building complex molecules, transporting materials, powering the motion of cilia or flagella, and contracting muscle fibers to create movement.
A living cell’s primary tasks of obtaining, transforming, and using energy to do work may seem simple. However, the second law of thermodynamics explains why these tasks are harder than they appear. All energy transfers and transformations are never completely efficient . In every energy transfer, some amount of energy is lost in a form that is unusable. In most cases, this form is heat energy. Thermodynamically, heat energy is defined as the energy transferred from one system to another that is not work. For example, when a light bulb is turned on, some of the energy being converted from electrical energy into light energy is lost as heat energy. Likewise, some energy is lost as heat energy during cellular metabolic reactions.
An important concept in physical systems is that of order and disorder. The more energy that is lost by a system to its surroundings, the less ordered and more random the system is. Scientists refer to the measure of randomness or disorder within a system as entropy . High entropy means high disorder and low energy. Molecules and chemical reactions have varying entropy as well. For example, entropy increases as molecules at a high concentration in one place diffuse and spread out. The second law of thermodynamics says that energy will always be lost as heat in energy transfers or transformations.
Living things are highly ordered, requiring constant energy input to be maintained in a state of low entropy.
Potential and kinetic energy
When an object is in motion, there is energy associated with that object. Think of a wrecking ball. Even a slow-moving wrecking ball can do a great deal of damage to other objects. The energy associated with objects in motion is called kinetic energy (Figure 5.4). A speeding bullet, a walking person, and the rapid movement of molecules in the air (which produces heat) all have kinetic energy.
Now, what if that same motionless wrecking ball is lifted two stories above ground with a crane? If the suspended wrecking ball is unmoving, is there energy associated with it? The answer is yes. The energy that was required to lift the wrecking ball did not disappear but is now stored in the wrecking ball by virtue of its position and the force of gravity acting on it. This type of energy is called potential energy ( Figure 5.4 ). If the ball were to fall, the potential energy would be transformed into kinetic energy until all of the potential energy was exhausted when the ball rested on the ground. Wrecking balls also swing like a pendulum; through the swing, there is a constant change of potential energy (highest at the top of the swing) to kinetic energy (highest at the bottom of the swing). Other examples of potential energy include the energy of water held behind a dam or a person about to skydive out of an airplane.
Potential energy is not only associated with the location of matter, but also with the structure of matter. Even a spring on the ground has potential energy if it is compressed; so does a rubber band that is pulled taut. On a molecular level, the bonds that hold the atoms of molecules together exist in a particular structure that has potential energy. Remember that anabolic cellular pathways require energy to synthesize complex molecules from simpler ones and catabolic pathways release energy when complex molecules are broken down. The fact that energy can be released by the breakdown of certain chemical bonds implies that those bonds have potential energy. In fact, there is potential energy stored within the bonds of all the food molecules we eat, which is eventually harnessed for use. This is because these bonds can release energy when broken. The type of potential energy that exists within chemical bonds, and is released when those bonds are broken, is called chemical energy. Chemical energy is responsible for providing living cells with energy from food. The release of energy occurs when the molecular bonds within food molecules are broken.
Watch this video about kilocalories to help you connect with cellular respiration and energy information you read about in this section of Chapter 5.
Free and activation energy
After learning that chemical reactions release energy when energy-storing bonds are broken, an important next question is the following: How is the energy associated with these chemical reactions quantified and expressed? How can the energy released from one reaction be compared to that of another reaction? A measurement of free energy is used to quantify these energy transfers. Recall that according to the second law of thermodynamics, all energy transfers involve the loss of some amount of energy in an unusable form such as heat. Free energy specifically refers to the energy associated with a chemical reaction that is available after the losses are accounted for. In other words, free energy is usable energy or energy that is available to do work.
If energy is released during a chemical reaction, then the change in free energy, signified as \(∆G\) (delta G) will be a negative number. A negative change in free energy also means that the products of the reaction have less free energy than the reactants because they release some free energy during the reaction. Reactions that have a negative change in free energy and consequently release free energy are called exergonic reactions. Think: ex ergonic means energy is ex iting the system. These reactions are also referred to as spontaneous reactions, and their products have less stored energy than the reactants. An important distinction must be drawn between the term spontaneous and the idea of a chemical reaction occurring immediately. Contrary to the everyday use of the term, a spontaneous reaction is not one that suddenly or quickly occurs. The rusting of iron is an example of a spontaneous reaction that occurs slowly, little by little, over time.
If a chemical reaction absorbs energy rather than releases energy on balance, then the ∆G for that reaction will be a positive value. In this case, the products have more free energy than the reactants. Thus, the products of these reactions can be thought of as energy-storing molecules. These chemical reactions are called endergonic reactions and they are non-spontaneous . An endergonic reaction will not take place on its own without the addition of free energy.
Look at each of the processes shown in Figure 5.5 and decide if it is endergonic or exergonic.
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TBA
Figure 5.5. Some examples of endergonic processes (ones that require energy) and exergonic processes (ones that release energy). (credit a: modification of work by Natalie Maynor; credit b: modification of work by USDA; credit c: modification of work by Cory Zanker; credit d: modification of work by Harry Malsch).
There is another important concept that must be considered regarding endergonic and exergonic reactions. Exergonic reactions require a small amount of energy input to get going before they can proceed with their energy-releasing steps. These reactions have a net release of energy, but still, require some energy input in the beginning. This small amount of energy input necessary for all chemical reactions to occur is called the activation energy.
Enzymes
A substance that helps a chemical reaction to occur is called a catalyst, and the molecules that catalyze biochemical reactions are called enzymes. Most enzymes are proteins and perform the critical task of lowering the activation energies of chemical reactions inside the cell. Most of the reactions critical to a living cell happen too slowly at normal temperatures to be of any use to the cell. Without enzymes to speed up these reactions , life could not persist. Enzymes do this by binding to the reactant molecules and holding them in such a way as to make the chemical bond-breaking and -forming processes take place more easily. It is important to remember that enzymes do not change whether a reaction is exergonic (spontaneous) or endergonic. This is because they do not change the free energy of the reactants or products. They only reduce the activation energy required for the reaction to go forward ( Figure 5.6 ). In addition, an enzyme itself is unchanged by the reaction it catalyzes. Once one reaction has been catalyzed, the enzyme is able to participate in other reactions.
The chemical reactants to which an enzyme binds are called the enzyme’s substrates. There may be one or more substrates, depending on the particular chemical reaction. In some reactions, a single reactant substrate is broken down into multiple products. In others, two substrates may come together to create one larger molecule. Two reactants might also enter a reaction and both become modified, but they leave the reaction as two products. The location within the enzyme where the substrate binds is called the enzyme’s active site . The active site is where the “action” happens. Since enzymes are proteins, there is a unique combination of amino acid side chains within the active site. Each side chain is characterized by different properties. They can be large or small, weakly acidic or basic, hydrophilic or hydrophobic, positively or negatively charged, or neutral. The unique combination of side chains creates a very specific chemical environment within the active site. This specific environment is suited to bind to one specific chemical substrate (or substrates).
Active sites are subject to influences of the local environment. Increasing the environmental temperature generally increases reaction rates, enzyme-catalyzed or otherwise. However, temperatures outside of an optimal range reduce the rate at which an enzyme catalyzes a reaction. Hot temperatures will eventually cause enzymes to denature, an irreversible change in the three-dimensional shape and therefore the function of the enzyme. Enzymes are also suited to function best within a certain pH and salt concentration range, and, as with temperature, extreme pH, and salt concentrations can cause enzymes to denature.
For many years, scientists thought that enzyme-substrate binding took place in a simple “lock and key” fashion. This model asserted that the enzyme and substrate fit together perfectly in one instantaneous step. However, current research supports a model called induced fit ( Figure 5.7 ). The induced-fit model expands on the lock-and-key model by describing a more dynamic binding between enzyme and substrate. As the enzyme and substrate come together, their interaction causes a mild shift in the enzyme’s structure that forms an ideal binding arrangement between enzyme and substrate.
When an enzyme binds its substrate, an enzyme-substrate complex is formed. This complex lowers the activation energy of the reaction and promotes its rapid progression in one of the multiple possible ways. On a basic level, enzymes promote chemical reactions that involve more than one substrate by bringing the substrates together in an optimal orientation for reaction. Another way in which enzymes promote the reaction of their substrates is by creating an optimal environment within the active site for the reaction to occur. The chemical properties that emerge from the particular arrangement of amino acid R groups within an active site create the perfect environment for an enzyme’s specific substrates to react.
The enzyme-substrate complex can also lower activation energy by compromising the bond structure so that it is easier to break. Finally, enzymes can also lower activation energies by taking part in the chemical reaction itself. In these cases, it is important to remember that the enzyme will always return to its original state upon the completion of the reaction. One of the hallmark properties of enzymes is that they remain ultimately unchanged by the reactions they catalyze. After an enzyme has catalyzed a reaction, it releases its product(s) and can catalyze a new reaction.
It would seem ideal to have a scenario in which all of an organism’s enzymes existed in abundant supply and functioned optimally under all cellular conditions, in all cells, at all times. However, a variety of mechanisms ensures that this does not happen. Cellular needs and conditions constantly vary from cell to cell and change within individual cells over time. The required enzymes of stomach cells differ from those of fat storage cells, skin cells, blood cells, and nerve cells. Furthermore, a digestive organ cell works much harder to process and break down nutrients during the time that closely follows a meal compared with many hours after a meal. As these cellular demands and conditions vary, so must the amounts and functionality of different enzymes.
Since the rates of biochemical reactions are controlled by activation energy, and enzymes lower and determine activation energies for chemical reactions, the relative amounts and functioning of the variety of enzymes within a cell ultimately determine which reactions will proceed and at what rates. This determination is tightly controlled in cells. In certain cellular environments, enzyme activity is partly controlled by environmental factors like pH, temperature, salt concentration, and, in some cases, cofactors or coenzymes.
Enzymes can also be regulated in ways that either promote or reduce enzyme activity. There are many kinds of molecules that inhibit or promote enzyme function, and various mechanisms by which they do so. In some cases of enzyme inhibition , an inhibitor molecule is similar enough to a substrate that it can bind to the active site and simply block the substrate from binding. When this happens, the enzyme is inhibited through competitive inhibition , because an inhibitor molecule competes with the substrate for binding to the active site.
On the other hand, in noncompetitive inhibition , an inhibitor molecule binds to the enzyme in a location other than the active site, called an allosteric site , but still manages to block substrate binding to the active site. Some inhibitor molecules bind to enzymes in a location where their binding induces a conformational change that reduces the affinity of the enzyme for its substrate. This type of inhibition is called allosteric inhibition (Figure 5.8). Most allosterically regulated enzymes are made up of more than one polypeptide, meaning that they have more than one protein subunit. When an allosteric inhibitor binds to a region on an enzyme, all active sites on the protein subunits are changed slightly such that they bind their substrates with less efficiency. There are allosteric activators as well as inhibitors. Allosteric activators bind to locations on an enzyme away from the active site, inducing a conformational change that increases the affinity of the enzyme’s active site(s) for its substrate(s) ( Figure 5.8 ).
Enzymes are key components of metabolic pathways. Understanding how enzymes work and how they can be regulated are key principles behind the development of many of the pharmaceutical drugs on the market today. Biologists working in this field collaborate with other scientists to design drugs ( Figure 5.9 ).
Consider statins for example—statins is the name given to one class of drugs that can reduce cholesterol levels. These compounds are inhibitors of the enzyme HMG-CoA reductase, which is the enzyme that synthesizes cholesterol from lipids in the body. By inhibiting this enzyme, the level of cholesterol synthesized in the body can be reduced. Similarly, acetaminophen, popularly marketed under the brand name Tylenol, is an inhibitor of the enzyme cyclooxygenase. While it is used to provide relief from fever and inflammation (pain), its mechanism of action is still not completely understood.
How are drugs discovered? One of the biggest challenges in drug discovery is identifying a drug target. A drug target is a molecule that is literally the target of the drug. In the case of statins, HMG-CoA reductase is the drug target. Drug targets are identified through painstaking research in the laboratory. Identifying the target alone is not enough; scientists also need to know how the target acts inside the cell and which reactions go awry in the case of a disease. Once the target and the pathway are identified, then the actual process of drug design begins. In this stage, chemists and biologists work together to design and synthesize molecules that can block or activate a particular reaction. However, this is only the beginning: If and when a drug prototype is successful in performing its function, then it is subjected to many tests from in vitro experiments to clinical trials before it can get approval from the U.S. Food and Drug Administration to be on the market.
Many enzymes do not work optimally, or even at all, unless bound to other specific non-protein helper molecules. They may bond either temporarily through ionic or hydrogen bonds, or permanently through stronger covalent bonds. Binding to these molecules promotes optimal shape and function of their respective enzymes. Two examples of these types of helper molecules are cofactors and coenzymes. Cofactors are inorganic ions such as ions of iron and magnesium. Coenzymes are organic helper molecules, those with a basic atomic structure made up of carbon and hydrogen. Like enzymes, these molecules participate in reactions without being changed themselves and are ultimately recycled and reused. Vitamins are the source of coenzymes. Some vitamins are the precursors of coenzymes and others act directly as coenzymes. Vitamin C is a direct coenzyme for multiple enzymes that take part in building the important connective tissue, collagen. Therefore, enzyme function is, in part, regulated by the abundance of various cofactors and coenzymes, which may be supplied by an organism’s diet or, in some cases, produced by the organism.
Feedback inhibition in metabolic pathways
Molecules can regulate enzyme function in many ways. The major question remains, however: What are these molecules and where do they come from? Some are cofactors and coenzymes, as you have learned. What other molecules in the cell provide enzymatic regulation such as allosteric modulation and competitive and non-competitive inhibition? Perhaps the most relevant sources of regulatory molecules, with respect to enzymatic cellular metabolism, are the products of the cellular metabolic reactions themselves. In a most efficient and elegant way, cells have evolved to use the products of their own reactions for feedback inhibition of enzyme activity. Feedback inhibition involves the use of a reaction product to regulate its own further production ( Figure 5.10 ). The cell responds to an abundance of the products by slowing down production during anabolic or catabolic reactions. Such reaction products may inhibit the enzymes that catalyzed their production through the mechanisms described above.
The production of both amino acids and nucleotides is controlled through feedback inhibition. Additionally, ATP is an allosteric regulator of some of the enzymes involved in the catabolic breakdown of sugar, the process that creates ATP. In this way, when ATP is in abundant supply, the cell can prevent the production of ATP. On the other hand, ADP serves as a positive allosteric regulator (an allosteric activator) for some of the same enzymes that are inhibited by ATP. Thus, when relative levels of ADP are high compared to ATP, the cell is triggered to produce more ATP through sugar catabolism.
What are different types of energy and which one is most commonly used by animals to sustain and maintain life?
How would energy production and needs be affected by body size of an animal? What about its environment? (aquatic versus terrestrial) | libretexts | 2025-03-17T22:26:52.689738 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/5%3A_Digestion_and_Energy/5.1%3A_Biological_Energy",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "5.1: Biological Energy",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/5%3A_Digestion_and_Energy/5.2%3A_Components_of_the_Digestive_System | 5.2: Components of the Digestive System
- Compare and contrast digestive system structures in sample animal systems
Animals maintain work and energy demands on their body through nutrition. They obtain their nutrition from the consumption of other organisms. Depending on their diet, animals can be classified into the following categories: plant eaters (herbivores), meat eaters (carnivores), and those that eat both plants and animals (omnivores). The nutrients and macromolecules present in food are not immediately accessible to the cells. There are a number of processes that modify food within the animal body in order to make the nutrients and organic molecules accessible for cellular function. As animals evolved in the complexity of form and function, their digestive systems have also evolved to accommodate their various dietary needs.
Herbivores, omnivores, and carnivores
Herbivores are animals whose primary food source is plant-based. Examples of herbivores, as shown in Figure 5.11 include vertebrates like deer, koalas, and some bird species, as well as invertebrates such as crickets and caterpillars. These animals have evolved digestive systems capable of handling large amounts of plant material. Herbivores can be further classified into frugivores (fruit-eaters), granivores (seed eaters), nectivores (nectar feeders), and folivores (leaf eaters).
Carnivores are animals that eat other animals. The word carnivore is derived from Latin and literally means “meat eater.” Wild cats such as lions, shown in Figure 5.12a and tigers are examples of vertebrate carnivores, as are snakes and sharks, while invertebrate carnivores include sea stars, spiders, and ladybugs, shown in Figure 5.12b. Obligate carnivores are those that rely entirely on animal flesh to obtain their nutrients; examples of obligate carnivores are members of the cat family, such as lions and cheetahs. Facultative carnivores are those that also eat non-animal food in addition to animal food. Note that there is no clear line that differentiates facultative carnivores from omnivores; dogs would be considered facultative carnivores.
Omnivores are animals that eat both plant- and animal-derived food. In Latin, omnivore means to eat everything. Humans, bears (shown in Figure 5.13a), and chickens are examples of vertebrate omnivores; invertebrate omnivores include cockroaches and crayfish (shown in Figure 5.13b).
Invertebrate digestive systems
Animals have evolved different types of digestive systems to aid in the digestion of the different foods they consume. The simplest example is that of a gastrovascular cavity and is found in organisms with only one opening for digestion. Platyhelminthes (flatworms), Ctenophora (comb jellies), and Cnidaria (coral, jellyfish, and sea anemones) use this type of digestion. Gastrovascular cavities, as shown in Figure 5.14a, are typically a blind tube or cavity with only one opening, the “mouth”, which also serves as an “anus”. Ingested material enters the mouth and passes through a hollow, tubular cavity. Cells within the cavity secrete digestive enzymes that break down the food. The food particles are engulfed by the cells lining the gastrovascular cavity.
The alimentary canal , shown in Figure 5.14b, is a more advanced system: it consists of one tube with a mouth at one end and an anus at the other. Earthworms are an example of an animal with an alimentary canal. Once the food is ingested through the mouth, it passes through the esophagus and is stored in an organ called the crop; then it passes into the gizzard where it is churned and digested. From the gizzard, the food passes through the intestine, the nutrients are absorbed, and the waste is eliminated as feces, called castings, through the anus.
Vertebrate Digestive Systems
Vertebrates have evolved more complex digestive systems to adapt to their dietary needs. Some animals have a single stomach, while others have multi-chambered stomachs. Birds have developed a digestive system adapted to eating unmasticated food.
Monogastric: single-chambered stomach
As the word monogastric suggests, this type of digestive system consists of one (“mono”) stomach chamber (“gastric”). Humans and many animals have a monogastric digestive system as illustrated in Figure 5.15. The process of digestion begins with the mouth and the intake of food. The teeth play an important role in masticating (chewing) or physically breaking down food into smaller particles. The enzymes present in saliva also begin to chemically break down food. The esophagus is a long tube that connects the mouth to the stomach. Using peristalsis, or wave-like smooth muscle contractions, the muscles of the esophagus push the food towards the stomach. In order to speed up the actions of enzymes in the stomach, the stomach is an extremely acidic environment, with a pH between 1.5 and 2.5. The gastric juices, which include enzymes in the stomach, act on the food particles and continue the process of digestion. The further breakdown of food takes place in the small intestine where enzymes produced by the liver, the small intestine, and the pancreas continue the process of digestion. The nutrients are absorbed into the bloodstream across the epithelial cells lining the walls of the small intestines. The waste material travels on to the large intestine where water is absorbed and the drier waste material is compacted into feces; it is stored until it is excreted through the rectum.
Avian Digestive System
Birds face special challenges when it comes to obtaining nutrition from food. They do not have teeth and so their digestive system, shown in Figure 5.16 , must be able to process un-masticated food. Birds have evolved a variety of beak types that reflect the vast variety in their diet, ranging from seeds and insects to fruits and nuts. Because most birds fly, their metabolic rates are high in order to efficiently process food and keep their body weight low. The stomach of birds has two chambers: the proventriculus , where gastric juices are produced to digest the food before it enters the stomach, and the gizzard , where the food is stored, soaked, and mechanically ground. The undigested material forms food pellets that are sometimes regurgitated. Most of the chemical digestion and absorption happens in the intestine and the waste is excreted through the cloaca.
Parts of the digestive system
The vertebrate digestive system is designed to facilitate the transformation of food matter into the nutrient components that sustain organisms.
Oral cavity
The oral cavity, or mouth, is the point of entry of food into the digestive system, illustrated in Figure 5.17 . The food consumed is broken into smaller particles by mastication, the chewing action of the teeth. All mammals have teeth and can chew their food
The extensive chemical process of digestion begins in the mouth. As food is being chewed, saliva, produced by the salivary glands, mixes with the food. Saliva is a watery substance produced in the mouths of many animals. There are three major glands that secrete saliva—the parotid, the submandibular, and the sublingual. Saliva contains mucus that moistens food and buffers the pH of the food. Saliva also contains immunoglobulins and lysozymes, which have an antibacterial action to reduce tooth decay by inhibiting the growth of some bacteria. Saliva also contains an enzyme called salivary amylase that begins the process of converting starches in the food into a disaccharide called maltose. Another enzyme called lipase is produced by the cells in the tongue. Lipases are a class of enzymes that can break down triglycerides. The lingual lipase begins the breakdown of fat components in the food. The chewing and wetting action provided by the teeth and saliva prepare the food into a mass called the bolus for swallowing. The tongue helps in swallowing—moving the bolus from the mouth into the pharynx. The pharynx opens to two passageways: the trachea, which leads to the lungs, and the esophagus, which leads to the stomach. The trachea has an opening called the glottis, which is covered by a cartilaginous flap called the epiglottis. When swallowing, the epiglottis closes the glottis and food passes into the esophagus and not the trachea. This arrangement allows food to be kept out of the trachea.
Esophagus
The esophagus is a tubular organ that connects the mouth to the stomach. The chewed and softened food passes through the esophagus after being swallowed. The smooth muscles of the esophagus undergo a series of wave-like movements called peristalsis that push the food toward the stomach, as illustrated in Figure 5.18. The peristalsis wave is unidirectional—it moves food from the mouth to the stomach, and reverse movement is not possible. The peristaltic movement of the esophagus is an involuntary reflex; it takes place in response to the act of swallowing.
A ring-like muscle called a sphincter forms valves in the digestive system. The gastro-esophageal sphincter is located at the stomach end of the esophagus. In response to swallowing and the pressure exerted by the bolus of food, this sphincter opens, and the bolus enters the stomach. When there is no swallowing action, this sphincter is shut and prevents the contents of the stomach from traveling up the esophagus. Many animals have a true sphincter; however, in humans, there is no true sphincter, but the esophagus remains closed when there is no swallowing action. Acid reflux or “heartburn” occurs when the acidic digestive juices escape into the esophagus.
Stomach
A large part of digestion occurs in the stomach, shown in Figure 5.19. The stomach is a sac-like organ that secretes gastric digestive juices. The pH in the stomach is between 1.5 and 2.5. This highly acidic environment is required for the chemical breakdown of food and the extraction of nutrients. When empty, the stomach is a rather small organ; however, it can expand to up to 20 times its resting size when filled with food. This characteristic is particularly useful for animals that need to eat when food is available.
Which of the following statements about the digestive system is false?
- Chyme is a mixture of food and digestive juices that is produced in the stomach.
- Food enters the large intestine before the small intestine.
- In the small intestine, chyme mixes with bile, which emulsifies fats.
- The stomach is separated from the small intestine by the pyloric sphincter.
The stomach is also the major site for protein digestion in animals other than ruminants. Protein digestion is mediated by an enzyme called pepsin in the stomach chamber. Pepsin is secreted by the chief cells in the stomach in an inactive form called pepsinogen . Pepsin breaks peptide bonds and cleaves proteins into smaller polypeptides; it also helps activate more pepsinogen, starting a positive feedback mechanism that generates more pepsin. Another cell type—parietal cells—secrete hydrogen and chloride ions, which combine in the lumen to form hydrochloric acid, the primary acidic component of the stomach juices. Hydrochloric acid helps to convert the inactive pepsinogen to pepsin. The highly acidic environment also kills many microorganisms in the food and, combined with the action of the enzyme pepsin, results in the hydrolysis of protein in the food. Chemical digestion is facilitated by the churning action of the stomach. Contraction and relaxation of smooth muscles mix the stomach contents about every 20 minutes. The partially digested food and gastric juice mixture are called chyme . Chyme passes from the stomach to the small intestine. Further protein digestion takes place in the small intestine. Gastric emptying occurs within two to six hours after a meal. Only a small amount of chyme is released into the small intestine at a time. The movement of chyme from the stomach into the small intestine is regulated by the pyloric sphincter.
When digesting protein and some fats, the stomach lining must be protected from getting digested by pepsin. There are two points to consider when describing how the stomach lining is protected. First, as previously mentioned, the enzyme pepsin is synthesized in the inactive form. This protects the chief cells because pepsinogen does not have the same enzyme functionality of pepsin. Second, the stomach has a thick mucus lining that protects the underlying tissue from the action of the digestive juices. When this mucus lining is ruptured, ulcers can form in the stomach. Ulcers are open wounds in or on an organ caused by bacteria ( Helicobacter pylori ) when the mucus lining is ruptured and fails to reform.
Small intestine
Chyme moves from the stomach to the small intestine. The small intestine is the organ where the digestion of protein, fats, and carbohydrates is completed. The small intestine is a long tube-like organ with a highly folded surface containing finger-like projections called the villi . The apical surface of each villus has many microscopic projections called microvilli. These structures, illustrated in Figure 5.20, are lined with epithelial cells on the luminal side and allow for the nutrients to be absorbed from the digested food and absorbed into the blood stream on the other side. The villi and microvilli, with their many folds, increase the surface area of the intestine and increase absorption efficiency of the nutrients. Absorbed nutrients in the blood are carried into the hepatic portal vein, which leads to the liver. There, the liver regulates the distribution of nutrients to the rest of the body and removes toxic substances, including drugs, alcohol, and some pathogens.
Which of the following statements about the small intestine is false?
- Absorptive cells that line the small intestine have microvilli, small projections that increase surface area and aid in the absorption of food.
- The inside of the small intestine has many folds, called villi.
- Microvilli are lined with blood vessels as well as lymphatic vessels.
- The inside of the small intestine is called the lumen.
The human small intestine is over 6m long and is divided into three parts: the duodenum, the jejunum, and the ileum. The “C-shaped,” fixed part of the small intestine is called the duodenum and is shown in Figure 5.19. The duodenum is separated from the stomach by the pyloric sphincter which opens to allow chyme to move from the stomach to the duodenum. In the duodenum, chyme is mixed with pancreatic juices in an alkaline solution rich in bicarbonate that neutralizes the acidity of chyme and acts as a buffer. Pancreatic juices also contain several digestive enzymes. Digestive juices from the pancreas, liver, and gallbladder, as well as from gland cells of the intestinal wall itself, enter the duodenum. Bile is produced in the liver and stored and concentrated in the gallbladder. Bile contains bile salts which emulsify lipids while the pancreas produces enzymes that catabolize starches, disaccharides, proteins, and fats. These digestive juices break down the food particles in the chyme into glucose, triglycerides, and amino acids. Some chemical digestion of food takes place in the duodenum. Absorption of fatty acids also takes place in the duodenum.
The second part of the small intestine is called the jejunum , shown in Figure 5.19. Here, hydrolysis of nutrients is continued while most of the carbohydrates and amino acids are absorbed through the intestinal lining. The bulk of chemical digestion and nutrient absorption occurs in the jejunum.
The ileum , also illustrated in Figure 5.19 is the last part of the small intestine and here the bile salts and vitamins are absorbed into bloodstream. The undigested food is sent to the colon from the ileum via peristaltic movements of the muscle. The ileum ends and the large intestine begins at the ileocecal valve. The vermiform, “worm-like,” appendix is located at the ileocecal valve. The appendix of humans secretes no enzymes and has an insignificant role in immunity.
Large intestine
The large intestine , illustrated in Figure 5.21, reabsorbs the water from the undigested food material and processes the waste material. The human large intestine is much smaller in length compared to the small intestine but larger in diameter. It has three parts: the cecum, the colon, and the rectum. The cecum joins the ileum to the colon and is the receiving pouch for the waste matter. The colon is home to many bacteria or “intestinal flora” that aid in the digestive processes. The colon can be divided into four regions, the ascending colon, the transverse colon, the descending colon and the sigmoid colon. The main functions of the colon are to extract the water and mineral salts from undigested food, and to store waste material. Carnivorous mammals have a shorter large intestine compared to herbivorous mammals due to their diet.
Rectum and anus
The rectum is the terminal end of the large intestine, as shown in Figure 5.21. The primary role of the rectum is to store the feces until defecation. The feces are propelled using peristaltic movements during elimination. The anus is an opening at the far-end of the digestive tract and is the exit point for the waste material. Two sphincters between the rectum and anus control elimination: the inner sphincter is involuntary and the outer sphincter is voluntary.
Accessory organs
The organs discussed above are the organs of the digestive tract through which food passes. Accessory organs are organs that add secretions (enzymes) that catabolize food into nutrients. Accessory organs include salivary glands, the liver, the pancreas, and the gallbladder. The liver, pancreas, and gallbladder are regulated by hormones in response to the food consumed.
The liver is the largest internal organ in humans and it plays a very important role in the digestion of fats and detoxifying blood. The liver produces bile, a digestive juice that is required for the breakdown of fatty components of the food in the duodenum. The liver also processes the vitamins and fats and synthesizes many plasma proteins.
The pancreas is another important gland that secretes digestive juices. The chyme produced from the stomach is highly acidic in nature; the pancreatic juices contain high levels of bicarbonate, an alkali that neutralizes the acidic chyme. Additionally, the pancreatic juices contain a large variety of enzymes that are required for the digestion of protein and carbohydrates.
The gallbladder is a small organ that aids the liver by storing bile and concentrating bile salts. When chyme containing fatty acids enters the duodenum, the bile is secreted from the gallbladder into the duodenum.
Compare and contrast components of the digestive system among nematodes, rabbit, and chicken.
If a structure contains thin epithelium, hydrogen pump, and many folds, what could it be good for during digestion? Explain your reasoning.
How do plants obtain nutrients? | libretexts | 2025-03-17T22:26:52.777480 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/5%3A_Digestion_and_Energy/5.2%3A_Components_of_the_Digestive_System",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "5.2: Components of the Digestive System",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/5%3A_Digestion_and_Energy/5.3%3A_Nutrient_Transport_and_Energy_Metabolism | 5.3: Nutrient Transport and Energy Metabolism
- Summarize chemical digestion of carbohydrates, proteins and fats and energy metabolism.
Food energy and ATP
Animals need food to obtain energy and maintain homeostasis. Homeostasis is the ability of a system to maintain a stable internal environment even in the face of external changes to the environment. For example, the normal body temperature of humans is 37°C (98.6°F). Humans maintain this temperature even when the external temperature is hot or cold. It takes energy to maintain this body temperature, and animals obtain this energy from food.
The primary source of energy for animals is carbohydrates, mainly glucose. Glucose is called the body’s fuel. The digestible carbohydrates in an animal’s diet are converted to glucose molecules through a series of catabolic chemical reactions.
Adenosine triphosphate, or ATP, is the primary energy currency in cells; ATP stores energy in phosphate ester bonds. ATP releases energy when the phosphodiester bonds are broken and ATP is converted to ADP and a phosphate group. ATP is produced by the oxidative reactions in the cytoplasm and mitochondrion of the cell, where carbohydrates, proteins, and fats undergo a series of metabolic reactions collectively called cellular respiration. For example, glycolysis is a series of reactions in which glucose is converted to pyruvic acid and some of its chemical potential energy is transferred to NADH and ATP.
ATP is required for all cellular functions. It is used to build the organic molecules that are required for cells and tissues; it provides energy for muscle contraction and for the transmission of electrical signals in the nervous system. When the amount of ATP is available in excess of the body’s requirements, the liver uses the excess ATP and excess glucose to produce molecules called glycogen. Glycogen is a polymeric form of glucose and is stored in the liver and skeletal muscle cells. When blood sugar drops, the liver releases glucose from stores of glycogen. Skeletal muscle converts glycogen to glucose during intense exercise. The process of converting glucose and excess ATP to glycogen and the storage of excess energy is an evolutionarily important step in helping animals deal with mobility, food shortages, and famine.
Digestive system processes
Obtaining nutrition and energy from food is a multi-step process. For true animals, the first step is ingestion, the act of taking in food. This is followed by digestion, absorption, and elimination. In the following sections, each of these steps will be discussed in detail.
Ingestion
The large molecules found in intact food cannot pass through the cell membranes. Food needs to be broken into smaller particles so that animals can harness the nutrients and organic molecules. The first step in this process is ingestion . Ingestion is the process of taking in food through the mouth. In vertebrates, the teeth, saliva, and tongue play important roles in mastication (preparing the food into bolus). While the food is being mechanically broken down, the enzymes in saliva begin to chemically process the food as well. The combined action of these processes modifies the food from large particles to a soft mass that can be swallowed and can travel the length of the esophagus.
Digestion and absorption
Digestion is the mechanical and chemical break down of food into small organic fragments. It is important to break down macromolecules into smaller fragments that are of suitable size for absorption across the digestive epithelium. Large, complex molecules of proteins, polysaccharides, and lipids must be reduced to simpler particles such as simple sugar before they can be absorbed by the digestive epithelial cells. Different organs play specific roles in the digestive process. The animal diet needs carbohydrates, protein, and fat, as well as vitamins and inorganic components for nutritional balance. How each of these components is digested is discussed in the following sections.
Carbohydrates
The digestion of carbohydrates begins in the mouth. The salivary enzyme amylase begins the breakdown of food starches into maltose, a disaccharide. As the bolus of food travels through the esophagus to the stomach, no significant digestion of carbohydrates takes place. The esophagus produces no digestive enzymes but does produce mucous for lubrication. The acidic environment in the stomach stops the action of the amylase enzyme.
The next step of carbohydrate digestion takes place in the duodenum. Recall that the chyme from the stomach enters the duodenum and mixes with the digestive secretion from the pancreas, liver, and gallbladder. Pancreatic juices also contain amylase, which continues the breakdown of starch and glycogen into maltose, a disaccharide. The disaccharides are broken down into monosaccharides by enzymes called maltases, sucrases, and lactases , which are also present in the brush border of the small intestinal wall. Maltase breaks down maltose into glucose. Other disaccharides, such as sucrose and lactose are broken down by sucrase and lactase, respectively. Sucrase breaks down sucrose (or “table sugar”) into glucose and fructose, and lactase breaks down lactose (or “milk sugar”) into glucose and galactose. The monosaccharides (glucose) thus produced are absorbed and then can be used in metabolic pathways to harness energy. The monosaccharides are transported across the intestinal epithelium into the bloodstream to be transported to the different cells in the body. The steps in carbohydrate digestion are summarized in Figure 5.22 and Table 5.1.
| Enzyme | Produced By | Site of Action | Substrate Acting On | End Products |
|---|---|---|---|---|
| Salivary amylase | Salivary glands | Mouth | Polysaccharides (Starch) | Disaccharides (maltose), oligosaccharides |
| Pancreatic amylase | Pancreas | Small intestine | Polysaccharides (starch) | Disaccharides (maltose), monosaccharides |
| Oligosaccharidases | Lining of the intestine; brush border membrane | Small intestine | Disaccharides | Monosaccharides (e.g., glucose, fructose, galactose) |
Protein
A large part of protein digestion takes place in the stomach. The enzyme pepsin plays an important role in the digestion of proteins by breaking down the intact protein to peptides, which are short chains of four to nine amino acids. In the duodenum, other enzymes— trypsin, elastase , and chymotrypsin —act on the peptides reducing them to smaller peptides. Trypsin elastase, carboxypeptidase, and chymotrypsin are produced by the pancreas and released into the duodenum where they act on the chyme. The further breakdown of peptides to single amino acids is aided by enzymes called peptidases (those that break down peptides). Specifically, carboxypeptidase, dipeptidase , and aminopeptidase play important roles in reducing the peptides to free amino acids. The amino acids are absorbed into the bloodstream through the small intestines. The steps in protein digestion are summarized in Figure 5.23 and Table 5.2.
| Enzyme | Produced By | Site of Action | Substrate Acting On | End Products |
|---|---|---|---|---|
| Pepsin | Stomach chief cells | Stomach | Proteins | Peptides |
|
Pancreas | Small intestine | Proteins | Peptides |
| Carboxypeptidase | Pancreas | Small intestine | Peptides | Amino acids and peptides |
|
Lining of intestine | Small intestine | Peptides | Amino acids |
Lipids
Lipid digestion begins in the stomach with the aid of lingual lipase and gastric lipase. However, the bulk of lipid digestion occurs in the small intestine due to pancreatic lipase. When chyme enters the duodenum, the hormonal responses trigger the release of bile, which is produced in the liver and stored in the gallbladder. Bile aids in the digestion of lipids, primarily triglycerides by emulsification. Emulsification is a process in which large lipid globules are broken down into several small lipid globules. These small globules are more widely distributed in the chyme rather than forming large aggregates. Lipids are hydrophobic substances: in the presence of water, they will aggregate to form globules to minimize exposure to water. Bile contains bile salts, which are amphipathic, meaning they contain hydrophobic and hydrophilic parts. Thus, the bile salts hydrophilic side can interface with water on one side and the hydrophobic side interfaces with lipids on the other. By doing so, bile salts emulsify large lipid globules into small lipid globules.
Why is emulsification important for digestion of lipids? Pancreatic juices contain enzymes called lipases (enzymes that break down lipids). If the lipid in the chyme aggregates into large globules, very little surface area of the lipids is available for the lipases to act on, leaving lipid digestion incomplete. By forming an emulsion, bile salts increase the available surface area of the lipids many folds. The pancreatic lipases can then act on the lipids more efficiently and digest them, as detailed in Figure 5.24. Lipases break down the lipids into fatty acids and glycerides. These molecules can pass through the plasma membrane of the cell and enter the epithelial cells of the intestinal lining. The bile salts surround long-chain fatty acids and monoglycerides forming tiny spheres called micelles. The micelles move into the brush border of the small intestine absorptive cells where the long-chain fatty acids and monoglycerides diffuse out of the micelles into the absorptive cells leaving the micelles behind in the chyme. The long-chain fatty acids and monoglycerides recombine in the absorptive cells to form triglycerides, which aggregate into globules and become coated with proteins. These large spheres are called chylomicrons . Chylomicrons contain triglycerides, cholesterol, and other lipids and have proteins on their surface. The surface is also composed of the hydrophilic phosphate “heads” of phospholipids. Together, they enable the chylomicron to move in an aqueous environment without exposing the lipids to water. Chylomicrons leave the absorptive cells via exocytosis. Chylomicrons enter the lymphatic vessels and then enter the blood in the subclavian vein.
Vitamins
Vitamins can be either water-soluble or lipid-soluble. Fat-soluble vitamins are absorbed in the same manner as lipids. It is important to consume some amount of dietary lipid to aid the absorption of lipid-soluble vitamins. Water-soluble vitamins can be directly absorbed into the bloodstream from the intestine.
- Review Figure 5.25 on your own.
Figure 5.25. Mechanical and chemical digestion of food takes place in many steps, beginning in the mouth and ending in the rectum.
Which of the following statements about digestive processes is true?
a. Amylase, maltase, and lactase in the mouth digest carbohydrates.
b. Trypsin and lipase in the stomach digest protein.
c. Bile emulsifies lipids in the small intestine.
d. No food is absorbed until the small intestine
Elimination
The final step in digestion is the elimination of undigested food content and waste products. The undigested food material enters the colon, where most of the water is reabsorbed. Recall that the colon is also home to the microflora called “intestinal flora” that aid in the digestion process. The semi-solid waste is moved through the colon by peristaltic movements of the muscle and is stored in the rectum. As the rectum expands in response to storage of fecal matter, it triggers the neural signals required to set up the urge to eliminate. The solid waste is eliminated through the anus using peristaltic movements of the rectum.
Diarrhea and constipation are some of the most common health concerns that affect digestion. Constipation is a condition where the feces are hardened because of excess water removal in the colon. In contrast, if enough water is not removed from the feces, it results in diarrhea. Many bacteria, including the ones that cause cholera, affect the proteins involved in water reabsorption in the colon and result in excessive diarrhea.
Emesis, or vomiting, is the elimination of food by forceful expulsion through the mouth. It is often in response to an irritant that affects the digestive tract, including but not limited to viruses, bacteria, emotions, sights, and food poisoning. This forceful expulsion of the food is due to the strong contractions produced by the stomach muscles. The process of emesis is regulated by the medulla.
Think about the digestion lab you had earlier this week. Summarize digestion of lipids, carbohydrates, and proteins in the digestive system based on your lab observations and results with your lab partners.
- Not all animals process different nutrients equally. Here is is a video about hummingbird digestion with a focus on work by Dr. Welch at the University of Toronto Scarborough.
Why and how are hummingbirds and humans different in carbohydrate digestion? | libretexts | 2025-03-17T22:26:52.860481 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/5%3A_Digestion_and_Energy/5.3%3A_Nutrient_Transport_and_Energy_Metabolism",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "5.3: Nutrient Transport and Energy Metabolism",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/6%3A_Locomotion | Skip to main content
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6.1: Fuels for Locomotion
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6.2: Types of Muscles
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6.3: Types of Locomotion
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6.4: Muscle Contraction
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For a muscle cell to contract, the sarcomere must shorten. However, thick and thin filaments—the components of sarcomeres—do not shorten. Instead, they slide by one another, causing the sarcomere to shorten while the filaments remain the same length. The sliding filament theory of muscle contraction was developed to fit the differences observed in the named bands on the sarcomere at different degrees of muscle contraction and relaxation. | libretexts | 2025-03-17T22:26:52.918587 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/6%3A_Locomotion",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "6: Locomotion",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/6%3A_Locomotion/6.1%3A_Fuels_for_Locomotion | 6.1: Fuels for Locomotion
- Compare carbohydrate absorption in hummingbirds to absorption of this and other nutrients.
Nutrition and energy production
Given the diversity of animal life on our planet, it is not surprising that the animal diet would also vary substantially. The animal diet is the source of materials needed for building DNA and other complex molecules needed for growth, maintenance, reproduction; collectively these processes are called biosynthesis. The diet is also the source of materials for ATP production in the cells, which fuels animal locomotion among many other processes. The diet must be balanced to provide the minerals and vitamins that are required for cellular function.
Food requirements
What are the fundamental requirements of the animal diet? The animal diet should be well balanced and provide nutrients required for bodily function and the minerals and vitamins required for maintaining structure and regulation necessary for good health and reproductive capability. These requirements for a human are illustrated graphically in Figure 6.1.
- The first step in ensuring that you are meeting the food requirements of your body is an awareness of the food groups and the nutrients they provide.
Organic precursors
The organic molecules required for building cellular material and tissues must come from food. Carbohydrates or sugars are the primary sources of organic carbons in the animal body. During digestion, digestible carbohydrates are ultimately broken down into glucose and used to provide energy through metabolic pathways. Complex carbohydrates, including polysaccharides, can be broken down into glucose through biochemical modification; however, humans do not produce the enzyme cellulase and lack the ability to derive glucose from the polysaccharide cellulose. In humans, these molecules provide the fiber required for moving waste through the large intestine and a healthy colon. The intestinal flora in the human gut are able to extract some nutrition from these plant fibers. The excess sugars in the body are converted into glycogen and stored in the liver and muscles for later use. Glycogen stores are used to fuel prolonged exertions, such as long-distance running, and to provide energy during a food shortage. Excess glycogen can be converted to fats, which are stored in the lower layer of the skin of mammals for insulation and energy storage. Excess digestible carbohydrates are stored by mammals in order to survive famine and aid in mobility.
Another important requirement is that of nitrogen. Protein catabolism provides a source of organic nitrogen. Amino acids are the building blocks of proteins and protein breakdown provides amino acids that are used for cellular function. The carbon and nitrogen derived from these become the building block for nucleotides, nucleic acids, proteins, cells, and tissues. Excess nitrogen must be excreted as it is toxic. Fats add flavor to food and promote a sense of satiety or fullness. Fatty foods are also significant sources of energy because one gram of fat contains nine calories. Fats are required in the diet to aid the absorption of fat-soluble vitamins and the production of fat-soluble hormones.
Essential nutrients
While the animal body can synthesize many of the molecules required for the function from the organic precursors, there are some nutrients that need to be consumed from food. These nutrients are termed essential nutrients , meaning they must be eaten, and the body cannot produce them.
The omega-3 alpha-linolenic acid and the omega-6 linoleic acid are essential fatty acids needed to make some membrane phospholipids. Vitamins are another class of essential organic molecules that are required in small quantities for many enzymes to function and, for this reason, are considered to be co-enzymes. Vitamins can be either water-soluble or lipid-soluble. Fat-soluble vitamins are absorbed in the same manner as lipids. It is important to consume some amount of dietary lipid to aid the absorption of lipid-soluble vitamins. Water-soluble vitamins can be directly absorbed into the bloodstream from the intestine. Absence or low levels of vitamins can have a dramatic effect on health. Both fat-soluble and water-soluble vitamins must be obtained from food. Minerals are inorganic essential nutrients that must be obtained from food. Among their many functions, minerals help in structure and regulation and are considered co-factors. Certain amino acids also must be procured from food and cannot be synthesized by the body. These amino acids are the “essential” amino acids. The human body can synthesize only 11 of the 20 required amino acids; the rest must be obtained from food. The essential amino acids are listed in Table 6.1.
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You should watch these videos to learn more about water-soluble vitamins, fat-soluble vitamins, and minerals.
| Amino acids that must be consumed | Amino acids anabolized by the body |
|---|---|
| isoleucine | alanine |
| leucine | selenocysteine |
| lysine | aspartate |
| methionine | cysteine |
| phenylalanine | glutamate |
| tryptophan | glycine |
| valine | proline |
| histidine* | serine |
| threonine | tyrosine |
| arginine* | asparagine |
| *The human body can synthesize histidine and arginine, but not in the quantities required, especially for growing children. |
Summarize roles different vitamins, minerals and essential amino acids play in animal nutrition and energy production. | libretexts | 2025-03-17T22:26:52.987378 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/6%3A_Locomotion/6.1%3A_Fuels_for_Locomotion",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "6.1: Fuels for Locomotion",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/6%3A_Locomotion/6.2%3A_Types_of_Muscles | 6.2: Types of Muscles
- List different types of muscles and briefly explain how they generate locomotion.
Types of muscle tissue
Nutrients that animals obtained are used for ATP production, which are fuels for the production of work by muscles, among other physiological processes in the animal bodies. Depending on what type of muscle tissues animals are using, they will use ATP differently to generate work. Overall, muscle cells are specialized for contraction. Muscles allow for motions such as walking, and they also facilitate bodily processes such as respiration and digestion. The body contains three types of muscle tissue: skeletal muscle, cardiac muscle, and smooth muscle (Figure 6.2).
Skeletal muscle tissue forms skeletal muscles, which attach to bones or skin and control locomotion and any movement that can be consciously controlled. Because it can be controlled by thought, skeletal muscle is also called voluntary muscle. Skeletal muscles are long and cylindrical in appearance; when viewed under a microscope, skeletal muscle tissue has a striped or striated appearance. The striations are caused by the regular arrangement of contractile proteins (actin and myosin). Actin is a globular contractile protein that interacts with myosin for muscle contraction. Skeletal muscle also has multiple nuclei present in a single cell.
Smooth muscle tissue occurs in the walls of hollow organs such as the intestines, stomach, and urinary bladder, and around passages such as the respiratory tract and blood vessels. Smooth muscle has no striations, is not under voluntary control, has only one nucleus per cell, is tapered at both ends, and is called involuntary muscle.
Cardiac muscle tissue is only found in the heart, and cardiac contractions pump blood throughout the body and maintain blood pressure. Like skeletal muscle, cardiac muscle is striated, but unlike skeletal muscle, cardiac muscle cannot be consciously controlled and is called involuntary muscle. It has one nucleus per cell, is branched, and is distinguished by the presence of intercalated disks.
Skeletal muscle fiber structure
Each skeletal muscle fiber is a skeletal muscle cell. These cells are incredibly large, with diameters of up to 100 µm and lengths of up to 30 cm. The plasma membrane of a skeletal muscle fiber is called the sarcolemma . The sarcolemma is the site of action potential conduction, which triggers muscle contraction. Within each muscle fiber are myofibrils —long cylindrical structures that lie parallel to the muscle fiber. Myofibrils run the entire length of the muscle fiber, and because they are only approximately 1.2 µm in diameter, hundreds to thousands can be found inside one muscle fiber. They attach to the sarcolemma at their ends, so that as myofibrils shorten, the entire muscle cell contracts ( Figure 6.3 ).
The striated appearance of skeletal muscle tissue is a result of repeating bands of the proteins actin and myosin that are present along the length of myofibrils. Dark A bands and light I bands repeat along myofibrils, and the alignment of myofibrils in the cell causes the entire cell to appear striated or banded.
Each I band has a dense line running vertically through the middle called a Z disc or Z line. The Z discs mark the border of units called sarcomeres , which are the functional units of skeletal muscle. One sarcomere is the space between two consecutive Z discs and contains one entire A band and two halves of an I band, one on either side of the A band. A myofibril is composed of many sarcomeres running along its length, and as the sarcomeres individually contract, the myofibrils and muscle cells shorten (Figure 6.4).
Myofibrils are composed of smaller structures called myofilaments . There are two main types of filaments: thick filaments and thin filaments; each has different compositions and locations. Thick filaments occur only in the A band of a myofibril. Thin filaments attach to a protein in the Z disc called alpha-actinin and occur across the entire length of the I band and partway into the A band. The region at which thick and thin filaments overlap has a dense appearance, as there is little space between the filaments. Thin filaments do not extend all the way into the A bands, leaving a central region of the A band that only contains thick filaments. This central region of the A band looks slightly lighter than the rest of the A band and is called the H zone. The middle of the H zone has a vertical line called the M line, at which accessory proteins hold together thick filaments. Both the Z disc and the M line hold myofilaments in place to maintain the structural arrangement and layering of the myofibril. Myofibrils are connected to each other by intermediate, or desmin, filaments that attach to the Z disc.
Thick and thin filaments are themselves composed of proteins. Thick filaments are composed of the protein myosin. The tail of a myosin molecule connects with other myosin molecules to form the central region of a thick filament near the M line, whereas the heads align on either side of the thick filament where the thin filaments overlap. The primary component of thin filaments is the actin protein. Two other components of the thin filament are tropomyosin and troponin. Actin has binding sites for myosin attachment. Strands of tropomyosin block the binding sites and prevent actin-myosin interactions when the muscles are at rest. Troponin consists of three globular subunits. One subunit binds to tropomyosin, one subunit binds to actin, and one subunit binds Ca2+ ions.
- View this animation showing the organization of muscle fibers.
In addition to three different types of muscles discussed above, there are also differences in skeletal muscle function. Different skeletal muscles in animals are described as white and red muscle. These different types of skeletal muscles are recruited depending on whether a fast and short versus steady and prolonged locomotion is needed by the animal.
Which is more efficient – red or white muscle? Why? | libretexts | 2025-03-17T22:26:53.049471 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/6%3A_Locomotion/6.2%3A_Types_of_Muscles",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "6.2: Types of Muscles",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/6%3A_Locomotion/6.3%3A_Types_of_Locomotion | 6.3: Types of Locomotion
- Explain different types of locomotion with reference to environmental limitations and/or hard structure support for muscle function
The muscular and skeletal systems provide support to the body and allow for movement. The bones of the skeleton protect the body’s internal organs and support the weight of the body. The muscles of the muscular system contract and pull on the bones, allowing for movements as diverse as standing, walking, running, and grasping items. We will focus on the connection between human skeletal and muscular system to illustrate how these two organ systems coordinate to generate locomotion.
Human Skeletal System
The human skeleton is an endoskeleton that consists of 206 bones in the adult. An endoskeleton develops within the body rather than outside like the exoskeleton of insects. The skeleton has five main functions: providing support to the body, storing minerals and lipids, producing blood cells, protecting internal organs, and allowing for movement. The skeletal system in vertebrates is divided into the axial skeleton (which consists of the skull, vertebral column, and rib cage), and the appendicular skeleton (which consists of limb bones, the pectoral or shoulder girdle, and the pelvic girdle).
Explore the human skeleton by viewing the following video with digital 3D sculpturing.
The axial skeleton forms the central axis of the body and includes the bones of the skull, ossicles of the middle ear, hyoid bone of the throat, vertebral column, and the thoracic cage (rib cage) (Figure 6.5).
The bones of the skull support the structures of the face and protect the brain. The skull consists of cranial bones and facial bones. The cranial bones form the cranial cavity, which encloses the brain and serves as an attachment site for muscles of the head and neck. In the adult, they are tightly jointed with connective tissue and adjoining bones do not move.
The auditory ossicles of the middle ear transmit sounds from the air as vibrations to the fluid-filled cochlea. The auditory ossicles consist of two malleus (hammer) bones, two incus (anvil) bones, and two stapes (stirrups), one on each side. Facial bones provide cavities for the sense organs (eyes, mouth, and nose), and serve as attachment points for facial muscles.
The hyoid bone lies below the mandible in the front of the neck. It acts as a movable base for the tongue and is connected to muscles of the jaw, larynx, and tongue. The mandible forms a joint with the base of the skull. The mandible controls the opening to the mouth and hence, the airway and gut.
The vertebral column, or spinal column, surrounds and protects the spinal cord, supports the head, and acts as an attachment point for ribs and muscles of the back and neck. It consists of 26 bones: the 24 vertebrae, the sacrum, and the coccyx. Each vertebral body has a large hole in the center through which the spinal cord passes down to the level of the first lumbar vertebra. Below this level, the hole contains spinal nerves which exit between the vertebrae. There is a notch on each side of the hole through which the spinal nerves, can exit from the spinal cord to serve different regions of the body. The vertebral column is approximately 70 cm (28 in) in adults and is curved, which can be seen from a side view.
Intervertebral discs composed of fibrous cartilage lie between adjacent vertebrae from the second cervical vertebra to the sacrum. Each disc helps form a slightly moveable joint and acts as a cushion to absorb shocks from movements such as walking and running.
The thoracic cage, also known as the rib cage consists of the ribs, sternum, thoracic vertebrae, and costal cartilages. The thoracic cage encloses and protects the organs of the thoracic cavity including the heart and lungs. It also provides support for the shoulder girdles and upper limbs and serves as the attachment point for the diaphragm, muscles of the back, chest, neck, and shoulders. Changes in the volume of the thorax enable breathing. The sternum, or breastbone, is a long flat bone located at the anterior of the chest. Like the skull, it is formed from many bones in the embryo, which fuses in the adult. The ribs are 12 pairs of long curved bones that attach to the thoracic vertebrae and curve toward the front of the body, forming the ribcage. Costal cartilages connect the anterior ends of most ribs to the sternum.
The appendicular skeleton is composed of the bones of the upper and lower limbs. It also includes the pectoral, or shoulder girdle, which attaches the upper limbs to the body, and the pelvic girdle, which attaches the lower limbs to the body (Figure 6.5).
The pectoral girdle bones transfer force generated by muscles acting on the upper limb to the thorax. It consists of the clavicles (or collarbones) in the anterior, and the scapulae (or shoulder blades) in the posterior.
The upper limb contains bones of the arm (shoulder to elbow), the forearm, and the hand. The humerus is the largest and longest bone of the upper limb. It forms a joint with the shoulder and with the forearm at the elbow. The forearm extends from the elbow to the wrist and consists of two bones. The hand includes the bones of the wrist, the palm, and the bones of the fingers.
The pelvic girdle attaches to the lower limbs of the axial skeleton. Since it is responsible for bearing the weight of the body and for locomotion, the pelvic girdle is securely attached to the axial skeleton by strong ligaments. It also has deep sockets with robust ligaments that securely attach to the femur. The pelvic girdle is mainly composed of two large hip bones. The hip bones join together in the anterior of the body at a joint called the pubic symphysis and with the bones of the sacrum at the posterior of the body.
The lower limb consists of the thigh, the leg, and the foot. The bones of the lower limbs are thicker and stronger than the bones of the upper limbs to support the entire weight of the body and the forces from locomotion. The femur, or thighbone, is the longest, heaviest, and strongest bone in the body. The femur and pelvis form the hip joint. At its other end, the femur, along with the shinbone and kneecap, form the knee joint.
Joints and skeletal movement
The point at which two or more bones meet is called a joint, or articulation. Joints are responsible for movement, such as the movement of limbs, and stability, such as the stability found in the bones of the skull.
There are two ways to classify joints: based on their structure or based on their function. The structural classification divides joints into fibrous, cartilaginous, and synovial joints depending on the material composing the joint and the presence or absence of a cavity in the joint. The bones of fibrous joints are held together by fibrous connective tissue. There is no cavity, or space, present between the bones, so most fibrous joints do not move at all, or are only capable of minor movements. The joints between the bones in the skull and between the teeth and the bone of their sockets are examples of fibrous joints (Figure 6.6a).
Cartilaginous joints are joints in which the bones are connected by cartilage (Figure 6.6b). An example is found at the joints between vertebrae, the so-called “disks” of the backbone. Cartilaginous joints allow for very little movement.
Synovial joints are the only joints that have a space between the adjoining bones (Figure 6.6c). This space is referred to as the joint cavity and is filled with fluid. The fluid lubricates the joint, reducing friction between the bones and allowing for greater movement. The ends of the bones are covered with cartilage and the entire joint is surrounded by a capsule. Synovial joints are capable of the greatest movement of the joint types. Knees, elbows, and shoulders are examples of synovial joints.
The wide range of movement allowed by synovial joints produces different types of movements. Angular movements are produced when the angle between the bones of a joint changes. Flexion, or bending, occurs when the angle between the bones decreases. Moving the forearm upward at the elbow is an example of flexion. Extension is the opposite of flexion in that the angle between the bones of a joint increases. Rotational movement is the movement of a bone as it rotates around its own longitudinal axis. Movement of the head as in saying “no” is an example of rotation.
Animal Locomotion
Humans are limited to terrestrial locomotion on two limbs or swimming with all four limbs. Other animals explore both the aquatic and aerial realm more extensively. For example, birds are excellent swimmers and fliers. However, some birds are not good at moving in both air and water. One example of birds with limited abilities in flight are penguins and another are members of an extinct group Hesperirnithformes, who, just like penguins, were not able to swim in water and only had limited locomotion on land.
Watch the video about swimming locomotion in extinct Hesperornithiformes to help you understand how skeleton and muscles coordinate to generate locomotion and why some animals may be more limited in locomotion in one type of environment.
Which mode of locomotion do you think is least efficient in terms of total energy expenditure if you are considering animals of different size? | libretexts | 2025-03-17T22:26:53.114381 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/6%3A_Locomotion/6.3%3A_Types_of_Locomotion",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "6.3: Types of Locomotion",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/6%3A_Locomotion/6.4%3A_Muscle_Contraction | 6.4: Muscle Contraction
- Describe steps involved in muscle contraction
Sliding Filament Model of Contraction
For a muscle cell to contract, the sarcomere must shorten. However, thick and thin filaments—the components of sarcomeres—do not shorten. Instead, they slide by one another, causing the sarcomere to shorten while the filaments remain the same length. The sliding filament theory of muscle contraction was developed to fit the differences observed in the named bands on the sarcomere at different degrees of muscle contraction and relaxation. The mechanism of contraction is the binding of myosin to actin, forming cross-bridges that generate filament movement (Figure 6.7).
When a sarcomere shortens, some regions shorten whereas others stay the same length. A sarcomere is defined as the distance between two consecutive Z discs or Z lines; when a muscle contracts, the distance between the Z discs is reduced. The H zone—the central region of the A zone—contains only thick filaments and is shortened during contraction. The I band contains only thin filaments and also shortens. The A band does not shorten—it remains the same length—but A bands of different sarcomeres move closer together during contraction, eventually disappearing. Thin filaments are pulled by the thick filaments toward the center of the sarcomere until the Z discs approach the thick filaments. The zone of overlap, in which thin filaments and thick filaments occupy the same area, increases as the thin filaments move inward.
ATP and Muscle Contraction
The motion of muscle shortening occurs as myosin heads bind to actin and pull the actin inwards. This action requires energy, which is provided by ATP. Myosin binds to actin at a binding site on the globular actin protein. Myosin has another binding site for ATP at which enzymatic activity hydrolyzes ATP to ADP, releasing an inorganic phosphate molecule and energy.
ATP binding causes myosin to release actin, allowing actin and myosin to detach from each other. After this happens, the newly bound ATP is converted to ADP and inorganic phosphate, Pi. The enzyme at the binding site on myosin is called ATPase. The energy released during ATP hydrolysis changes the angle of the myosin head into a “cocked” position. The myosin head is then in a position for further movement, possessing potential energy, but ADP and Pi are still attached. If actin binding sites are covered and unavailable, the myosin will remain in the high energy configuration with ATP hydrolyzed but still attached.
If the actin binding sites are uncovered, a cross-bridge will form; that is, the myosin head spans the distance between the actin and myosin molecules. Pi is then released, allowing myosin to expend the stored energy as a conformational change. The myosin head moves toward the M line, pulling the actin along with it. As the actin is pulled, the filaments move approximately 10 nm toward the M line. This movement is called the power stroke, as it is the step at which force is produced. As the actin is pulled toward the M line, the sarcomere shortens and the muscle contracts.
When the myosin head is “cocked,” it contains energy and is in a high-energy configuration. This energy is expended as the myosin head moves through the power stroke; at the end of the power stroke, the myosin head is in a low-energy position. After the power stroke, ADP is released; however, the cross-bridge formed is still in place, and actin and myosin are bound together. ATP can then attach to myosin, which allows the cross-bridge cycle to start again and further muscle contraction can occur (Figure 6.8).
Watch this video explaining how a muscle contraction is signaled.
Which of the following statements about muscle contraction is true?
- The power stroke occurs when ATP is hydrolyzed to ADP and phosphate.
- The power stroke occurs when ADP and phosphate dissociate from the myosin head.
- The power stroke occurs when ADP and phosphate dissociate from the actin active site.
- The power stroke occurs when Ca2+ binds the calcium head.
View this video animation of the cross-bridge muscle contraction.
Regulatory proteins
When a muscle is in a resting state, actin and myosin are separated. To keep actin from binding to the active site on myosin, regulatory proteins block the molecular binding sites. Tropomyosin blocks myosin binding sites on actin molecules, preventing cross-bridge formation and preventing contraction in a muscle without nervous input. Troponin binds to tropomyosin and helps to position it on the actin molecule; it also binds calcium ions.
To enable a muscle contraction, tropomyosin must change conformation, uncovering the myosin-binding site on an actin molecule and allowing cross-bridge formation. This can only happen in the presence of calcium, which is kept at extremely low concentrations in the sarcoplasm. If present, calcium ions bind to troponin, causing conformational changes in troponin that allow tropomyosin to move away from the myosin binding sites on actin. Once the tropomyosin is removed, a cross-bridge can form between actin and myosin, triggering contraction. Cross-bridge cycling continues until Ca2+ ions and ATP are no longer available and tropomyosin again covers the binding sites on actin.
Excitation-contraction coupling
Excitation-contraction coupling is the link (transduction) between the action potential generated in the sarcolemma and the start of a muscle contraction. The trigger for calcium release from the sarcoplasmic reticulum into the sarcoplasm is a neural signal. Each skeletal muscle fiber is controlled by a motor neuron, which conducts signals from the brain or spinal cord to the muscle. The area of the sarcolemma on the muscle fiber that interacts with the neuron is called the motor end plate. The end of the neuron’s axon is called the synaptic terminal, and it does not actually contact the motor end plate. A small space called the synaptic cleft separates the synaptic terminal from the motor end plate. Electrical signals travel along the neuron’s axon, which branches through the muscle and connects to individual muscle fibers at a neuromuscular junction.
The ability of cells to communicate electrically requires that the cells expend energy to create an electrical gradient across their cell membranes. This charge gradient is carried by ions, which are differentially distributed across the membrane. Each ion exerts an electrical influence and a concentration influence. Just as milk will eventually mix with coffee without the need to stir, ions also distribute themselves evenly, if they are permitted to do so. In this case, they are not permitted to return to an evenly mixed state.
The sodium-potassium ATPase uses cellular energy to move K+ ions inside the cell and Na+ ions outside. This alone accumulates a small electrical charge, but a big concentration gradient. There is lots of K+ in the cell and lots of Na+ outside the cell. Potassium is able to leave the cell through K+ channels that are open 90% of the time, and it does. However, Na+ channels are rarely open, so Na+remains outside the cell. When K+ leaves the cell, obeying its concentration gradient, that effectively leaves a negative charge behind. So at rest, there is a large concentration gradient for Na+ to enter the cell, and there is an accumulation of negative charges left behind in the cell. This is the resting membrane potential. The potential in this context means a separation of electrical charge that is capable of doing work. It is measured in volts, just like a battery. However, the transmembrane potential is considerably smaller (0.07 V); therefore, the small value is expressed as millivolts (mV) or 70 mV. Because the inside of a cell is negative compared with the outside, a minus sign signifies the excess of negative charges inside the cell, −70 mV.
If an event changes the permeability of the membrane to Na+ ions, they will enter the cell. That will change the voltage. This is an electrical event, called an action potential, that can be used as a cellular signal. Communication occurs between nerves and muscles through neurotransmitters. Neuron action potentials cause the release of neurotransmitters from the synaptic terminal into the synaptic cleft, where they can then diffuse across the synaptic cleft and bind to a receptor molecule on the motor end plate. The motor end plate possesses junctional folds—folds in the sarcolemma that create a large surface area for the neurotransmitter to bind to receptors. The receptors are actually sodium channels that open to allow the passage of Na+ into the cell when they receive neurotransmitter signal.
Acetylcholine (ACh) is a neurotransmitter released by motor neurons that binds to receptors in the motor end plate. Neurotransmitter release occurs when an action potential travels down the motor neuron’s axon, resulting in altered permeability of the synaptic terminal membrane and an influx of calcium. The Ca2+ ions allow synaptic vesicles to move to and bind with the presynaptic membrane (on the neuron) and release neurotransmitter from the vesicles into the synaptic cleft. Once released by the synaptic terminal, ACh diffuses across the synaptic cleft to the motor end plate, where it binds with ACh receptors. As a neurotransmitter binds, these ion channels open, and Na+ ions cross the membrane into the muscle cell. This reduces the voltage difference between the inside and outside of the cell, which is called depolarization. As ACh binds at the motor end plate, this depolarization is called an end-plate potential. The depolarization then spreads along the sarcolemma, creating an action potential as sodium channels adjacent to the initial depolarization site detect the change in voltage and open. The action potential moves across the entire cell, creating a wave of depolarization.
ACh is broken down by the enzyme acetylcholinesterase (AChE) into acetyl and choline. AChE resides in the synaptic cleft, breaking down ACh so that it does not remain bound to ACh receptors, which would cause unwanted extended muscle contraction (Figure 6.9).
The deadly nerve gas Sarin irreversibly inhibits acetylcholinesterase. What effect would Sarin have on muscle contraction?
After depolarization, the membrane returns to its resting state. This is called repolarization, during which voltage-gated sodium channels close. Potassium channels continue at 90% conductance. Because the plasma membrane sodium–potassium ATPase always transports ions, the resting state (negatively charged inside relative to the outside) is restored. The period immediately following the transmission of an impulse in a nerve or muscle, in which a neuron or muscle cell regains its ability to transmit another impulse, is called the refractory period. During the refractory period, the membrane cannot generate another action potential. The refractory period allows the voltage-sensitive ion channels to return to their resting configurations. The sodium-potassium ATPase continually moves Na+ back out of the cell and K+ back into the cell, and the K+ leaks out leaving negative charge behind. Very quickly, the membrane repolarizes, so that it can again be depolarized.
Control of muscle tension
Neural control initiates the formation of actin-myosin cross-bridges, leading to the sarcomere shortening involved in muscle contraction. These contractions extend from the muscle fiber through connective tissue to pull on bones, causing skeletal movement. The pull exerted by a muscle is called tension, and the amount of force created by this tension can vary. This enables the same muscles to move very light objects and very heavy objects. In individual muscle fibers, the amount of tension produced depends on the cross-sectional area of the muscle fiber and the frequency of neural stimulation.
The number of cross-bridges formed between actin and myosin determines the amount of tension that a muscle fiber can produce. Cross-bridges can only form where thick and thin filaments overlap, allowing myosin to bind to actin. If more cross-bridges are formed, more myosin will pull on actin, and more tension will be produced.
The ideal length of a sarcomere during the production of maximal tension occurs when thick and thin filaments overlap to the greatest degree. If a sarcomere at rest is stretched past an ideal resting length, thick and thin filaments do not overlap to the greatest degree, and fewer cross-bridges can form. This results in fewer myosin heads pulling on actin, and less tension is produced. As a sarcomere is shortened, the zone of overlap is reduced as the thin filaments reach the H zone, which is composed of myosin tails. Because it is myosin heads that form cross-bridges, actin will not bind to myosin in this zone, reducing the tension produced by this myofiber. If the sarcomere is shortened, even more, thin filaments begin to overlap with each other—reducing cross-bridge formation even further and producing even less tension. Conversely, if the sarcomere is stretched to the point at which thick and thin filaments do not overlap at all, no cross-bridges are formed and no tension is produced. This amount of stretching does not usually occur because accessory proteins, internal sensory nerves, and connective tissue oppose extreme stretching.
The primary variable determining force production is the number of myofibers within the muscle that receive an action potential from the neuron that controls that fiber. When using the biceps to pick up a pencil, the motor cortex of the brain only signals a few neurons of the biceps, and only a few myofibers respond. In vertebrates, each myofiber responds fully if stimulated. When picking up a piano, the motor cortex signals all of the neurons in the biceps and every myofiber participates. This is close to the maximum force the muscle can produce. As mentioned above, increasing the frequency of action potentials (the number of signals per second) can increase the force a bit more, because the tropomyosin is flooded with calcium.
Teach your peer about the events during muscle contraction, from the arrival of the neural signal to generation of motion powered by the muscle. When you are done, ask your peer what terms or steps you missed or did not explain well. Let your peer fill the gaps. If there were no gaps, your peer can challenge you with some questions about your explanation. Remember that one way that you can test whether you are learning is to be able to transmit your knowledge to another person. | libretexts | 2025-03-17T22:26:53.186209 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/6%3A_Locomotion/6.4%3A_Muscle_Contraction",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "6.4: Muscle Contraction",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/7%3A_Electrical_Signals | 7: Electrical Signals Last updated Save as PDF Page ID 9582 Sanja Hinić-Frlog University of Toronto Mississauga 7.1: Electrical Potential 7.2: Resting, Graded and Action Potential 7.3: Synapses 7.4: Neuroglia | libretexts | 2025-03-17T22:26:53.263357 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/7%3A_Electrical_Signals",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "7: Electrical Signals",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/7%3A_Electrical_Signals/7.1%3A_Electrical_Potential | 7.1: Electrical Potential
- Describe membrane potential in terms of variables in Ohm’s Law
Ohm’s Law
What drives current? We can think of various devices—such as batteries, generators, wall outlets, and so on—which are necessary to maintain a current. All such devices create a potential difference and are loosely referred to as voltage sources. When a voltage source is connected to a conductor, it applies a potential difference \(V\) that creates an electric field. The electric field in turn exerts force on charges, causing current.
The current that flows through most substances is directly proportional to the voltage \(V\) applied to it. The German physicist Georg Simon Ohm (1787–1854) was the first to demonstrate experimentally that the current in a metal wire is directly proportional to the voltage applied :
\[T \propto V\]
This important relationship is known as Ohm’s law. It can be viewed as a cause-and-effect relationship, with voltage the cause and current the effect. This is an empirical law like that for friction—an experimentally observed phenomenon. Such a linear relationship doesn’t always occur.
Resistance and Simple Circuits
If voltage drives current, what impedes it? The electric property that impedes current (crudely similar to friction and air resistance) is called resistance \(R\) Collisions of moving charges with atoms and molecules in a substance transfer energy to the substance and limit current. Resistance is defined as inversely proportional to current, or
\[ I \propto \dfrac{1}{R}\]
Thus, for example, current is cut in half if resistance doubles. Combining the relationships of current to voltage and current to resistance gives
\[I = \dfrac{V}{R}\]
This relationship is also called Ohm’s law. Ohm’s law in this form really defines resistance for certain materials. Ohm’s law (like Hooke’s law) is not universally valid. The many substances for which Ohm’s law holds are called ohmic. These include good conductors like copper and aluminum, and some poor conductors under certain circumstances. Ohmic materials have a resistance that is independent of voltage and current . An object that has simple resistance is called a resistor , even if its resistance is small. The unit for resistance is an ohm and is given the symbol (upper case Greek omega). Rearranging gives , and so the units of resistance are 1 ohm = 1 volt per ampere:
\[1 = 1 \dfrac{V}{A}\]
Figure 7.1 shows the schematic for a simple circuit. A simple circuit has a single voltage source and a single resistor. The wires connecting the voltage source to the resistor can be assumed to have negligible resistance, or their resistance can be included in .
A simple electric circuit in which a closed path for current to flow is supplied by conductors (usually metal wires) connecting a load to the terminals of a battery, represented by the red parallel lines. The zigzag symbol represents the single resistor and includes any resistance in the connections to the voltage source.
Resistances range over many orders of magnitude. Some ceramic insulators, such as those used to support power lines, have resistances of or more. A dry person may have a hand-to-foot resistance of , whereas the resistance of the human heart is about . A meter-long piece of large-diameter copper wire may have a resistance of , and superconductors have no resistance at all (they are non-ohmic). Resistance is related to the shape of an object and the material of which it is composed.
Additional insight is gained by solving for yielding
\[V = IR\]
Figure 7.2. The voltage drop across a resistor in a simple circuit equals the voltage output of the battery.
In a simple electrical circuit, the sole resistor converts energy supplied by the source into another form. Conservation of energy is evidenced here by the fact that all of the energy supplied by the source is converted to another form by the resistor alone. We will find that conservation of energy has other important applications in circuits and is a powerful tool in circuit analysis.
Nerve Conduction
Electric currents in the vastly complex system of billions of nerves in our body allow us to sense the world, control parts of our body, and think. These are representative of the three major functions of nerves. First, nerves carry messages from our sensory organs and others to the central nervous system, consisting of the brain and spinal cord. Second, nerves carry messages from the central nervous system to muscles and other organs. Third, nerves transmit and process signals within the central nervous system. The sheer number of nerve cells and the incredibly greater number of connections between them makes this system the subtle wonder that it is. Nerve conduction is a general term for electrical signals carried by nerve cells. It is one aspect of bioelectricity, or electrical effects in and created by biological systems.
Nerve cells, properly called neurons , look different from other cells—they have tendrils, some of them many centimeters long, connecting them with other cells ( Figure 7.3 ). Signals arrive at the cell body across synapses or through dendrites , stimulating the neuron to generate its own signal, sent along its long axon to other nerve or muscle cells. Signals may arrive from many other locations and be transmitted to yet others, conditioning the synapses by use, giving the system its complexity and its ability to learn.
A neuron with its dendrites and long axon. Signals in the form of electric currents reach the cell body through dendrites and across synapses, stimulating the neuron to generate its own signal sent down the axon. The number of interconnections can be far greater than shown here.
The method by which these electric currents are generated and transmitted is more complex than the simple movement of free charges in a conductor, but it can be understood with principles already discussed in this text. The most important of these are the Coulomb force and diffusion.
Figure 7.4 illustrates how a voltage (potential difference) is created across the cell membrane of a neuron in its resting state. This thin membrane separates electrically neutral fluids having differing concentrations of ions, the most important varieties being , , and (these are sodium, potassium, and chlorine ions with single plus or minus charges as indicated). Free ions will diffuse from a region of high concentration to one of low concentration. But the cell membrane is semipermeable, meaning that some ions may cross it while others cannot. In its resting state, the cell membrane is permeable to and , and impermeable to . Diffusion of and thus creates the layers of positive and negative charge on the outside and inside of the membrane. The Coulomb force prevents the ions from diffusing across in their entirety. Once the charge layer has built up, the repulsion of like charges prevents more from moving across, and the attraction of unlike charges prevents more from leaving either side. The result is two layers of charge right on the membrane, with diffusion being balanced by the Coulomb force. A tiny fraction of the charges move across and the fluids remain neutral (other ions are present), while a separation of charge and a voltage have been created across the membrane.
Figure 7.4. The process of how a voltage (potential difference) is created across the cell membrane of a neuron in its resting state.
Describe electrical signal in a cell using the following terms: resistance, flow, and potential difference
Which cells are involved in maintaining electrical potential of an animal cell and how | libretexts | 2025-03-17T22:26:53.333393 | 2018-10-18T16:30:14 | {
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"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "7.1: Electrical Potential",
"author": "Sanja Hinić-Frlog"
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/7%3A_Electrical_Signals/7.2%3A_Resting_Graded_and_Action_Potential | 7.2: Resting, Graded and Action Potential
- Differentiate between resting, graded and action potential with reference to structures of the neuron as well as ion movements and membrane potentials
The nervous system of the common laboratory fly, Drosophila melanogaster , contains around 100,000 neurons, the same number as a lobster. This number compares to 75 million in the mouse and 300 million in the octopus. A human brain contains around 86 billion neurons. Despite these very different numbers, the nervous systems of these animals control many of the same behaviors—from basic reflexes to more complicated behaviors like finding food and courting mates. The ability of neurons to communicate with each other as well as with other types of cells underlies all of these behaviors. Most neurons share the same cellular components. But neurons are also highly specialized—different types of neurons have different sizes and shapes that relate to their functional roles.
Parts of a neuron
Like other cells, each neuron has a cell body (or soma) that contains a nucleus, smooth and rough endoplasmic reticulum, Golgi apparatus, mitochondria, and other cellular components. Neurons also contain unique structures, illustrated in Figure 7.5 for receiving and sending the electrical signals that make neuronal communication possible. Dendrites are tree-like structures that extend away from the cell body to receive messages from other neurons at specialized junctions called synapses . Although some neurons do not have any dendrites, some types of neurons have multiple dendrites. Dendrites can have small protrusions called dendritic spines, which further increase surface area for possible synaptic connections.
Once a signal is received by the dendrite, it then travels passively to the cell body. The cell body contains a specialized structure, the axon hillock that integrates signals from multiple synapses and serves as a junction between the cell body and an axon . An axon is a tube-like structure that propagates the integrated signal to specialized endings called axon terminals . These terminals in turn synapse on other neurons, muscle, or target organs. Chemicals released at axon terminals allow signals to be communicated to these other cells. Neurons usually have one or two axons, but some neurons, like amacrine cells in the retina, do not contain any axons. Some axons are covered with myelin , which acts as an insulator to minimize dissipation of the electrical signal as it travels down the axon, greatly increasing the speed on conduction. This insulation is important as the axon from a human motor neuron can be as long as a meter—from the base of the spine to the toes. The myelin sheath is not actually part of the neuron. Myelin is produced by glial cells. Along the axon there are periodic gaps in the myelin sheath. These gaps are called nodes of Ranvier and are sites where the signal is “recharged” as it travels along the axon.
It is important to note that a single neuron does not act alone—neuronal communication depends on the connections that neurons make with one another (as well as with other cells, like muscle cells). Dendrites from a single neuron may receive synaptic contact from many other neurons. For example, dendrites from a Purkinje cell in the cerebellum are thought to receive contact from as many as 200,000 other neurons.
Which of the following statements is false?
- The soma is the cell body of a nerve cell.
- Myelin sheath provides an insulating layer to the dendrites.
- Axons carry the signal from the soma to the target.
- Dendrites carry the signal to the soma.
Types of neurons
There are different types of neurons, and the functional role of a given neuron is intimately dependent on its structure. There is an amazing diversity of neuron shapes and sizes found in different parts of the nervous system (and across species), as illustrated by the neurons shown in Figure 7.6.
While there are many defined neuron cell subtypes, neurons are broadly divided into four basic types: unipolar, bipolar, multipolar, and pseudounipolar. Figure 7.7 illustrates these four basic neuron types. Unipolar neurons have only one structure that extends away from the soma. These neurons are not found in vertebrates but are found in insects where they stimulate muscles or glands. A bipolar neuron has one axon and one dendrite extending from the soma. An example of a bipolar neuron is a retinal bipolar cell, which receives signals from photoreceptor cells that are sensitive to light and transmits these signals to ganglion cells that carry the signal to the brain. Multipolar neurons are the most common type of neuron. Each multipolar neuron contains one axon and multiple dendrites. Multipolar neurons can be found in the central nervous system (brain and spinal cord). An example of a multipolar neuron is a Purkinje cell in the cerebellum, which has many branching dendrites but only one axon. Pseudounipolar cells share characteristics with both unipolar and bipolar cells. A pseudounipolar cell has a single process that extends from the soma, like a unipolar cell, but this process later branches into two distinct structures, like a bipolar cell. Most sensory neurons are pseudounipolar and have an axon that branches into two extensions: one connected to dendrites that receive sensory information and another that transmits this information to the spinal cord.
Neurogenesis
At one time, scientists believed that people were born with all the neurons they would ever have. Research performed during the last few decades indicates that neurogenesis, the birth of new neurons, continues into adulthood. Neurogenesis was first discovered in songbirds that produce new neurons while learning songs. For mammals, new neurons also play an important role in learning: about 1000 new neurons develop in the hippocampus (a brain structure involved in learning and memory) each day. While most of the new neurons will die, researchers found that an increase in the number of surviving new neurons in the hippocampus correlated with how well rats learned a new task. Interestingly, both exercise and some antidepressant medications also promote neurogenesis in the hippocampus. Stress has the opposite effect. While neurogenesis is quite limited compared to regeneration in other tissues, research in this area may lead to new treatments for disorders such as Alzheimer’s, stroke, and epilepsy.
How do scientists identify new neurons? A researcher can inject a compound called bromodeoxyuridine (BrdU) into the brain of an animal. While all cells will be exposed to BrdU, BrdU will only be incorporated into the DNA of newly generated cells that are in S phase. A technique called immunohistochemistry can be used to attach a fluorescent label to the incorporated BrdU, and a researcher can use fluorescent microscopy to visualize the presence of BrdU, and thus new neurons, in brain tissue. Figure 7.8 is a micrograph which shows fluorescently labeled neurons in the hippocampus of a rat.
Nerve impulse transmission within a neuron
For the nervous system to function, neurons must be able to send and receive signals. These signals are possible because each neuron has a charged cellular membrane (a voltage difference between the inside and the outside), and the charge of this membrane can change in response to neurotransmitter molecules released from other neurons and environmental stimuli. To understand how neurons communicate, one must first understand the basis of the baseline or ‘resting’ membrane charge.
Neuronal charged membranes
The lipid bilayer membrane that surrounds a neuron is impermeable to charged molecules or ions. To enter or exit the neuron, ions must pass through special proteins called ion channels that span the membrane. Ion channels have different configurations: open, closed, and inactive, as illustrated in Figure 7.9. Some ion channels need to be activated in order to open and allow ions to pass into or out of the cell. These ion channels are sensitive to the environment and can change their shape accordingly. Ion channels that change their structure in response to voltage changes are called voltage-gated ion channels. Voltage-gated ion channels regulate the relative concentrations of different ions inside and outside the cell. The difference in total charge between the inside and outside of the cell is called the membrane potential .
This video discusses the basis of the resting membrane potential.
Resting Membrane Potential
A neuron at rest is negatively charged: the inside of a cell is approximately 70 millivolts more negative than the outside (−70 mV, note that this number varies by neuron type and by species). This voltage is called the resting membrane potential; it is caused by differences in the concentrations of ions inside and outside the cell. If the membrane were equally permeable to all ions, each type of ion would flow across the membrane and the system would reach equilibrium. Because ions cannot simply cross the membrane at will, there are different concentrations of several ions inside and outside the cell, as shown in Table 7.1.
| Ion | Extracellular concentration (mM) | Intracellular concentration (mM) | Ratio outside/inside |
|---|---|---|---|
| Na+ | 145 | 12 | 12 |
| K+ | 4 | 155 | 0.026 |
| Cl− | 120 | 4 | 30 |
| Organic anions (A−) | — | 100 |
The difference in the number of positively charged potassium ions (K+) inside and outside the cell dominates the resting membrane potential (Figure 7.10). When the membrane is at rest, K+ions accumulate inside the cell due to a net movement with the concentration gradient. The negative resting membrane potential is created and maintained by increasing the concentration of cations outside the cell (in the extracellular fluid) relative to inside the cell (in the cytoplasm). The negative charge within the cell is created by the cell membrane being more permeable to potassium ion movement than sodium ion movement. In neurons, potassium ions are maintained at high concentrations within the cell while sodium ions are maintained at high concentrations outside of the cell. The cell possesses potassium and sodium leakage channels that allow the two cations to diffuse down their concentration gradient. However, the neurons have far more potassium leakage channels than sodium leakage channels. Therefore, potassium diffuses out of the cell at a much faster rate than sodium leaks in. Because more cations are leaving the cell than are entering, this causes the interior of the cell to be negatively charged relative to the outside of the cell. The actions of the sodium-potassium pump help to maintain the resting potential, once established. Recall that sodium potassium pumps bring two K+ ions into the cell while removing three Na+ ions per ATP consumed. As more cations are expelled from the cell than taken in, the inside of the cell remains negatively charged relative to the extracellular fluid. It should be noted that calcium ions (Cl–) tend to accumulate outside of the cell because they are repelled by negatively-charged proteins within the cytoplasm.
Action potential
A neuron can receive input from other neurons and, if this input is strong enough, send the signal to downstream neurons. Transmission of a signal between neurons is generally carried by a chemical called a neurotransmitter. Transmission of a signal within a neuron (from dendrite to axon terminal) is carried by a brief reversal of the resting membrane potential called an action potential . When neurotransmitter molecules bind to receptors located on a neuron’s dendrites, ion channels open. At excitatory synapses, this opening allows positive ions to enter the neuron and results in depolarization of the membrane—a decrease in the difference in voltage between the inside and outside of the neuron. A stimulus from a sensory cell or another neuron depolarizes the target neuron to its threshold potential (-55 mV). Na+ channels in the axon hillock open, allowing positive ions to enter the cell (Figure 7.10 and Figure 7.11). Once the sodium channels open, the neuron completely depolarizes to a membrane potential of about +40 mV. Action potentials are considered an “all-or-nothing” event, in that, once the threshold potential is reached, the neuron always completely depolarizes. Once depolarization is complete, the cell must now “reset” its membrane voltage back to the resting potential. To accomplish this, the Na+ channels close and cannot be opened. This begins the neuron’s refractory period , in which it cannot produce another action potential because its sodium channels will not open. At the same time, voltage-gated K+channels open, allowing K+ to leave the cell. As K+ ions leave the cell, the membrane potential once again becomes negative. The diffusion of K+ out of the cell actually hyperpolarizes the cell, in that the membrane potential becomes more negative than the cell’s normal resting potential. At this point, the sodium channels will return to their resting state, meaning they are ready to open again if the membrane potential again exceeds the threshold potential. Eventually, the extra K+ ions diffuse out of the cell through the potassium leakage channels, bringing the cell from its hyperpolarized state, back to its resting membrane potential.
In summary, an action potential is caused by movements of ions across the cell membrane as shown. Depolarization occurs when a stimulus makes the membrane permeable to ions. Repolarization follows as the membrane again becomes impermeable to and moves from high to low concentration. In the long term, active transport slowly maintains the concentration differences, but the cell may fire hundreds of times in rapid succession without seriously depleting them.
The action potential is a voltage pulse at one location on a cell membrane. How does it get transmitted along the cell membrane, and in particular down an axon, as a nerve impulse? The answer is that the changing voltage and electric fields affect the permeability of the adjacent cell membrane so that the same process takes place there. The adjacent membrane depolarizes, affecting the membrane further down, and so on, as illustrated in Figure 7.6. Thus the action potential stimulated at one location triggers a nerve impulse that moves slowly (about 1 m/s) along the cell membrane.
Potassium channel blockers, such as amiodarone and procainamide, which are used to treat abnormal electrical activity in the heart, called cardiac dysrhythmia, impede the movement of K+ through voltage-gated K+ channels. Which part of the action potential would you expect potassium channels to affect? Explain why.
- Answer
-
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Figure 7.13. The action potential is conducted down the axon as the axon membrane depolarizes, then repolarizes.
- This video presents an overview of an action potential.
Myelin and the propagation of the action potential
For an action potential to communicate information to another neuron, it must travel along the axon and reach the axon terminals where it can initiate neurotransmitter release. The speed of conduction of an action potential along an axon is influenced by both the diameter of the axon and the axon’s resistance to current leak. Myelin acts as an insulator that prevents current from leaving the axon; this increases the speed of action potential conduction. In demyelinating diseases like multiple sclerosis, action potential conduction slows because the current leaks from previously insulated axon areas. The nodes of Ranvier, illustrated in Figure 7.15 are gaps in the myelin sheath along the axon. These unmyelinated spaces are about one micrometer long and contain voltage-gated Na+ and K+ channels. The flow of ions through these channels, particularly the Na+ channels, regenerates the action potential over and over again along the axon. This ‘jumping’ of the action potential from one node to the next is called saltatory conduction . If nodes of Ranvier were not present along an axon, the action potential would propagate very slowly since Na+ and K+ channels would have to continuously regenerate action potentials at every point along the axon instead of at specific points. Nodes of Ranvier also save energy for the neuron since the channels only need to be present at the nodes and not along the entire axon.
Figure 7.16 shows an enlarged view of an axon having myelin sheaths characteristically separated by unmyelinated gaps (called nodes of Ranvier). This arrangement gives the axon a number of interesting properties. Since myelin is an insulator, it prevents signals from jumping between adjacent nerves (crosstalk). Additionally, the myelinated regions transmit electrical signals at a very high speed, as an ordinary conductor or resistor would. There is no action potential in the myelinated regions so that no cell energy is used in them. There is an signal loss in the myelin, but the signal is regenerated in the gaps, where the voltage pulse triggers the action potential at full voltage. So a myelinated axon transmits a nerve impulse faster, with less energy consumption, and is better protected from cross talk than an unmyelinated one. Not all axons are myelinated so that crosstalk and slow signal transmission are a characteristic of the normal operation of these axons, another variable in the nervous system.
The degeneration or destruction of the myelin sheaths that surround the nerve fibers impairs signal transmission and can lead to numerous neurological effects. One of the most prominent of these diseases comes from the body’s own immune system attacking the myelin in the central nervous system—multiple sclerosis. MS symptoms include fatigue, vision problems, weakness of arms and legs, loss of balance, and tingling or numbness in one’s extremities (neuropathy). It is more apt to strike younger adults, especially females. Causes might come from infection, environmental or geographic effects, or genetics. At the moment there is no known cure for MS.
Most animal cells can fire or create their own action potential. Muscle cells contract when they fire and are often induced to do so by a nerve impulse. In fact, nerve and muscle cells are physiologically similar, and there are even hybrid cells, such as in the heart, that have characteristics of both nerves and muscles. Some animals, like the infamous electric eel (Figure 7.17) use muscles ganged so that their voltages add in order to create a shock great enough to stun prey.
Propagation of a nerve impulse down a myelinated axon, from left to right. The signal travels very fast and without energy input in the myelinated regions, but it loses voltage. It is regenerated in the gaps. The signal moves faster than in unmyelinated axons and is insulated from signals in other nerves, limiting crosstalk.
Which of the following statements is false?
a. The soma is the cell body of a nerve cell.
b. Myelin sheath provides an insulating layer to the dendrites.
c. Axons carry the signal from the soma to the target.
d. Dendrites carry the signal to the soma.
Neurons contain ________, which can receive signals from other neurons.
a. axons
b. mitochondria
c. dendrites
d. Golgi bodies
A(n) ________ neuron has one axon and one dendrite extending directly from the cell body.
a. unipolar
b. bipolar
c. multipolar
d. pseudounipolar
Glia that provide myelin for neurons in the brain are called ________.
a. Schwann cells
b. oligodendrocytes
c. microglia
d. astrocytes
How are neurons similar to other cells? How are they unique?
Compare and contrast resting, graded and action potential? In your answer, make sure you have included channels and voltage reference as well as relevant structures of the neurons. Once you have come up with an answer, give it to another student to review. Based on the review by your peer, is there something you need to work on in terms of your understanding of the resting, graded and action potential.
Multiple sclerosis causes demyelination of axons in the brain and spinal cord. Why is this problematic? | libretexts | 2025-03-17T22:26:53.425581 | 2018-10-18T16:30:14 | {
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"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/7%3A_Electrical_Signals/7.2%3A_Resting_Graded_and_Action_Potential",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "7.2: Resting, Graded and Action Potential",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/7%3A_Electrical_Signals/7.3%3A_Synapses | 7.3: Synapses
- Explain how sample signals cross the synapse
Synaptic Transmission
The synapse or “gap” is the place where information is transmitted from one neuron to another. Synapses usually form between axon terminals and dendritic spines, but this is not universally true. There are also axon-to-axon, dendrite-to-dendrite, and axon-to-cell body synapses. The neuron transmitting the signal is called the presynaptic neuron, and the neuron receiving the signal is called the postsynaptic neuron. Note that these designations are relative to a particular synapse—most neurons are both presynaptic and postsynaptic. There are two types of synapses: chemical and electrical.
Chemical Synapses
When an action potential reaches the axon terminal it depolarizes the membrane and opens voltage-gated \(\ce{Na^{+}}\) channels. \(\ce{Na^{+}}\) ions enter the cell, further depolarizing the presynaptic membrane. This depolarization causes voltage-gated \(\ce{Ca^{3+}}\) channels to open. Calcium ions entering the cell initiate a signaling cascade that causes small membrane-bound vesicles, called synaptic vesicles, containing neurotransmitter molecules to fuse with the presynaptic membrane. Synaptic vesicles are shown in Figure 7.18, which is an image from a scanning electron microscope.
Fusion of a vesicle with the presynaptic membrane causes the neurotransmitter to be released into the synaptic cleft, the extracellular space between the presynaptic and postsynaptic membranes, as illustrated in Figure 7.19 . The neurotransmitter diffuses across the synaptic cleft and binds to receptor proteins on the postsynaptic membrane.
The binding of a specific neurotransmitter causes particular ion channels, in this case, ligand-gated channels, on the postsynaptic membrane to open. Neurotransmitters can either have excitatory or inhibitory effects on the postsynaptic membrane, as detailed in Table 7.2. For example, when acetylcholine is released at the synapse between a nerve and muscle (called the neuromuscular junction) by a presynaptic neuron, it causes postsynaptic Na+ channels to open. Na+ enters the postsynaptic cell and causes the postsynaptic membrane to depolarize. This depolarization is called an excitatory postsynaptic potential (EPSP) and makes the postsynaptic neuron more likely to fire an action potential. The release of neurotransmitter at inhibitory synapses causes inhibitory postsynaptic potentials (IPSPs) , a hyperpolarization of the presynaptic membrane. For example, when the neurotransmitter GABA (gamma-aminobutyric acid) is released from a presynaptic neuron, it binds to and opens Cl– channels. Cl– ions enter the cell and hyperpolarizes the membrane, making the neuron less likely to fire an action potential.
Once neurotransmission has occurred, the neurotransmitter must be removed from the synaptic cleft so that the postsynaptic membrane can “reset” and be ready to receive another signal. This can be accomplished in three ways: the neurotransmitter can diffuse away from the synaptic cleft, it can be degraded by enzymes in the synaptic cleft, or it can be recycled (sometimes called reuptake) by the presynaptic neuron. Several drugs act at this step of neurotransmission. For example, some drugs that are given to Alzheimer’s patients work by inhibiting acetylcholinesterase, the enzyme that degrades acetylcholine. This inhibition of the enzyme essentially increases neurotransmission at synapses that release acetylcholine. Once released, the acetylcholine stays in the cleft and can continually bind and unbind to postsynaptic receptors.
| Neurotransmitter | Example | Location |
|---|---|---|
| Acetylcholine | — | CNS and/or PNS |
| Biogenic amine | Dopamine, serotonin, norepinephrine | CNS and/or PNS |
| Amino acid | Glycine, glutamate, aspartate, gamma-aminobutyric acid | CNS |
| Neuropeptide | Substance P, endorphins | CNS and/or PNS |
Design and/or act out events at the synapse! Here are some hints to help you design your review activity:
- You will need to think about how many peers you need to participate.
- Who will be which structure?
- How will your peers and you move with respect to each other?
- What is the overall goal of each movement you design (what is the function of each structure)?
Electrical Synapses
While electrical synapses are fewer in number than chemical synapses, they are found in all nervous systems and play important and unique roles. The mode of neurotransmission in electrical synapses is quite different from that in chemical synapses. In an electrical synapse, the presynaptic and postsynaptic membranes are very close together and are actually physically connected by channel proteins forming gap junctions. Gap junctions allow current to pass directly from one cell to the next. In addition to the ions that carry this current, other molecules, such as ATP, can diffuse through the large gap junction pores.
There are key differences between chemical and electrical synapses. Because chemical synapses depend on the release of neurotransmitter molecules from synaptic vesicles to pass on their signal, there is an approximately one-millisecond delay between when the axon potential reaches the presynaptic terminal and when the neurotransmitter leads to opening of postsynaptic ion channels. Additionally, this signaling is unidirectional. Signaling in electrical synapses, in contrast, is virtually instantaneous (which is important for synapses involved in key reflexes), and some electrical synapses are bidirectional. Electrical synapses are also more reliable as they are less likely to be blocked, and they are important for synchronizing the electrical activity of a group of neurons. For example, electrical synapses in the thalamus are thought to regulate slow-wave sleep, and disruption of these synapses can cause seizures.
Signal summation
Sometimes a single EPSP is strong enough to induce an action potential in the postsynaptic neuron, but often multiple presynaptic inputs must create EPSPs around the same time for the postsynaptic neuron to be sufficiently depolarized to fire an action potential. This process is called summation and occurs at the axon hillock, as illustrated in Figure 7.20. Additionally, one neuron often has inputs from many presynaptic neurons—some excitatory and some inhibitory—so IPSPs can cancel out EPSPs and vice versa. It is the net change in postsynaptic membrane voltage that determines whether the postsynaptic cell has reached its threshold of excitation needed to fire an action potential. Together, synaptic summation and the threshold for excitation act as a filter so that random “noise” in the system is not transmitted as important information.
Synaptic plasticity
Synapses are not static structures. They can be weakened or strengthened. They can be broken, and new synapses can be made. Synaptic plasticity allows for these changes, which are all needed for a functioning nervous system. In fact, synaptic plasticity is the basis of learning and memory. Two processes in particular, long-term potentiation (LTP) and long-term depression (LTD) are important forms of synaptic plasticity that occur in synapses in the hippocampus, a brain region that is involved in storing memories.
Long-term potentiation (LTP)
Long-term potentiation (LTP) is a persistent strengthening of a synaptic connection. LTP is based on the Hebbian principle: cells that fire together wire together. There are various mechanisms, none fully understood, behind the synaptic strengthening seen with LTP. One known mechanism involves a type of postsynaptic glutamate receptor, called NMDA (N-Methyl-D-aspartate) receptors, shown in Figure 7.21. These receptors are normally blocked by magnesium ions; however, when the postsynaptic neuron is depolarized by multiple presynaptic inputs in quick succession (either from one neuron or multiple neurons), the magnesium ions are forced out allowing Ca ions to pass into the postsynaptic cell. Next, Ca2+ions entering the cell initiate a signaling cascade that causes a different type of glutamate receptor, called AMPA (α-amino-3-hydroxy-5-methyl-4-isoxazolepropionic acid) receptors, to be inserted into the postsynaptic membrane, since activated AMPA receptors allow positive ions to enter the cell. So, the next time glutamate is released from the presynaptic membrane, it will have a larger excitatory effect (EPSP) on the postsynaptic cell because the binding of glutamate to these AMPA receptors will allow more positive ions into the cell. The insertion of additional AMPA receptors strengthens the synapse and means that the postsynaptic neuron is more likely to fire in response to presynaptic neurotransmitter release. Some drugs of abuse co-opt the LTP pathway, and this synaptic strengthening can lead to addiction.
Long-term depression (LTD)
Long-term depression (LTD) is essentially the reverse of LTP: it is a long-term weakening of a synaptic connection. One mechanism known to cause LTD also involves AMPA receptors. In this situation, calcium that enters through NMDA receptors initiates a different signaling cascade, which results in the removal of AMPA receptors from the postsynaptic membrane, as illustrated in Figure 7.21. The decrease in AMPA receptors in the membrane makes the postsynaptic neuron less responsive to glutamate released from the presynaptic neuron. While it may seem counterintuitive, LTD may be just as important for learning and memory as LTP. The weakening and pruning of unused synapses allow for unimportant connections to be lost and makes the synapses that have undergone LTP that much stronger by comparison.
For a neuron to fire an action potential, its membrane must reach ________.
a. hyperpolarization
b. the threshold of excitation
c. the refractory period
d. inhibitory postsynaptic potentia
After an action potential, the opening of additional voltage-gated ________ channels and the inactivation of sodium channels, cause the membrane to return to its resting membrane potential.
a. sodium
b. potassium
c. calcium
d. chloride
What is the term for protein channels that connect two neurons at an electrical synapse?
a. synaptic vesicles
b. voltage-gated ion channels
c. gap junction protein
d. sodium-potassium exchange pumps
How does myelin aid propagation of an action potential along an axon? How do the nodes of Ranvier help this process?
What are the main steps in chemical neurotransmission? | libretexts | 2025-03-17T22:26:53.502568 | 2018-10-18T16:30:14 | {
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"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/7%3A_Electrical_Signals/7.3%3A_Synapses",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "7.3: Synapses",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/7%3A_Electrical_Signals/7.4%3A_Neuroglia | 7.4: Neuroglia
- Describe the roles of different neuroglia
Glia
While glia are often thought of as the supporting cast of the nervous system, the number of glial cells in the brain actually outnumbers the number of neurons by a factor of ten. Neurons would be unable to function without the vital roles that are fulfilled by these glial cells. Glia guide developing neurons to their destinations, buffer ions and chemicals that would otherwise harm neurons and provide myelin sheaths around axons. Scientists have recently discovered that they also play a role in responding to nerve activity and modulating communication between nerve cells. When glia do not function properly, the result can be disastrous—most brain tumors are caused by mutations in glia.
Glia have different roled in both the central nervous system (CNS) and the peripheral nervous system (PNS). CNS is comprised of the brain and the spinal chord while PNS is the connection between the central nervous system and the rest of the body. The CNS is like the power plant of the nervous system. It creates the signals that control the functions of the body. The PNS is like the wires that go to individual houses. Without those “wires,” the signals produced by the CNS could not control the body (and the CNS would not be able to receive sensory information from the body either). The PNS can be broken down into the autonomic nervous system, which controls bodily functions without conscious control, and the sensory-somatic nervous system, which transmits sensory information from the skin, muscles, and sensory organs to the CNS and sends motor commands from the CNS to the muscles.
Types of Glia
There are several different types of glia with different functions, two of which are shown in Figure 7.22. Astrocytes , shown in Figure 7.23a make contact with both capillaries and neurons in the central nervous system (CNS). They provide nutrients and other substances to neurons, regulate the concentrations of ions and chemicals in the extracellular fluid, and provide structural support for synapses. Astrocytes also form the blood-brain barrier—a structure that blocks the entrance of toxic substances into the brain. Astrocytes, in particular, have been shown through calcium imaging experiments to become active in response to nerve activity, transmit calcium waves between astrocytes, and modulate the activity of surrounding synapses. Satellite glia provide nutrients and structural support for neurons in the PNS. Microglia scavenge and degrade dead cells and protect the brain from invading microorganisms.
Oligodendrocytes , shown in Figure 7.23b form myelin sheaths around axons in the CNS. One axon can be myelinated by several oligodendrocytes, and one oligodendrocyte can provide myelin for multiple neurons. This is distinctive from the PNS where a single Schwann cell provides myelin for only one axon as the entire Schwann cell surrounds the axon. Radial glia serve as scaffolds for developing neurons as they migrate to their end destinations. Ependymal cells line fluid-filled ventricles of the brain and the central canal of the spinal cord. They are involved in the production of cerebrospinal fluid, which serves as a cushion for the brain, moves the fluid between the spinal cord and the brain, and is a component for the choroid plexus.
What is a functionally equivalent cell in CNS to a Schwann cell in PNS? What other types of glia exist in CNS and what is their function. Try drawing all the glia of the CNS this cell in addition to answering the above questions | libretexts | 2025-03-17T22:26:53.561850 | 2018-10-18T16:30:14 | {
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"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "7.4: Neuroglia",
"author": "Sanja Hinić-Frlog"
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/8%3A_Sensory_Systems | 8: Sensory Systems Last updated Save as PDF Page ID 9589 Sanja Hinić-Frlog University of Toronto Mississauga 8.1: Neuronal Organization 8.2: Brain Evolution 8.3: Special Senses | libretexts | 2025-03-17T22:26:53.637619 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/8%3A_Sensory_Systems",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "8: Sensory Systems",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/8%3A_Sensory_Systems/8.1%3A_Neuronal_Organization | 8.1: Neuronal Organization
- Describe the general organization of the neurons within and outside of the central nervous system.
Central nervous system
The central nervous system (CNS) is made up of the brain, a part of which is shown in Figure 8.1 and spinal cord and is covered with three layers of protective coverings called meninges (from the Greek word for membrane). The outermost layer is the dura mater (Latin for “hard mother”). As the Latin suggests, the primary function of this thick layer is to protect the brain and spinal cord. The dura mater also contains vein-like structures that carry blood from the brain back to the heart. The middle layer is the web-like arachnoid mater . The last layer is the pia mater (Latin for “soft mother”), which directly contacts and covers the brain and spinal cord like plastic wrap. The space between the arachnoid and pia maters is filled with cerebrospinal fluid (CSF) . CSF is produced by a tissue called choroid plexus in fluid-filled compartments in the CNS called ventricles . The brain floats in CSF, which acts as a cushion and shock absorber and makes the brain neutrally buoyant. CSF also functions to circulate chemical substances throughout the brain and into the spinal cord.
The entire brain contains only about 8.5 tablespoons of CSF, but CSF is constantly produced in the ventricles. This creates a problem when a ventricle is blocked—the CSF builds up and creates swelling and the brain is pushed against the skull. This swelling condition is called hydrocephalus (“water head”) and can cause seizures, cognitive problems, and even death if a shunt is not inserted to remove the fluid and pressure.
Autonomic nervous system
Which of the following statements is false?
a. The parasympathetic pathway is responsible for resting the body, while the sympathetic pathway is responsible for preparing for an emergency.
b. Most preganglionic neurons in the sympathetic pathway originate in the spinal cord.
c. Slowing of the heartbeat is a parasympathetic response.
d. Parasympathetic neurons are responsible for releasing norepinephrine on the target organ, while sympathetic neurons are responsible for releasing acetylcholine.
Sympathetic nervous system
The
sympathetic nervous system
is responsible for the “fight or flight” response that occurs when an animal encounters a dangerous situation. One way to remember this is to think of the surprise a person feels when encountering a snake (“snake” and “sympathetic” both begin with “s”). Examples of functions controlled by the sympathetic nervous system include an accelerated heart rate and inhibited digestion. These functions help prepare an organism’s body for the physical strain required to escape a potentially dangerous situation or to fend off a predator.
Most preganglionic neurons in the sympathetic nervous system originate in the spinal cord, as illustrated in Figure 8.3. The axons of these neurons release acetylcholine on postganglionic neurons within the sympathetic ganglia (the sympathetic ganglia form a chain that extends alongside the spinal cord). The acetylcholine activates the postganglionic neurons. Postganglionic neurons then release norepinephrine onto target organs. As anyone who has ever felt a rush before a big test, speech, or athletic event can attest, the effects of the sympathetic nervous system are quite pervasive. This is both because one preganglionic neuron synapses on multiple postganglionic neurons, amplifying the effect of the original synapse, and because the adrenal gland also releases norepinephrine (and the closely related hormone epinephrine) into the bloodstream. The physiological effects of this norepinephrine release include dilating the trachea and bronchi (making it easier for the animal to breathe), increasing heart rate, and moving blood from the skin to the heart, muscles, and brain (so the animal can think and run). The strength and speed of the sympathetic response help an organism avoid danger, and scientists have found evidence that it may also increase LTP—allowing the animal to remember the dangerous situation and avoid it in the future.
Parasympathetic nervous system
While the sympathetic nervous system is activated in stressful situations, the parasympathetic nervous system allows an animal to “rest and digest.” One way to remember this is to think that during a restful situation like a picnic, the parasympathetic nervous system is in control (“picnic” and “parasympathetic” both start with “p”). Parasympathetic preganglionic neurons have cell bodies located in the brainstem and in the sacral (toward the bottom) spinal cord, as shown in Figure 8.3. The axons of the preganglionic neurons release acetylcholine on the postganglionic neurons, which are generally located very near the target organs. Most postganglionic neurons release acetylcholine onto target organs, although some release nitric oxide.
The parasympathetic nervous system resets organ function after the sympathetic nervous system is activated (the common adrenaline dump you feel after a ‘fight-or-flight’ event). Effects of acetylcholine release on target organs include slowing of heart rate, lowered blood pressure, and stimulation of digestion.
Sensory-somatic nervous system
The sensory-somatic nervous system is made up of cranial and spinal nerves and contains both sensory and motor neurons. Sensory neurons transmit sensory information from the skin, skeletal muscle, and sensory organs to the CNS. Motor neurons transmit messages about the desired movement from the CNS to the muscles to make them contract. Without its sensory-somatic nervous system, an animal would be unable to process any information about its environment (what it sees, feels, hears, and so on) and could not control motor movements. Unlike the autonomic nervous system, which has two synapses between the CNS and the target organ, sensory and motor neurons have only one synapse—one ending of the neuron is at the organ and the other directly contacts a CNS neuron. Acetylcholine is the main neurotransmitter released at these synapses.
Humans have 12 cranial nerves , nerves that emerge from or enter the skull (cranium), as opposed to the spinal nerves, which emerge from the vertebral column. Each cranial nerve is accorded a name, which is detailed in Figure 8.4. Some cranial nerves transmit only sensory information. For example, the olfactory nerve transmits information about smells from the nose to the brainstem. Other cranial nerves transmit almost solely motor information. For example, the oculomotor nerve controls the opening and closing of the eyelid and some eye movements. Other cranial nerves contain a mix of sensory and motor fibers. For example, the glossopharyngeal nerve has a role in both taste (sensory) and swallowing (motor).
Spinal nerves transmit sensory and motor information between the spinal cord and the rest of the body. Each of the 31 spinal nerves (in humans) contains both sensory and motor axons. The sensory neuron cell bodies are grouped in structures called dorsal root ganglia and are shown in Figure 8.5. Each sensory neuron has one projection—with a sensory receptor ending in the skin, muscle, or sensory organs—and another that synapses with a neuron in the dorsal spinal cord. Motor neurons have cell bodies in the ventral gray matter of the spinal cord that project to muscle through the ventral root. These neurons are usually stimulated by interneurons within the spinal cord but are sometimes directly stimulated by sensory neurons.
The peripheral nervous system contains both the autonomic and sensory-somatic nervous systems. The autonomic nervous system provides unconscious control over visceral functions and has two divisions: the sympathetic and parasympathetic nervous systems. The sympathetic nervous system is activated in stressful situations to prepare the animal for a “fight or flight” response. The parasympathetic nervous system is active during restful periods. The sensory-somatic nervous system is made of cranial and spinal nerves that transmit sensory information from skin and muscle to the CNS and motor commands from the CNS to the muscles.
When you use a word nerve what does it refer to? What about neuron? How are these terms different and how are they the same in terms of their function | libretexts | 2025-03-17T22:26:53.703637 | 2018-10-18T16:30:14 | {
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"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "8.1: Neuronal Organization",
"author": "Sanja Hinić-Frlog"
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/8%3A_Sensory_Systems/8.2%3A_Brain_Evolution | 8.2: Brain Evolution
- Summarize evolution of key brain features among animals.
Organization of the Central Nervous System
When you’re reading this book, your nervous system is performing several functions simultaneously. The visual system is processing what is seen on the page; the motor system controls the turn of the pages (or click of the mouse); the prefrontal cortex maintains attention. Even fundamental functions, like breathing and regulation of body temperature, are controlled by the nervous system. A nervous system is an organism’s control center: it processes sensory information from outside (and inside) the body and controls all behaviors—from eating to sleeping to finding a mate.
Nervous systems throughout the animal kingdom vary in structure and complexity, as illustrated by the variety of animals shown in Figure 8.6. Some organisms, like sea sponges, lack a true nervous system. Others, like jellyfish, lack a true brain and instead have a system of separate but connected nerve cells (neurons) called a “nerve net.” Echinoderms such as sea stars have nerve cells that are bundled into fibers called nerves. Flatworms of the phylum Platyhelminthes have both a central nervous system (CNS), made up of a small “brain” and two nerve cords and a peripheral nervous system (PNS) containing a system of nerves that extend throughout the body. The insect nervous system is more complex but also fairly decentralized. It contains a brain, ventral nerve cord, and ganglia (clusters of connected neurons). These ganglia can control movements and behaviors without input from the brain. Octopi may have the most complicated of invertebrate nervous systems—they have neurons that are organized in specialized lobes and eyes that are structurally similar to vertebrate species.
Compared to invertebrates, vertebrate nervous systems are more complex, centralized, and specialized. While there is great diversity among different vertebrate nervous systems, they all share a basic structure: a CNS that contains a brain and spinal cord and a PNS made up of peripheral sensory and motor nerves. One interesting difference between the nervous systems of invertebrates and vertebrates is that the nerve cords of many invertebrates are located ventrally whereas the vertebrate spinal cords are located dorsally. There is debate among evolutionary biologists as to whether these different nervous system plans evolved separately or whether the invertebrate body plan arrangement somehow “flipped” during the evolution of vertebrates.
Watch this video of biologist Mark Kirschner discussing the “flipping” phenomenon of vertebrate evolution.
The nervous system is made up of neurons , specialized cells that can receive and transmit chemical or electrical signals, and glia , cells that provide support functions for the neurons by playing an information processing role that is complementary to neurons. A neuron can be compared to an electrical wire—it transmits a signal from one place to another. Glia can be compared to the workers at the electric company who make sure wires go to the right places, maintain the wires, and take down wires that are broken. Although glia have been compared to workers, recent evidence suggests that also usurp some of the signaling functions of neurons.
There is great diversity in the types of neurons and glia that are present in different parts of the nervous system. There are four major types of neurons, and they share several important cellular components.
Brain
The brain is the part of the central nervous system that is contained in the cranial cavity of the skull. It includes the cerebral cortex, limbic system, basal ganglia, thalamus, hypothalamus, and cerebellum. There are three different ways that a brain can be sectioned in order to view internal structures: a sagittal section cuts the brain left to right, as shown in Figure 8.7, a coronal section cuts the brain front to back, as shown in Figure 8.8, and a horizontal section cuts the brain top to bottom.
Cerebral cortex
The outermost part of the brain is a thick piece of nervous system tissue called the cerebral cortex , which is folded into hills called gyri (singular: gyrus) and valleys called sulci (singular: sulcus). The cortex is made up of two hemispheres—right and left—which are separated by a large sulcus. A thick fiber bundle called the corpus callosum (Latin: “tough body”) connects the two hemispheres and allows information to be passed from one side to the other. Although there are some brain functions that are localized more to one hemisphere than the other, the functions of the two hemispheres are largely redundant. In fact, sometimes (very rarely) an entire hemisphere is removed to treat severe epilepsy. While patients do suffer some deficits following the surgery, they can have surprisingly few problems, especially when the surgery is performed on children who have very immature nervous systems.
In other surgeries to treat severe epilepsy, the corpus callosum is cut instead of removing an entire hemisphere. This causes a condition called split-brain, which gives insights into unique functions of the two hemispheres. For example, when an object is presented to patients’ left visual field, they may be unable to verbally name the object (and may claim to not have seen an object at all). This is because the visual input from the left visual field crosses and enters the right hemisphere and cannot then signal to the speech center, which generally is found in the left side of the brain. Remarkably, if a split-brain patient is asked to pick up a specific object out of a group of objects with the left hand, the patient will be able to do so but will still be unable to vocally identify it.
Each cortical hemisphere contains regions called lobes that are involved in different functions. Scientists use various techniques to determine what brain areas are involved in different functions: they examine patients who have had injuries or diseases that affect specific areas and see how those areas are related to functional deficits. They also conduct animal studies where they stimulate brain areas and see if there are any behavioral changes. They use a technique called transmagnetic stimulation (TMS) to temporarily deactivate specific parts of the cortex using strong magnets placed outside the head, and they use functional magnetic resonance imaging (fMRI) to look at changes in oxygenated blood flow in particular brain regions that correlate with specific behavioral tasks. These techniques have given great insight into the functions of different brain regions but have also shown that any given brain area can be involved in more than one behavior or process, and any given behavior or process generally involves neurons in multiple brain areas. That being said, each hemisphere of the mammalian cerebral cortex can be broken down into four functionally and spatially defined lobes: frontal, parietal, temporal, and occipital. Figure 8.9 illustrates these four lobes of the human cerebral cortex.
The frontal lobe is located at the front of the brain, over the eyes. This lobe contains the olfactory bulb, which processes smells. The frontal lobe also contains the motor cortex, which is important for planning and implementing movement. Areas within the motor cortex map to different muscle groups and there is some organization to this map, as shown in Figure 8.10. For example, the neurons that control the movement of the fingers are next to the neurons that control the movement of the hand. Neurons in the frontal lobe also control cognitive functions like maintaining attention, speech, and decision-making. Studies of humans who have damaged their frontal lobes show that parts of this area are involved in personality, socialization, and assessing risk.
The parietal lobe is located at the top of the brain. Neurons in the parietal lobe are involved in speech and also reading. Two of the parietal lobe’s main functions are processing somatosensation —touch sensations like pressure, pain, heat, cold—and processing proprioception —the sense of how parts of the body are oriented in space. The parietal lobe contains a somatosensory map of the body similar to the motor cortex.
The occipital lobe is located at the back of the brain. It is primarily involved in vision—seeing, recognizing, and identifying the visual world.
The temporal lobe is located at the base of the brain by your ears and is primarily involved in processing and interpreting sounds. It also contains the hippocampus (Greek for “seahorse”)—a structure that processes memory formation. The hippocampus is illustrated in Figure 8.9. The role of the hippocampus in memory was partially determined by studying one famous epileptic patient, HM, who had both sides of his hippocampus removed in an attempt to cure his epilepsy. His seizures went away, but he could no longer form new memories (although he could remember some facts from before his surgery and could learn new motor tasks).
Compared to other vertebrates, mammals have exceptionally large brains for their body size. An entire alligator’s brain, for example, would fill about one and a half teaspoons. This increase in brain to body size ratio is especially pronounced in apes, whales, and dolphins. While this increase in overall brain size doubtlessly played a role in the evolution of complex behaviors unique to mammals, it does not tell the whole story. Scientists have found a relationship between the relatively high surface area of the cortex and the intelligence and complex social behaviors exhibited by some mammals. This increased surface area is due, in part, to increased folding of the cortical sheet (more sulci and gyri). For example, a rat cortex is very smooth with very few sulci and gyri ( Figure 8.11 ). Cat and sheep cortices have more sulci and gyri. Chimps, humans, and dolphins have even more.
Basal ganglia
Interconnected brain areas called the basal ganglia (or basal nuclei), shown in Figure 8.7, play important roles in movement control and posture. Damage to the basal ganglia, as in Parkinson’s disease, leads to motor impairments like a shuffling gait when walking. The basal ganglia also regulate motivation. For example, when a wasp sting led to bilateral basal ganglia damage in a 25-year-old businessman, he began to spend all his days in bed and showed no interest in anything or anybody. But when he was externally stimulated—as when someone asked to play a card game with him—he was able to function normally. Interestingly, he and other similar patients do not report feeling bored or frustrated by their state.
Thalamus
The thalamus (Greek for “inner chamber”), illustrated in Figure 8.12, acts as a gateway to and from the cortex. It receives sensory and motor inputs from the body and also receives feedback from the cortex. This feedback mechanism can modulate conscious awareness of sensory and motor inputs depending on the attention and arousal state of the animal. The thalamus helps regulate consciousness, arousal, and sleep states. A rare genetic disorder called fatal familial insomnia causes the degeneration of thalamic neurons and glia. This disorder prevents affected patients from being able to sleep, among other symptoms, and is eventually fatal.
Hypothalamus
Below the thalamus is the hypothalamus , shown in Figure 8.12. The hypothalamus controls the endocrine system by sending signals to the pituitary gland, a pea-sized endocrine gland that releases several different hormones that affect other glands as well as other cells. This relationship means that the hypothalamus regulates important behaviors that are controlled by these hormones. The hypothalamus is the body’s thermostat—it makes sure key functions like food and water intake, energy expenditure, and body temperature are kept at appropriate levels. Neurons within the hypothalamus also regulate circadian rhythms, sometimes called sleep cycles.
Limbic system
The l imbic system is a connected set of structures that regulates emotion, as well as behaviors related to fear and motivation. It plays a role in memory formation and includes parts of the thalamus and hypothalamus as well as the hippocampus. One important structure within the limbic system is a temporal lobe structure called the amygdala (Greek for “almond”), illustrated in Figure 8.12. The two amygdala are important both for the sensation of fear and for recognizing fearful faces. The cingulate gyrus helps regulate emotions and pain.
Cerebellum
The cerebellum (Latin for “little brain”), shown in Figure 8.9, sits at the base of the brain on top of the brainstem. The cerebellum controls balance and aids in coordinating movement and learning new motor tasks.
Brainstem
The brainstem , illustrated in Figure 8.9, connects the rest of the brain with the spinal cord. It consists of the midbrain, medulla oblongata, and the pons. Motor and sensory neurons extend through the brainstem allowing for the relay of signals between the brain and spinal cord. Ascending neural pathways cross in this section of the brain allowing the left hemisphere of the cerebrum to control the right side of the body and vice versa. The brainstem coordinates motor control signals sent from the brain to the body. The brainstem controls several important functions of the body including alertness, arousal, breathing, blood pressure, digestion, heart rate, swallowing, walking, and sensory and motor information integration.
Spinal cord
Connecting to the brainstem and extending down the body through the spinal column is the spinal cord , shown in Figure 8.9. The spinal cord is a thick bundle of nerve tissue that carries information about the body to the brain and from the brain to the body. The spinal cord is contained within the bones of the vertebrate column but is able to communicate signals to and from the body through its connections with spinal nerves (part of the peripheral nervous system). A cross-section of the spinal cord looks like a white oval containing a gray butterfly-shape, as illustrated in Figure 8.13. Myelinated axons make up the “white matter” and neuron and glial cell bodies make up the “gray matter.” Gray matter is also composed of interneurons, which connect two neurons each located in different parts of the body. Axons and cell bodies in the dorsal (facing the back of the animal) spinal cord convey mostly sensory information from the body to the brain. Axons and cell bodies in the ventral (facing the front of the animal) spinal cord primarily transmit signals controlling movement from the brain to the body.
The spinal cord also controls motor reflexes. These reflexes are quick, unconscious movements—like automatically removing a hand from a hot object. Reflexes are so fast because they involve local synaptic connections. For example, the knee reflex that a doctor tests during a routine physical is controlled by a single synapse between a sensory neuron and a motor neuron. While a reflex may only require the involvement of one or two synapses, synapses with interneurons in the spinal column transmit information to the brain to convey what happened (the knee jerked, or the hand was hot).
In the United States, there around 10,000 spinal cord injuries each year. Because the spinal cord is the information superhighway connecting the brain with the body, damage to the spinal cord can lead to paralysis. The extent of the paralysis depends on the location of the injury along the spinal cord and whether the spinal cord was completely severed. For example, if the spinal cord is damaged at the level of the neck, it can cause paralysis from the neck down, whereas damage to the spinal column further down may limit paralysis to the legs. Spinal cord injuries are notoriously difficult to treat because spinal nerves do not regenerate, although ongoing research suggests that stem cell transplants may be able to act as a bridge to reconnect severed nerves. Researchers are also looking at ways to prevent the inflammation that worsens nerve damage after injury. One such treatment is to pump the body with cold saline to induce hypothermia. This cooling can prevent swelling and other processes that are thought to worsen spinal cord injuries.
The ________ lobe contains the visual cortex.
a. frontal
b. parietal
c. temporal
d. occipital
The ________ connects the two cerebral hemispheres.
a. limbic system
b. corpus callosum
c. cerebellum
d. pituitary
Neurons in the ________ control motor reflexes.
a. thalamus
b. spinal cord
c. parietal lobe
d. hippocampus
What methods can be used to determine the function of a particular brain region?
What are the main functions of the spinal cord?
Which brain structure is the oldest on the evolutionary tree of animals? Which animal groups have this structure? | libretexts | 2025-03-17T22:26:53.784758 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/8%3A_Sensory_Systems/8.2%3A_Brain_Evolution",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "8.2: Brain Evolution",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/8%3A_Sensory_Systems/8.3%3A_Special_Senses | 8.3: Special Senses
- Explain how olfactory, gustatory, equilibrium, auditory, and visual sensations are interpreted by animals with reference to signal processing and/or specific ion movements.
Sensory information processing in animals
In more advanced animals, the senses are constantly at work, making the animal aware of stimuli—such as light, or sound, or the presence of a chemical substance in the external environment—and monitoring information about the organism’s internal environment. All bilaterally symmetric animals have a sensory system, and the development of any species’ sensory system has been driven by natural selection; thus, sensory systems differ among species according to the demands of their environments. The shark, unlike most fish predators, is electrosensitive—that is, sensitive to electrical fields produced by other animals in its environment. While it is helpful to this underwater predator, electrosensitivity is a sense not found in most land animals.
Senses provide information about the body and its environment. Humans have five special senses: olfaction (smell), gustation (taste), equilibrium (balance and body position), vision, and hearing. Additionally, we possess general senses, also called somatosensation, which respond to stimuli like temperature, pain, pressure, and vibration. Vestibular sensation , which is an organism’s sense of spatial orientation and balance, proprioception (position of bones, joints, and muscles), and the sense of limb position that is used to track kinesthesia (limb movement) are part of somatosensation. Although the sensory systems associated with these senses are very different, all share a common function: to convert a stimulus (such as light, or sound, or the position of the body) into an electrical signal in the nervous system. This process is called sensory transduction .
There are two broad types of cellular systems that perform sensory transduction. In one, a neuron works with a sensory receptor , a cell, or cell process that is specialized to engage with and detect a specific stimulus. Stimulation of the sensory receptor activates the associated afferent neuron, which carries information about the stimulus to the central nervous system. In the second type of sensory transduction, a sensory nerve ending responds to a stimulus in the internal or external environment: this neuron constitutes the sensory receptor. Free nerve endings can be stimulated by several different stimuli, thus showing little receptor specificity. For example, pain receptors in your gums and teeth may be stimulated by temperature changes, chemical stimulation, or pressure.
Reception
The first step in sensation is reception , which is the activation of sensory receptors by stimuli such as mechanical stimuli (being bent or squished, for example), chemicals, or temperature. The receptor can then respond to the stimuli. The region in space in which a given sensory receptor can respond to a stimulus, be it far away or in contact with the body, is that receptor’s receptive field. Think for a moment about the differences in receptive fields for the different senses. For the sense of touch, a stimulus must come into contact with body. For the sense of hearing, a stimulus can be a moderate distance away (some baleen whale sounds can propagate for many kilometers). For vision, a stimulus can be very far away; for example, the visual system perceives light from stars at enormous distances.
Transduction
The most fundamental function of a sensory system is the translation of a sensory signal to an electrical signal in the nervous system. This takes place at the sensory receptor, and the change in electrical potential that is produced is called the receptor potential . How is sensory input, such as pressure on the skin, changed to a receptor potential? In this example, a type of receptor called a mechanoreceptor (as shown in Figure 8.14) possesses specialized membranes that respond to pressure. Disturbance of these dendrites by compressing them or bending them opens gated ion channels in the plasma membrane of the sensory neuron, changing its electrical potential. Recall that in the nervous system, a positive change of a neuron’s electrical potential (also called the membrane potential), depolarizes the neuron. Receptor potentials are graded potentials: the magnitude of these graded (receptor) potentials varies with the strength of the stimulus. If the magnitude of depolarization is sufficient (that is, if membrane potential reaches a threshold), the neuron will fire an action potential. In most cases, the correct stimulus impinging on a sensory receptor will drive membrane potential in a positive direction, although for some receptors, such as those in the visual system, this is not always the case.
Sensory receptors for different senses are very different from each other, and they are specialized according to the type of stimulus they sense: they have receptor specificity. For example, touch receptors, light receptors, and sound receptors are each activated by different stimuli. Touch receptors are not sensitive to light or sound; they are sensitive only to touch or pressure. However, stimuli may be combined at higher levels in the brain, as happens with olfaction, contributing to our sense of taste.
Encoding and transmission of sensory information
Four aspects of sensory information are encoded by sensory systems: the type of stimulus, the location of the stimulus in the receptive field, the duration of the stimulus, and the relative intensity of the stimulus. Thus, action potentials transmitted over a sensory receptor’s afferent axons encode one type of stimulus, and this segregation of the senses is preserved in other sensory circuits. For example, auditory receptors transmit signals over their own dedicated system, and electrical activity in the axons of the auditory receptors will be interpreted by the brain as an auditory stimulus—a sound.
The intensity of a stimulus is often encoded in the rate of action potentials produced by the sensory receptor. Thus, an intense stimulus will produce a more rapid train of action potentials, and reducing the stimulus will likewise slow the rate of production of action potentials. A second way in which intensity is encoded is by the number of receptors activated. An intense stimulus might initiate action potentials in a large number of adjacent receptors, while a less intense stimulus might stimulate fewer receptors. Integration of sensory information begins as soon as the information is received in the CNS, and the brain will further process incoming signals.
Perception
Perception is an individual’s interpretation of a sensation. Although perception relies on the activation of sensory receptors, perception happens not at the level of the sensory receptor, but at higher levels in the nervous system, in the brain. The brain distinguishes sensory stimuli through a sensory pathway: action potentials from sensory receptors travel along neurons that are dedicated to a particular stimulus. These neurons are dedicated to that particular stimulus and synapse with particular neurons in the brain or spinal cord.
All sensory signals, except those from the olfactory system, are transmitted though the central nervous system and are routed to the thalamus and to the appropriate region of the cortex. Recall that the thalamus is a structure in the forebrain that serves as a clearinghouse and relay station for sensory (as well as motor) signals. When the sensory signal exits the thalamus, it is conducted to the specific area of the cortex (Figure 8.15) dedicated to processing that particular sense.
How are neural signals interpreted? Interpretation of sensory signals between individuals of the same species is largely similar, owing to the inherited similarity of their nervous systems; however, there are some individual differences. A good example of this is individual tolerances to a painful stimulus, such as dental pain, which certainly differ.
Where does perception occur?
a. spinal cord
b. cerebral cortex
c. receptors
d. thalamus
If a person’s cold receptors no longer convert cold stimuli into sensory signals, that person has a problem with the process of ________.
a. reception
b. transmission
c. perception
d. transduction
After somatosensory transduction, the sensory signal travels through the brain as a(n) _____ signal.
a. electrical
b. pressure
c. optical
d. thermal
Somatosensation
Somatosensation is a mixed sensory category and includes all sensation received from the skin and mucous membranes, as well from as the limbs and joints. Somatosensation is also known as tactile sense, or more familiarly, as the sense of touch. Somatosensation occurs all over the exterior of the body and at some interior locations as well. A variety of receptor types—embedded in the skin, mucous membranes, muscles, joints, internal organs, and cardiovascular system—play a role.
Somatosensory receptors
Sensory receptors are classified into five categories: mechanoreceptors, thermoreceptors, proprioceptors, pain receptors, and chemoreceptors. These categories are based on the nature of stimuli each receptor class transduces. What is commonly referred to as “touch” involves more than one kind of stimulus and more than one kind of receptor. Mechanoreceptors in the skin are described as encapsulated (that is, surrounded by a capsule) or unencapsulated (a group that includes free nerve endings). A free nerve ending , as its name implies, is an unencapsulated dendrite of a sensory neuron. Free nerve endings are the most common nerve endings in skin, and they extend into the middle of the epidermis. Free nerve endings are sensitive to painful stimuli, to hot and cold, and to light touch. They are slow to adjust to a stimulus and so are less sensitive to abrupt changes in stimulation.
There are three classes of mechanoreceptors: tactile, proprioceptors, and baroreceptors. Mechanoreceptors sense stimuli due to physical deformation of their plasma membranes. They contain mechanically gated ion channels whose gates open or close in response to pressure, touch, stretching, and sound.” There are four primary tactile mechanoreceptors in human skin: Merkel’s disks, Meissner’s corpuscles, Ruffini endings, and Pacinian corpuscle; two are located toward the surface of the skin and two are located deeper. A fifth type of mechanoreceptor, Krause end bulbs, are found only in specialized regions. Merkel’s disks (shown in Figure 8.16) are found in the upper layers of skin near the base of the epidermis, both in skin that has hair and on glabrous skin, that is, the hairless skin found on the palms and fingers, the soles of the feet, and the lips of humans and other primates. Merkel’s disks are densely distributed in the fingertips and lips. They are slow-adapting, unencapsulated nerve endings, and they respond to light touch. Light touch, also known as discriminative touch, is a light pressure that allows the location of a stimulus to be pinpointed. The receptive fields of Merkel’s disks are small with well-defined borders. That makes them finely sensitive to edges and they come into use in tasks such as typing on a keyboard.
Which of the following statements about mechanoreceptors is false?
a. Pacini corpuscles are found in both glabrous and hairy skin.
b. Merkel’s disks are abundant on the fingertips and lips.
c. Ruffini endings are encapsulated mechanoreceptors.
d. Meissner’s corpuscles extend into the lower dermis.
In proprioception, proprioceptive and kinesthetic signals travel through myelinated afferent neurons running from the spinal cord to the medulla. Neurons are not physically connected, but communicate via neurotransmitters secreted into synapses or “gaps” between communicating neurons. Once in the medulla, the neurons continue carrying the signals to the thalamus.
Muscle spindles are stretch receptors that detect the amount of stretch, or lengthening of muscles. Related to these are Golgi tendon organs , which are tension receptors that detect the force of muscle contraction. Proprioceptive and kinesthetic signals come from limbs. Unconscious proprioceptive signals run from the spinal cord to the cerebellum, the brain region that coordinates muscle contraction, rather than to the thalamus, like most other sensory information.
Barorecptors detect pressure changes in an organ. They are found in the walls of the carotid artery and the aorta where they monitor blood pressure, and in the lungs where they detect the degree of lung expansion. Stretch receptors are found at various sites in the digestive and urinary systems.
In addition to these two types of deeper receptors, there are also rapidly adapting hair receptors, which are found on nerve endings that wrap around the base of hair follicles. There are a few types of hair receptors that detect slow and rapid hair movement, and they differ in their sensitivity to movement. Some hair receptors also detect skin deflection, and certain rapidly adapting hair receptors allow detection of stimuli that have not yet touched the skin.
Integration of signals from mechanoreceptors
The configuration of the different types of receptors working in concert in human skin results in a very refined sense of touch. The nociceptive receptors—those that detect pain—are located near the surface. Small, finely calibrated mechanoreceptors—Merkel’s disks and Meissner’s corpuscles—are located in the upper layers and can precisely localize even gentle touch. The large mechanoreceptors—Pacinian corpuscles and Ruffini endings—are located in the lower layers and respond to deeper touch. (Consider that the deep pressure that reaches those deeper receptors would not need to be finely localized.) Both the upper and lower layers of the skin hold rapidly and slowly adapting receptors. Both primary somatosensory cortex and secondary cortical areas are responsible for processing the complex picture of stimuli transmitted from the interplay of mechanoreceptors.
Density of mechanoreceptors
The distribution of touch receptors in human skin is not consistent over the body. In humans, touch receptors are less dense in skin covered with any type of hair, such as the arms, legs, torso, and face. Touch receptors are denser in glabrous skin (the type found on human fingertips and lips, for example), which is typically more sensitive and is thicker than hairy skin (4 to 5 mm versus 2 to 3 mm).
How is receptor density estimated in a human subject? The relative density of pressure receptors in different locations on the body can be demonstrated experimentally using a two-point discrimination test. In this demonstration, two sharp points, such as two thumbtacks, are brought into contact with the subject’s skin (though not hard enough to cause pain or break the skin). The subject reports if he or she feels one point or two points. If the two points are felt as one point, it can be inferred that the two points are both in the receptive field of a single sensory receptor. If two points are felt as two separate points, each is in the receptive field of two separate sensory receptors. The points could then be moved closer and re-tested until the subject reports feeling only one point, and the size of the receptive field of a single receptor could be estimated from that distance.
Thermoreception
In addition to Krause end bulbs that detect cold and Ruffini endings that detect warmth, there are different types of cold receptors on some free nerve endings: thermoreceptors, located in the dermis, skeletal muscles, liver, and hypothalamus, that are activated by different temperatures. Their pathways into the brain run from the spinal cord through the thalamus to the primary somatosensory cortex. Warmth and cold information from the face travels through one of the cranial nerves to the brain. You know from experience that a tolerably cold or hot stimulus can quickly progress to a much more intense stimulus that is no longer tolerable. Any stimulus that is too intense can be perceived as pain because temperature sensations are conducted along the same pathways that carry pain sensations
Pain
Pain is the name given to nociception , which is the neural processing of injurious stimuli in response to tissue damage. Pain is caused by true sources of injury, such as contact with a heat source that causes a thermal burn or contact with a corrosive chemical. But pain also can be caused by harmless stimuli that mimic the action of damaging stimuli, such as contact with capsaicins, the compounds that cause peppers to taste hot and which are used in self-defense pepper sprays and certain topical medications. Peppers taste “hot” because the protein receptors that bind capsaicin open the same calcium channels that are activated by warm receptors.
Nociception starts at the sensory receptors, but pain, inasmuch as it is the perception of nociception, does not start until it is communicated to the brain. There are several nociceptive pathways to and through the brain. Most axons carrying nociceptive information into the brain from the spinal cord project to the thalamus (as do other sensory neurons) and the neural signal undergoes final processing in the primary somatosensory cortex. Interestingly, one nociceptive pathway projects not to the thalamus but directly to the hypothalamus in the forebrain, which modulates the cardiovascular and neuroendocrine functions of the autonomic nervous system. Recall that threatening—or painful—stimuli stimulate the sympathetic branch of the visceral sensory system, readying a fight-or-flight response.
- View this video that animates the five phases of nociceptive pain.
Taste and smell
Taste, also called gustation , and smell, also called olfaction , are the most interconnected senses in that both involve molecules of the stimulus entering the body and bonding to receptors. Smell lets an animal sense the presence of food or other animals—whether potential mates, predators, or prey—or other chemicals in the environment that can impact their survival. Similarly, the sense of taste allows animals to discriminate between types of foods. While the value of a sense of smell is obvious, what is the value of a sense of taste? Different tasting foods have different attributes, both helpful and harmful. For example, sweet-tasting substances tend to be highly caloric, which could be necessary for survival in lean times. Bitterness is associated with toxicity, and sourness is associated with spoiled food. Salty foods are valuable in maintaining homeostasis by helping the body retain water and by providing ions necessary for cells to function.
Tastes and odors
Both taste and odor stimuli are molecules taken in from the environment. The primary tastes detected by humans are sweet, sour, bitter, salty and umami. The first four tastes need little explanation. The identification of umami as a fundamental taste occurred fairly recently—it was identified in 1908 by Japanese scientist Kikunae Ikeda while he worked with seaweed broth, but it was not widely accepted as a taste that could be physiologically distinguished until many years later. The taste of umami, also known as savoriness, is attributable to the taste of the amino acid L-glutamate. In fact, monosodium glutamate, or MSG, is often used in cooking to enhance the savory taste of certain foods. What is the adaptive value of being able to distinguish umami? Savory substances tend to be high in protein.
All odors that we perceive are molecules in the air we breathe. If a substance does not release molecules into the air from its surface, it has no smell. And if a human or other animal does not have a receptor that recognizes a specific molecule, then that molecule has no smell. Humans have about 350 olfactory receptor subtypes that work in various combinations to allow us to sense about 10,000 different odors. Compare that to mice, for example, which have about 1,300 olfactory receptor types, and therefore probably sense more odors. Both odors and tastes involve molecules that stimulate specific chemoreceptors. Although humans commonly distinguish taste as one sense and smell as another, they work together to create the perception of flavor. A person’s perception of flavor is reduced if he or she has congested nasal passages.
Reception and transduction
Odorants (odor molecules) enter the nose and dissolve in the olfactory epithelium, the mucosa at the back of the nasal cavity (as illustrated in Figure 8.17). The olfactory epithelium is a collection of specialized olfactory receptors in the back of the nasal cavity that spans an area about 5 cm 2 in humans. Recall that sensory cells are neurons. An olfactory receptor , which is a dendrite of a specialized neuron, responds when it binds certain molecules inhaled from the environment by sending impulses directly to the olfactory bulb of the brain. Humans have about 12 million olfactory receptors, distributed among hundreds of different receptor types that respond to different odors. Twelve million seems like a large number of receptors, but compare that to other animals: rabbits have about 100 million, most dogs have about 1 billion, and bloodhounds—dogs selectively bred for their sense of smell—have about 4 billion. The overall size of the olfactory epithelium also differs between species, with that of bloodhounds, for example, being many times larger than that of humans.
Olfactory neurons are bipolar neurons (neurons with two processes from the cell body). Each neuron has a single dendrite buried in the olfactory epithelium, and extending from this dendrite are 5 to 20 receptor-laden, hair-like cilia that trap odorant molecules. The sensory receptors on the cilia are proteins, and it is the variations in their amino acid chains that make the receptors sensitive to different odorants. Each olfactory sensory neuron has only one type of receptor on its cilia, and the receptors are specialized to detect specific odorants, so the bipolar neurons themselves are specialized. When an odorant binds with a receptor that recognizes it, the sensory neuron associated with the receptor is stimulated. Olfactory stimulation is the only sensory information that directly reaches the cerebral cortex, whereas other sensations are relayed through the thalamus.
Pheromones
A pheromone is a chemical released by an animal that affects the behavior or physiology of animals of the same species. Pheromonal signals can have profound effects on animals that inhale them, but pheromones apparently are not consciously perceived in the same way as other odors. There are several different types of pheromones, which are released in urine or as glandular secretions. Certain pheromones are attractants to potential mates, others are repellants to potential competitors of the same sex, and still others play roles in mother-infant attachment. Some pheromones can also influence the timing of puberty, modify reproductive cycles, and even prevent embryonic implantation. While the roles of pheromones in many nonhuman species are important, pheromones have become less important in human behavior over evolutionary time compared to their importance to organisms with more limited behavioral repertoires.
The vomeronasal organ (VNO, or Jacobson’s organ) is a tubular, fluid-filled, olfactory organ present in many vertebrate animals that sits adjacent to the nasal cavity. It is very sensitive to pheromones and is connected to the nasal cavity by a duct. When molecules dissolve in the mucosa of the nasal cavity, they then enter the VNO where the pheromone molecules among them bind with specialized pheromone receptors. Upon exposure to pheromones from their own species or others, many animals, including cats, may display the flehmen response (shown in Figure 8.18), a curling of the upper lip that helps pheromone molecules enter the VNO.
Pheromonal signals are sent, not to the main olfactory bulb, but to a different neural structure that projects directly to the amygdala (recall that the amygdala is a brain center important in emotional reactions, such as fear). The pheromonal signal then continues to areas of the hypothalamus that are key to reproductive physiology and behavior. While some scientists assert that the VNO is apparently functionally vestigial in humans, eve though there is a similar structure located near human nasal cavities, others are researching it as a possible functional system that may, for example, contribute to synchronization of menstrual cycles in women living in close proximity.
Taste
Detecting a taste (gustation) is fairly similar to detecting an odor (olfaction), given that both taste and smell rely on chemical receptors being stimulated by certain molecules. The primary organ of taste is the taste bud. A taste bud is a cluster of gustatory receptors (taste cells) that are located within the bumps on the tongue called papillae (singular: papilla) (illustrated in Figure 8.19). There are several structurally distinct papillae. Filiform papillae, which are located across the tongue, are tactile, providing friction that helps the tongue move substances, and contain no taste cells. In contrast, fungiform papillae, which are located mainly on the anterior two-thirds of the tongue, each contain one to eight taste buds and also have receptors for pressure and temperature. The large circumvallate papillae contain up to 100 taste buds and form a V near the posterior margin of the tongue.
In addition to those two types of chemically and mechanically sensitive papillae are foliate papillae—leaf-like papillae located in parallel folds along the edges and toward the back of the tongue, as seen in the Figure 8.18b micrograph. Foliate papillae contain about 1,300 taste buds within their folds. Finally, there are circumvallate papillae, which are wall-like papillae in the shape of an inverted “V” at the back of the tongue. Each of these papillae is surrounded by a groove and contains about 250 taste buds.
Each taste bud’s taste cells are replaced every 10 to 14 days. These are elongated cells with hair-like processes called microvilli at the tips that extend into the taste bud pore (illustrate in Figure 8.20). Food molecules ( tastants ) are dissolved in saliva, and they bind with and stimulate the receptors on the microvilli. The receptors for tastants are located across the outer portion and front of the tongue, outside of the middle area where the filiform papillae are most prominent.
In humans, there are five primary tastes, and each taste has only one corresponding type of receptor. Thus, like olfaction, each receptor is specific to its stimulus (tastant). Transduction of the five tastes happens through different mechanisms that reflect the molecular composition of the tastant. A salty tastant (containing NaCl) provides the sodium ions (Na + ) that enter the taste neurons and excite them directly. Sour tastants are acids and belong to the thermoreceptor protein family. Binding of an acid or other sour-tasting molecule triggers a change in the ion channel and these increase hydrogen ion (H + ) concentrations in the taste neurons, thus depolarizing them. Sweet, bitter, and umami tastants require a G-protein coupled receptor. These tastants bind to their respective receptors, thereby exciting the specialized neurons associated with them. Both tasting abilities and sense of smell change with age. In humans, the senses decline dramatically by age 50 and continue to decline. A child may find a food to be too spicy, whereas an elderly person may find the same food to be bland and unappetizing.
Smell and taste in the brain
Olfactory neurons project from the olfactory epithelium to the olfactory bulb as thin, unmyelinated axons. The olfactory bulb is composed of neural clusters called glomeruli , and each glomerulus receives signals from one type of olfactory receptor, so each glomerulus is specific to one odorant. From glomeruli, olfactory signals travel directly to the olfactory cortex and then to the frontal cortex and the thalamus. Recall that this is a different path from most other sensory information, which is sent directly to the thalamus before ending up in the cortex. Olfactory signals also travel directly to the amygdala, thereafter reaching the hypothalamus, thalamus, and frontal cortex. The last structure that olfactory signals directly travel to is a cortical center in the temporal lobe structure important in spatial, autobiographical, declarative, and episodic memories. Olfaction is finally processed by areas of the brain that deal with memory, emotions, reproduction, and thought.
Taste neurons project from taste cells in the tongue, esophagus, and palate to the medulla, in the brainstem. From the medulla, taste signals travel to the thalamus and then to the primary gustatory cortex. Information from different regions of the tongue is segregated in the medulla, thalamus, and cortex.
Salty foods activate the taste cells by ______.
a. exciting the taste cell directly
b. causing hydrogen ions to enter the cell
d. causing sodium channels to close
e. binding directly to the receptors
All sensory signals except _____ travel to the _____ in the brain before the cerebral cortex.
a. vision; thalamus
b. olfaction; thalmus
c. vision; cranial nerves
d. olfaction; cranial nerves
From the perspective of the recipient of the signal, in what ways do pheromones differ from other odorants?
What might be the effect on an animal of not being able to perceive taste?
Hearing and vestibular information
Audition , or hearing, is important to humans and to other animals for many different interactions. It enables an organism to detect and receive information about danger, such as an approaching predator, and to participate in communal exchanges like those concerning territories or mating. On the other hand, although it is physically linked to the auditory system, the vestibular system is not involved in hearing. Instead, an animal’s vestibular system detects its own movement, both linear and angular acceleration and deceleration, and balance.
Sound
Auditory stimuli are sound waves, which are mechanical, pressure waves that move through a medium, such as air or water. There are no sound waves in a vacuum since there are no air molecules to move in waves. The speed of sound waves differs, based on altitude, temperature, and medium, but at sea level and a temperature of 20º C (68º F), sound waves travel in the air at about 343 meters per second.
As is true for all waves, there are four main characteristics of a sound wave: frequency, wavelength, period, and amplitude. Frequency is the number of waves per unit of time, and in sound is heard as pitch. High-frequency (≥15.000Hz) sounds are higher-pitched (short wavelength) than low-frequency (long wavelengths; ≤100Hz) sounds. Frequency is measured in cycles per second, and for sound, the most commonly used unit is hertz (Hz), or cycles per second. Most humans can perceive sounds with frequencies between 30 and 20,000 Hz. Women are typically better at hearing high frequencies, but everyone’s ability to hear high frequencies decreases with age. Dogs detect up to about 40,000 Hz; cats, 60,000 Hz; bats, 100,000 Hz; and dolphins 150,000 Hz, and American shad ( Alosa sapidissima ), a fish, can hear 180,000 Hz. Those frequencies above the human range are called ultrasound .
Amplitude, or the dimension of a wave from peak to trough, in sound is heard as volume and is illustrated in Figure 8.21. The sound waves of louder sounds have greater amplitude than those of softer sounds. For sound, volume is measured in decibels (dB). The softest sound that a human can hear is the zero point. Humans speak normally at 60 decibels.
Reception of sound
In mammals, sound waves are collected by the external, cartilaginous part of the ear called the pinna , then travel through the auditory canal and cause vibration of the thin diaphragm called the tympanum or ear drum, the innermost part of the outer ear (illustrated in Figure 8.22). Interior to the tympanum is the middle ear . The middle ear holds three small bones called the ossicles, which transfer energy from the moving tympanum to the inner ear. The three ossicles are the malleus (also known as the hammer), the incus (the anvil), and stapes (the stirrup). The aptly named stapes looks very much like a stirrup. The three ossicles are unique to mammals, and each plays a role in hearing. The malleus attaches at three points to the interior surface of the tympanic membrane. The incus attaches the malleus to the stapes. In humans, the stapes is not long enough to reach the tympanum. If we did not have the malleus and the incus, then the vibrations of the tympanum would never reach the inner ear. These bones also function to collect force and amplify sounds. The ear ossicles are homologous to bones in a fish mouth: the bones that support gills in fish are thought to be adapted for use in the vertebrate ear over evolutionary time. Many animals (frogs, reptiles, and birds, for example) use the stapes of the middle ear to transmit vibrations to the middle ear.
Transduction of Sound
Vibrating objects, such as vocal cords, create sound waves or pressure waves in the air. When these pressure waves reach the ear, the ear transduces this mechanical stimulus (pressure wave) into a nerve impulse (electrical signal) that the brain perceives as sound. The pressure waves strike the tympanum, causing it to vibrate. The mechanical energy from the moving tympanum transmits the vibrations to the three bones of the middle ear. The stapes transmits the vibrations to a thin diaphragm called the oval window , which is the outermost structure of the inner ear . The structures of the inner ear are found in the labyrinth , a bony, hollow structure that is the most interior portion of the ear. Here, the energy from the sound wave is transferred from the stapes through the flexible oval window and to the fluid of the cochlea. The vibrations of the oval window create pressure waves in the fluid (perilymph) inside the cochlea. The cochlea is a whorled structure, like the shell of a snail, and it contains receptors for transduction of the mechanical wave into an electrical signal (as illustrated in Figure 8.23). Inside the cochlea, the basilar membrane is a mechanical analyzer that runs the length of the cochlea, curling toward the cochlea’s center.
The mechanical properties of the basilar membrane change along its length, such that it is thicker, tauter, and narrower at the outside of the whorl (where the cochlea is largest), and thinner, floppier, and broader toward the apex, or center, of the whorl (where the cochlea is smallest). Different regions of the basilar membrane vibrate according to the frequency of the sound wave conducted through the fluid in the cochlea. For these reasons, the fluid-filled cochlea detects different wave frequencies (pitches) at different regions of the membrane. When the sound waves in the cochlear fluid contact the basilar membrane, it flexes back and forth in a wave-like fashion. Above the basilar membrane is the tectorial membrane .
Cochlear implants can restore hearing in people who have a nonfunctional cochlear. The implant consists of a microphone that picks up sound. A speech processor selects sounds in the range of human speech, and a transmitter converts these sounds to electrical impulses, which are then sent to the auditory nerve. Which of the following types of hearing loss would not be restored by a cochlear implant?
a. Hearing loss resulting from absence or loss of hair cells in the organ of Corti.
b. Hearing loss resulting from an abnormal auditory nerve.
c. Hearing loss resulting from fracture of the cochlea.
d. Hearing loss resulting from damage to bones of the middle ear.
The site of transduction is in the organ of Corti (spiral organ). It is composed of hair cells held in place above the basilar membrane like flowers projecting up from soil, with their exposed short, hair-like stereocilia contacting or embedded in the tectorial membrane above them. The inner hair cells are the primary auditory receptors and exist in a single row, numbering approximately 3,500. The stereocilia from inner hair cells extend into small dimples on the tectorial membrane’s lower surface. The outer hair cells are arranged in three or four rows. They number approximately 12,000, and they function to fine tune incoming sound waves. The longer stereocilia that project from the outer hair cells actually attach to the tectorial membrane. All of the stereocilia are mechanoreceptors, and when bent by vibrations they respond by opening a gated ion channel (refer to Figure 8.22). As a result, the hair cell membrane is depolarized, and a signal is transmitted to the chochlear nerve. Intensity (volume) of sound is determined by how many hair cells at a particular location are stimulated.
The hair cells are arranged on the basilar membrane in an orderly way. The basilar membrane vibrates in different regions, according to the frequency of the sound waves impinging on it. Likewise, the hair cells that lay above it are most sensitive to a specific frequency of sound waves. Hair cells can respond to a small range of similar frequencies, but they require stimulation of greater intensity to fire at frequencies outside of their optimal range. The difference in response frequency between adjacent inner hair cells is about 0.2 percent. Compare that to adjacent piano strings, which are about six percent different. Place theory, which is the model for how biologists think pitch detection works in the human ear, states that high frequency sounds selectively vibrate the basilar membrane of the inner ear near the entrance port (the oval window). Lower frequencies travel farther along the membrane before causing appreciable excitation of the membrane. The basic pitch-determining mechanism is based on the location along the membrane where the hair cells are stimulated. The place theory is the first step toward an understanding of pitch perception. Considering the extreme pitch sensitivity of the human ear, it is thought that there must be some auditory “sharpening” mechanism to enhance the pitch resolution.
When sound waves produce fluid waves inside the cochlea, the basilar membrane flexes, bending the stereocilia that attach to the tectorial membrane. Their bending results in action potentials in the hair cells, and auditory information travels along the neural endings of the bipolar neurons of the hair cells (collectively, the auditory nerve) to the brain. When the hairs bend, they release an excitatory neurotransmitter at a synapse with a sensory neuron, which then conducts action potentials to the central nervous system. The cochlear branch of the vestibulocochlear cranial nerve sends information on hearing. The auditory system is very refined, and there is some modulation or “sharpening” built in. The brain can send signals back to the cochlea, resulting in a change of length in the outer hair cells, sharpening or dampening the hair cells’ response to certain frequencies.
Higher processing
The inner hair cells are most important for conveying auditory information to the brain. About 90 percent of the afferent neurons carry information from inner hair cells, with each hair cell synapsing with 10 or so neurons. Outer hair cells connect to only 10 percent of the afferent neurons, and each afferent neuron innervates many hair cells. The afferent, bipolar neurons that convey auditory information travel from the cochlea to the medulla, through the pons and midbrain in the brainstem, finally reaching the primary auditory cortex in the temporal lobe.
Vestibular information
The stimuli associated with the vestibular system are linear acceleration (gravity) and angular acceleration and deceleration. Gravity, acceleration, and deceleration are detected by evaluating the inertia on receptive cells in the vestibular system. Gravity is detected through head position. Angular acceleration and deceleration are expressed through turning or tilting of the head.
The vestibular system has some similarities with the auditory system. It utilizes hair cells just like the auditory system, but it excites them in different ways. There are five vestibular receptor organs in the inner ear: the utricle, the saccule, and three semicircular canals. Together, they make up what’s known as the vestibular labyrinth that is shown in Figure 8.24. The utricle and saccule respond to acceleration in a straight line, such as gravity. The roughly 30,000 hair cells in the utricle and 16,000 hair cells in the saccule lie below a gelatinous layer, with their stereocilia projecting into the gelatin. Embedded in this gelatin are calcium carbonate crystals—like tiny rocks. When the head is tilted, the crystals continue to be pulled straight down by gravity, but the new angle of the head causes the gelatin to shift, thereby bending the stereocilia. The bending of the stereocilia stimulates the neurons, and they signal to the brain that the head is tilted, allowing the maintenance of balance. It is the vestibular branch of the vestibulocochlear cranial nerve that deals with balance.
The fluid-filled semicircular canals are tubular loops set at oblique angles. They are arranged in three spatial planes. The base of each canal has a swelling that contains a cluster of hair cells. The hairs project into a gelatinous cap called the cupula and monitor angular acceleration and deceleration from rotation. They would be stimulated by driving your car around a corner, turning your head, or falling forward. One canal lies horizontally, while the other two lie at about 45 degree angles to the horizontal axis, as illustrated in Figure 8.24. When the brain processes input from all three canals together, it can detect angular acceleration or deceleration in three dimensions. When the head turns, the fluid in the canals shifts, thereby bending stereocilia and sending signals to the brain. Upon cessation accelerating or decelerating—or just moving—the movement of the fluid within the canals slows or stops. For example, imagine holding a glass of water. When moving forward, water may splash backwards onto the hand, and when motion has stopped, water may splash forward onto the fingers. While in motion, the water settles in the glass and does not splash. Note that the canals are not sensitive to velocity itself, but to changes in velocity, so moving forward at 60mph with your eyes closed would not give the sensation of movement, but suddenly accelerating or braking would stimulate the receptors.
Hair cells from the utricle, saccule, and semicircular canals also communicate through bipolar neurons to the cochlear nucleus in the medulla. Cochlear neurons send descending projections to the spinal cord and ascending projections to the pons, thalamus, and cerebellum. Connections to the cerebellum are important for coordinated movements. There are also projections to the temporal cortex, which account for feelings of dizziness; projections to autonomic nervous system areas in the brainstem, which account for motion sickness; and projections to the primary somatosensory cortex, which monitors subjective measurements of the external world and self-movement. People with lesions in the vestibular area of the somatosensory cortex see vertical objects in the world as being tilted. Finally, the vestibular signals project to certain optic muscles to coordinate eye and head movements.
Auditory hair cells are indirectly anchored to the _____.
a. basilar membrane
b. oval window
c. tectorial membrane
d. ossicles
Which of the following are found both in the auditory system and the vestibular system?
a. basilar membrane
b. hair cells
c. semicircular canals
d. ossicles
How would a rise in altitude likely affect the speed of a sound transmitted through air?
How might being in a place with less gravity than Earth has (such as Earth’s moon) affect vestibular sensation, and why?
Vision
Vision is the ability to detect light patterns from the outside environment and interpret them into images. Animals are bombarded with sensory information, and the sheer volume of visual information can be problematic. Fortunately, the visual systems of species have evolved to attend to the most-important stimuli. The importance of vision to humans is further substantiated by the fact that about one-third of the human cerebral cortex is dedicated to analyzing and perceiving visual information.
Light
As with auditory stimuli, light travels in waves. The compression waves that compose sound must travel in a medium—a gas, a liquid, or a solid. In contrast, light is composed of electromagnetic waves and needs no medium; light can travel in a vacuum (Figure 8.25). The behavior of light can be discussed in terms of the behavior of waves and also in terms of the behavior of the fundamental unit of light—a packet of electromagnetic radiation called a photon. A glance at the electromagnetic spectrum shows that visible light for humans is just a small slice of the entire spectrum, which includes radiation that we cannot see as light because it is below the frequency of visible red light and above the frequency of visible violet light.
Certain variables are important when discussing perception of light. Wavelength (which varies inversely with frequency) manifests itself as hue. Light at the red end of the visible spectrum has longer wavelengths (and is lower frequency), while light at the violet end has shorter wavelengths (and is higher frequency). The wavelength of light is expressed in nanometers (nm); one nanometer is one billionth of a meter. Humans perceive light that ranges between approximately 380 nm and 740 nm. Some other animals, though, can detect wavelengths outside of the human range. For example, bees see near-ultraviolet light in order to locate nectar guides on flowers, and some non-avian reptiles sense infrared light (heat that prey gives off).
Wave amplitude is perceived as luminous intensity, or brightness. The standard unit of intensity of light is the candela , which is approximately the luminous intensity of a one common candle.
Light waves travel 299,792 km per second in a vacuum, (and somewhat slower in various media such as air and water), and those waves arrive at the eye as long (red), medium (green), and short (blue) waves. What is termed “white light” is light that is perceived as white by the human eye. This effect is produced by light that stimulates equally the color receptors in the human eye. The apparent color of an object is the color (or colors) that the object reflects. Thus a red object reflects the red wavelengths in mixed (white) light and absorbs all other wavelengths of light.
Anatomy of the eye
The photoreceptive cells of the eye, where transduction of light to nervous impulses occurs, are located in the retina (Figure 8.26) on the inner surface of the back of the eye. But light does not impinge on the retina unaltered. It passes through other layers that process it so that it can be interpreted by the retina (Figure 8.25b). The cornea, the front transparent layer of the eye, and the crystalline lens , a transparent convex structure behind the cornea, both refract (bend) light to focus the image on the retina. The iris , which is conspicuous as the colored part of the eye, is a circular muscular ring lying between the lens and cornea that regulates the amount of light entering the eye. In conditions of high ambient light, the iris contracts, reducing the size of the pupil at its center. In conditions of low light, the iris relaxes and the pupil enlarges.
Which of the following statements about the human eye is false?
a. Rods detect color, while cones detect only shades of gray.
b. When light enters the retina, it passes the ganglion cells and bipolar cells before reaching photoreceptors at the rear of the eye.
c. The iris adjusts the amount of light coming into the eye.
d. The cornea is a protective layer on the front of the eye.
There are two types of photoreceptors in the retina: rods and cones , named for their general appearance as illustrated in Figure 8.27. Rods are strongly photosensitive and are located in the outer edges of the retina. They detect dim light and are used primarily for peripheral and nighttime vision. Cones are weakly photosensitive and are located near the center of the retina. They respond to bright light, and their primary role is in daytime, color vision.
Figure 8.27. Rods and cones are photoreceptors in the retina. Rods respond in low light and can detect only shades of gray. Cones respond in intense light and are responsible for color vision. (credit: modification of work by Piotr Sliwa)
The fovea is the region in the center back of the eye that is responsible for acute vision. The fovea has a high density of cones. When you bring your gaze to an object to examine it intently in bright light, the eyes orient so that the object’s image falls on the fovea. However, when looking at a star in the night sky or other object in dim light, the object can be better viewed by the peripheral vision because it is the rods at the edges of the retina, rather than the cones at the center, that operate better in low light. In humans, cones far outnumber rods in the fovea.
Transduction of light
The rods and cones are the site of transduction of light to a neural signal. Both rods and cones contain photopigments. In vertebrates, the main photopigment, rhodopsin , has two main parts: an opsin, which is a membrane protein (in the form of a cluster of α-helices that span the membrane), and retinal—a molecule that absorbs light. When light hits a photoreceptor, it causes a shape change in the retinal, altering its structure from a bent ( cis ) form of the molecule to its linear ( trans ) isomer. This isomerization of retinal activates the rhodopsin, starting a cascade of events that ends with the closing of Na + channels in the membrane of the photoreceptor. Thus, unlike most other sensory neurons (which become depolarized by exposure to a stimulus) visual receptors become hyperpolarized and thus driven away from threshold (Figure 8.28).
Higher processing
The myelinated axons of ganglion cells make up the optic nerves. Within the nerves, different axons carry different qualities of the visual signal. Some axons constitute the magnocellular (big cell) pathway, which carries information about form, movement, depth, and differences in brightness. Other axons constitute the parvocellular (small cell) pathway, which carries information on color and fine detail. Some visual information projects directly back into the brain, while other information crosses to the opposite side of the brain. This crossing of optical pathways produces the distinctive optic chiasma (Greek, for “crossing”) found at the base of the brain and allows us to coordinate information from both eyes.
Once in the brain, visual information is processed in several places, and its routes reflect the complexity and importance of visual information to humans and other animals. One route takes the signals to the thalamus, which serves as the routing station for all incoming sensory impulses except olfaction. In the thalamus, the magnocellular and parvocellular distinctions remain intact, and there are different layers of the thalamus dedicated to each. When visual signals leave the thalamus, they travel to the primary visual cortex at the rear of the brain. From the visual cortex, the visual signals travel in two directions. One stream that projects to the parietal lobe, in the side of the brain, carries magnocellular (“where”) information. A second stream projects to the temporal lobe and carries both magnocellular (“where”) and parvocellular (“what”) information.
Another important visual route is a pathway from the retina to the superior colliculus in the midbrain, where eye movements are coordinated and integrated with auditory information. Finally, there is the pathway from the retina to the suprachiasmatic nucleus (SCN) of the hypothalamus. The SCN is a cluster of cells that is considered to be the body’s internal clock, which controls our circadian (day-long) cycle. The SCN sends information to the pineal gland, which is important in sleep/wake patterns and annual cycles.
How could the pineal gland, the brain structure that plays a role in annual cycles, use visual information from the suprachiasmatic nucleus of the hypothalamus?
Why is it easier to see images at night using peripheral, rather than the central, vision?
a. Cones are denser in the periphery of the retina.
b. Bipolar cells are denser in the periphery of the retina.
c. Rods are denser in the periphery of the retina.
d. The optic nerve exits at the periphery of the retina.
What sensory information do animals use that human cannot? Provide three different examples and briefly describe each. | libretexts | 2025-03-17T22:26:53.993594 | 2018-10-18T16:30:14 | {
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"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/8%3A_Sensory_Systems/8.3%3A_Special_Senses",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "8.3: Special Senses",
"author": "Sanja Hinić-Frlog"
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/9%3A_Maintaining_Internal_Balance | 9: Maintaining Internal Balance Last updated Save as PDF Page ID 9596 Sanja Hinić-Frlog University of Toronto Mississauga 9.1: Interpreting External Cues 9.2: Signaling Pathways, Hormones and Endocrine System 9.3: Examples of Internal Balance | libretexts | 2025-03-17T22:26:54.074220 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/9%3A_Maintaining_Internal_Balance",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "9: Maintaining Internal Balance",
"author": "Sanja Hinić-Frlog"
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/9%3A_Maintaining_Internal_Balance/9.1%3A_Interpreting_External_Cues | 9.1: Interpreting External Cues
- Explain various external cues and their relevance for animal internal balance
An animal’s endocrine system controls body processes through the production, secretion, and regulation of hormones, which serve as chemical “messengers” functioning in cellular and organ activity and, ultimately, maintaining the body’s homeostasis. The endocrine system plays a role in growth, metabolism, and sexual development. In humans, common endocrine system diseases include thyroid disease and diabetes mellitus. In organisms that undergo metamorphosis, the process is controlled by the endocrine system. The transformation from tadpole to frog, for example, is complex and nuanced to adapt to specific environments and ecological circumstances (Figure \(\PageIndex{1}\)). The endocrine system also plays an important role in the timing of migration in animals that live in different regions in summer than in winter.
How does interpretation of external environment impact animal survival or internal environment? Provide at least one example | libretexts | 2025-03-17T22:26:54.130387 | 2018-10-18T16:30:14 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/9%3A_Maintaining_Internal_Balance/9.1%3A_Interpreting_External_Cues",
"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "9.1: Interpreting External Cues",
"author": "Sanja Hinić-Frlog"
} |
https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/9%3A_Maintaining_Internal_Balance/9.2%3A_Signaling_Pathways_Hormones_and_Endocrine_System | 9.2: Signaling Pathways, Hormones and Endocrine System
- Describe general cell signaling strategies and explain types of hormones and their signaling pathways in endocrine communication
Chemical signaling
There are two kinds of communication in the world of living cells. Communication between cells is called intercellular signaling, and communication within a cell is called intracellular signaling. An easy way to remember the distinction is by understanding the Latin origin of the prefixes: inter- means “between” (for example, intersecting lines are those that cross each other) and intra- means “inside” (like intravenous).
Chemical signals are released by signaling cells in the form of small, usually volatile or soluble molecules called ligands. A ligand is a molecule that binds another specific molecule, in some cases, delivering a signal in the process. Ligands can thus be thought of as signaling molecules. Ligands interact with proteins in target cells, which are cells that are affected by chemical signals; these proteins are also called receptors. Ligands and receptors exist in several varieties; however, a specific ligand will have a specific receptor that typically binds only that ligand.
Forms of Signaling
There are four categories of chemical signaling found in multicellular organisms: paracrine signaling, endocrine signaling, autocrine signaling, and direct signaling across gap junctions (Figure 9.2). The main difference between the different categories of signaling is the distance that the signal travels through the organism to reach the target cell. Not all cells are affected by the same signals.
In chemical signaling, a cell may target itself (autocrine signaling), a cell connected by gap junctions, a nearby cell (paracrine signaling), or a distant cell (endocrine signaling). Paracrine signaling acts on nearby cells, endocrine signaling uses the circulatory system to transport ligands, and autocrine signaling acts on the signaling cell. Signaling via gap junctions involves signaling molecules moving directly between adjacent cells.
Paracrine signaling
Signals that act locally between cells that are close together are called paracrine signals. Paracrine signals move by diffusion through the extracellular matrix. These types of signals usually elicit quick responses that last only a short amount of time. In order to keep the response localized, paracrine ligand molecules are normally quickly degraded by enzymes or removed by neighboring cells. Removing the signals will reestablish the concentration gradient for the signal, allowing them to quickly diffuse through the intracellular space if released again.
One example of paracrine signaling is the transfer of signals across synapses between nerve cells. A nerve cell consists of a cell body, several short, branched extensions called dendrites that receive stimuli, and a long extension called an axon, which transmits signals to other nerve cells or muscle cells. The junction between nerve cells where signal transmission occurs is called a synapse. A synaptic signal is a chemical signal that travels between nerve cells. Signals within the nerve cells are propagated by fast-moving electrical impulses. When these impulses reach the end of the axon, the signal continues on to a dendrite of the next cell by the release of chemical ligands called neurotransmitters by the presynaptic cell (the cell emitting the signal). The neurotransmitters are transported across the very small distances between nerve cells, which are called chemical synapses (Figure 9.3). The small distance between nerve cells allows the signal to travel quickly; this enables an immediate response, such as, Take your hand off the stove!
When the neurotransmitter binds the receptor on the surface of the postsynaptic cell, the electrochemical potential of the target cell changes, and the next electrical impulse is launched. The neurotransmitters that are released into the chemical synapse are degraded quickly or get reabsorbed by the presynaptic cell so that the recipient nerve cell can recover quickly and be prepared to respond rapidly to the next synaptic signal.
The distance between the presynaptic cell and the postsynaptic cell—called the synaptic gap—is very small and allows for rapid diffusion of the neurotransmitter. Enzymes in the synapatic cleft degrade some types of neurotransmitters to terminate the signal.
Endocrine signaling
Signals from distant cells are called endocrine signals, and they originate from endocrine cells. (In the body, many endocrine cells are located in endocrine glands, such as the thyroid gland, the hypothalamus, and the pituitary gland.) These types of signals usually produce a slower response but have a longer-lasting effect. The ligands released in endocrine signaling are called hormones, signaling molecules that are produced in one part of the body but affect other body regions some distance away.
Hormones travel the large distances between endocrine cells and their target cells via the bloodstream, which is a relatively slow way to move throughout the body. Because of their form of transport, hormones get diluted and are present in low concentrations when they act on their target cells. This is different from paracrine signaling, in which local concentrations of ligands can be very high.
Autocrine signaling
Autocrine signals are produced by signaling cells that can also bind to the ligand that is released. This means the signaling cell and the target cell can be the same or a similar cell (the prefix auto- means self, a reminder that the signaling cell sends a signal to itself). This type of signaling often occurs during the early development of an organism to ensure that cells develop into the correct tissues and take on the proper function. Autocrine signaling also regulates pain sensation and inflammatory responses. Further, if a cell is infected with a virus, the cell can signal itself to undergo programmed cell death, killing the virus in the process. In some cases, neighboring cells of the same type are also influenced by the released ligand. In embryological development, this process of stimulating a group of neighboring cells may help to direct the differentiation of identical cells into the same cell type, thus ensuring the proper developmental outcome.
Direct signaling across gap junctions
Gap junctions in animals and plasmodesmata in plants are connections between the plasma membranes of neighboring cells. These water-filled channels allow small signaling molecules, called intracellular mediators, to diffuse between the two cells. Small molecules, such as calcium ions (Ca2+), are able to move between cells, but large molecules like proteins and DNA cannot fit through the channels. The specificity of the channels ensures that the cells remain independent but can quickly and easily transmit signals. The transfer of signaling molecules communicates the current state of the cell that is directly next to the target cell; this allows a group of cells to coordinate their response to a signal that only one of them may have received. In plants, plasmodesmata are ubiquitous, making the entire plant into a giant, communication network.
Types of Receptors
Receptors are protein molecules in the target cell or on its surface that bind ligand. There are two types of receptors, internal receptors and cell-surface receptors.
Internal receptors
Internal receptors, also known as intracellular or cytoplasmic receptors, are found in the cytoplasm of the cell and respond to hydrophobic ligand molecules that are able to travel across the plasma membrane. Once inside the cell, many of these molecules bind to proteins that act as regulators of mRNA synthesis (transcription) to mediate gene expression. Gene expression is the cellular process of transforming the information in a cell’s DNA into a sequence of amino acids, which ultimately forms a protein. When the ligand binds to the internal receptor, a conformational change is triggered that exposes a DNA-binding site on the protein. The ligand-receptor complex moves into the nucleus, then binds to specific regulatory regions of the chromosomal DNA and promotes the initiation of transcription (Figure 9.4). Transcription is the process of copying the information in a cells DNA into a special form of RNA called messenger RNA (mRNA); the cell uses information in the mRNA (which moves out into the cytoplasm and associates with ribosomes) to link specific amino acids in the correct order, producing a protein. Internal receptors can directly influence gene expression without having to pass the signal on to other receptors or messengers.
Hydrophobic signaling molecules typically diffuse across the plasma membrane and interact with intracellular receptors in the cytoplasm. Many intracellular receptors are transcription factors that interact with DNA in the nucleus and regulate gene expression.
Cell-surface receptors
Cell-surface receptors, also known as transmembrane receptors, are cell surface, membrane-anchored (integral) proteins that bind to external ligand molecules. This type of receptor spans the plasma membrane and performs signal transduction, in which an extracellular signal is converted into an intercellular signal. Ligands that interact with cell-surface receptors do not have to enter the cell that they affect. Cell-surface receptors are also called cell-specific proteins or markers because they are specific to individual cell types.
Because cell-surface receptor proteins are fundamental to normal cell functioning, it should come as no surprise that a malfunction in any one of these proteins could have severe consequences. Errors in the protein structures of certain receptor molecules have been shown to play a role in hypertension (high blood pressure), asthma, heart disease, and cancer.
Each cell-surface receptor has three main components: an external ligand-binding domain, a hydrophobic membrane-spanning region, and an intracellular domain inside the cell. The ligand-binding domain is also called the extracellular domain. The size and extent of each of these domains vary widely, depending on the type of receptor. Cell-surface receptors are involved in most of the signaling in multicellular organisms. There are three general categories of cell-surface receptors: ion channel-linked receptors, G-protein-linked receptors, and enzyme-linked receptors.
Ion channel-linked receptors bind a ligand and open a channel through the membrane that allows specific ions to pass through. To form a channel, this type of cell-surface receptor has an extensive membrane-spanning region. In order to interact with the phospholipid fatty acid tails that form the center of the plasma membrane, many of the amino acids in the membrane-spanning region are hydrophobic in nature. Conversely, the amino acids that line the inside of the channel are hydrophilic to allow for the passage of water or ions. When a ligand binds to the extracellular region of the channel, there is a conformational change in the proteins structure that allows ions such as sodium, calcium, magnesium, and hydrogen to pass through (Figure 9.5)
G-protein-linked receptors bind a ligand and activate a membrane protein called a G-protein. The activated G-protein then interacts with either an ion channel or an enzyme in the membrane (Figure 9.6). All G-protein-linked receptors have seven transmembrane domains, but each receptor has its own specific extracellular domain and G-protein-binding site.
Cell signaling using G-protein-linked receptors occurs as a cyclic series of events. Before the ligand binds, the inactive G-protein can bind to a newly revealed site on the receptor specific for its binding. Once the G-protein binds to the receptor, the resultant shape change activates the G-protein, which releases GDP and picks up GTP. The subunits of the G-protein then split into the α subunit and the βγ subunit. One or both of these G-protein fragments may be able to activate other proteins as a result. After awhile, the GTP on the active α subunit of the G-protein is hydrolyzed to GDP and the βγ subunit is deactivated. The subunits reassociate to form the inactive G-protein and the cycle begins anew.
G-protein-linked receptors have been extensively studied and much has been learned about their roles in maintaining health. Bacteria that are pathogenic to humans can release poisons that interrupt specific G-protein-linked receptor function, leading to illnesses such as pertussis, botulism, and cholera. In cholera (Figure 9.7), for example, the water-borne bacterium Vibrio cholerae produces a toxin, choleragen, that binds to cells lining the small intestine. The toxin then enters these intestinal cells, where it modifies a G-protein that controls the opening of a chloride channel and causes it to remain continuously active, resulting in large losses of fluids from the body and potentially fatal dehydration as a result.
Enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme. In some cases, the intracellular domain of the receptor itself is an enzyme. Other enzyme-linked receptors have a small intracellular domain that interacts directly with an enzyme. The enzyme-linked receptors normally have large extracellular and intracellular domains, but the membrane-spanning region consists of a single alpha-helical region of the peptide strand. When a ligand binds to the extracellular domain, a signal is transferred through the membrane, activating the enzyme. Activation of the enzyme sets off a chain of events within the cell that eventually leads to a response. One example of this type of enzyme-linked receptor is the tyrosine kinase receptor (Figure 9.8). A kinase is an enzyme that transfers phosphate groups from ATP to another protein. The tyrosine kinase receptor transfers phosphate groups to tyrosine molecules (tyrosine residues). First, signaling molecules bind to the extracellular domain of two nearby tyrosine kinase receptors. The two neighboring receptors then bond together, or dimerize. Phosphates are then added to tyrosine residues on the intracellular domain of the receptors (phosphorylation). The phosphorylated residues can then transmit the signal to the next messenger within the cytoplasm.
Signaling molecules
Produced by signaling cells and the subsequent binding to receptors in target cells, ligands act as chemical signals that travel to the target cells to coordinate responses. The types of molecules that serve as ligands are incredibly varied and range from small proteins to small ions like calcium (Ca2+).
Small hydrophobic ligands
Small hydrophobic ligands can directly diffuse through the plasma membrane and interact with internal receptors. Important members of this class of ligands are the steroid hormones. Steroids are lipids that have a hydrocarbon skeleton with four fused rings; different steroids have different functional groups attached to the carbon skeleton. Steroid hormones include the female sex hormone, estradiol, which is a type of estrogen; the male sex hormone, testosterone; and cholesterol, which is an important structural component of biological membranes and a precursor of steriod hormones ( Figure 9.9 ). Other hydrophobic hormones include thyroid hormones and vitamin D. In order to be soluble in blood, hydrophobic ligands must bind to carrier proteins while they are being transported through the bloodstream.
Figure 9.9. Steroid hormones have similar chemical structures to their precursor, cholesterol. Because these molecules are small and hydrophobic, they can diffuse directly across the plasma membrane into the cell, where they interact with internal receptors.
Water-soluble ligands
Water-soluble ligands are polar and therefore cannot pass through the plasma membrane unaided; sometimes, they are too large to pass through the membrane at all. Instead, most water-soluble ligands bind to the extracellular domain of cell-surface receptors. This group of ligands is quite diverse and includes small molecules, peptides, and proteins.
Other ligands
Nitric oxide (NO) is a gas that also acts as a ligand. It is able to diffuse directly across the plasma membrane, and one of its roles is to interact with receptors in smooth muscle and induce relaxation of the tissue. NO has a very short half-life and therefore only functions over short distances. Nitroglycerin, a treatment for heart disease, acts by triggering the release of NO, which causes blood vessels to dilate (expand), thus restoring blood flow to the heart. NO has become better known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction, such as Viagra (erection involves dilated blood vessels).
Hormones
Hormones mediate changes in target cells by binding to specific hormone receptors . In this way, even though hormones circulate throughout the body and come into contact with many different cell types, they only affect cells that possess the necessary receptors. Receptors for a specific hormone may be found on many different cells or may be limited to a small number of specialized cells. For example, thyroid hormones act on many different tissue types, stimulating metabolic activity throughout the body. Cells can have many receptors for the same hormone but often also possess receptors for different types of hormones. The number of receptors that respond to a hormone determines the cell’s sensitivity to that hormone, and the resulting cellular response. Additionally, the number of receptors that respond to a hormone can change over time, resulting in increased or decreased cell sensitivity. In up-regulation , the number of receptors increases in response to rising hormone levels, making the cell more sensitive to the hormone and allowing for more cellular activity. When the number of receptors decreases in response to rising hormone levels, called down-regulation , cellular activity is reduced.
Receptor binding alters cellular activity and results in an increase or decrease in normal body processes. Depending on the location of the protein receptor on the target cell and the chemical structure of the hormone, hormones can mediate changes directly by binding to intracellular hormone receptors and modulating gene transcription, or indirectly by binding to cell surface receptors and stimulating signaling pathways.
Intracellular hormone receptors
Lipid-derived (soluble) hormones such as steroid hormones diffuse across the membranes of the endocrine cell. Once outside the cell, they bind to transport proteins that keep them soluble in the bloodstream. At the target cell, the hormones are released from the carrier protein and diffuse across the lipid bilayer of the plasma membrane of cells. The steroid hormones pass through the plasma membrane of a target cell and adhere to intracellular receptors residing in the cytoplasm or in the nucleus. The cell signaling pathways induced by the steroid hormones regulate specific genes on the cell’s DNA. The hormones and receptor complex act as transcription regulators by increasing or decreasing the synthesis of mRNA molecules of specific genes. This, in turn, determines the amount of corresponding protein that is synthesized by altering gene expression. This protein can be used either to change the structure of the cell or to produce enzymes that catalyze chemical reactions. In this way, the steroid hormone regulates specific cell processes as illustrated in Figure 9.1.
Other lipid-soluble hormones that are not steroid hormones, such as vitamin D and thyroxine, have receptors located in the nucleus. The hormones diffuse across both the plasma membrane and the nuclear envelope, then bind to receptors in the nucleus. The hormone-receptor complex stimulates transcription of specific genes.
Plasma membrane hormone receptors
Amino acid derived hormones and polypeptide hormones are not lipid-derived (lipid-soluble) and therefore cannot diffuse through the plasma membrane of cells. Lipid insoluble hormones bind to receptors on the outer surface of the plasma membrane, via plasma membrane hormone receptors . Unlike steroid hormones, lipid insoluble hormones do not directly affect the target cell because they cannot enter the cell and act directly on DNA. Binding of these hormones to a cell surface receptor results in activation of a signaling pathway; this triggers intracellular activity and carries out the specific effects associated with the hormone. In this way, nothing passes through the cell membrane; the hormone that binds at the surface remains at the surface of the cell while the intracellular product remains inside the cell. The hormone that initiates the signaling pathway is called a first messenger , which activates a second messenger in the cytoplasm, as illustrated in Figure 9.6.
One very important second messenger is cyclic AMP (cAMP). When a hormone binds to its membrane receptor, a G-protein that is associated with the receptor is activated; G-proteins are proteins separate from receptors that are found in the cell membrane. When a hormone is not bound to the receptor, the G-protein is inactive and is bound to guanosine diphosphate, or GDP. When a hormone binds to the receptor, the G-protein is activated by binding guanosine triphosphate, or GTP, in place of GDP. After binding, GTP is hydrolysed by the G-protein into GDP and becomes inactive.The activated G-protein in turn activates a membrane-bound enzyme called adenylyl cyclase . Adenylyl cyclase catalyzes the conversion of ATP to cAMP. cAMP, in turn, activates a group of proteins called protein kinases, which transfer a phosphate group from ATP to a substrate molecule in a process called phosphorylation. The phosphorylation of a substrate molecule changes its structural orientation, thereby activating it. These activated molecules can then mediate changes in cellular processes.The effect of a hormone is amplified as the signaling pathway progresses. The binding of a hormone at a single receptor causes the activation of many G-proteins, which activates adenylyl cyclase. Each molecule of adenylyl cyclase then triggers the formation of many molecules of cAMP. Further amplification occurs as protein kinases, once activated by cAMP, can catalyze many reactions. In this way, a small amount of hormone can trigger the formation of a large amount of cellular product. To stop hormone activity, cAMP is deactivated by the cytoplasmic enzyme phosphodiesterase , or PDE. PDE is always present in the cell and breaks down cAMP to control hormone activity, preventing overproduction of cellular products.The specific response of a cell to a lipid insoluble hormone depends on the type of receptors that are present on the cell membrane and the substrate molecules present in the cell cytoplasm. Cellular responses to hormone binding of a receptor include altering membrane permeability and metabolic pathways, stimulating synthesis of proteins and enzymes, and activating hormone release.
A new antagonist molecule has been discovered that binds to and blocks plasma membrane receptors. What effect will this antagonist have on testosterone, a steroid hormone?
a. It will block testosterone from binding to its receptor.
b. It will block testosterone from activating cAMP signaling.
c. It will increase testosterone-mediated signaling.
d. It will not affect testosterone-mediated signaling.
What effect will a cAMP inhibitor have on a peptide hormone-mediated signaling pathway?
a. It will prevent the hormone from binding its receptor.
b. It will prevent activation of a G-protein.
c. It will prevent activation of adenylate cyclase.
d. It will prevent the activation of protein kinases.
Name two important functions of hormone receptors.
How can hormones mediate changes?
List all the structures that are involved in general hormonal signaling. Once you have all the structures listed, arrange them from small to large. Next, arrange them in order of events during hormonal signaling, from first to last. | libretexts | 2025-03-17T22:26:54.217330 | 2018-10-18T16:30:14 | {
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"title": "9.2: Signaling Pathways, Hormones and Endocrine System",
"author": "Sanja Hinić-Frlog"
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https://med.libretexts.org/Bookshelves/Veterinary_Medicine/Book%3A_Introductory_Animal_Physiology_(Hinic-Frlog)/9%3A_Maintaining_Internal_Balance/9.3%3A_Examples_of_Internal_Balance | 9.3: Examples of Internal Balance
- Use different hormone examples (molting, digestion and circadian rhythm-related) to explain their role in maintaining homeostasis
Maintaining homeostasis within the body requires the coordination of many different systems and organs. Communication between neighboring cells, and between cells and tissues in distant parts of the body, occurs through the release of chemicals called hormones. Hormones are released into body fluids (usually blood) that carry these chemicals to their target cells. At the target cells, which are cells that have a receptor for a signal or ligand from a signal cell, the hormones elicit a response. The cells, tissues, and organs that secrete hormones make up the endocrine system. Examples of glands of the endocrine system include the adrenal glands, which produce hormones such as epinephrine and norepinephrine that regulate responses to stress, and the thyroid gland, which produces thyroid hormones that regulate metabolic rates.
Pick a figure from this paper that explains this hormone effect the best. Which figure did you choose and why?
Find a paper that documents the effect of a hormone on an animal reproductive system. Which paper did you choose?
Show this figure to a peer in your class and ask them the following question: based on this figure what are the main results from this study? How was the data collected? What was the hypothesis that was tested / study question that was asked based on the figure? You should accept the same challenge from your peer and the figure from the paper they have selected. This type of activity is called Reverse Journal Club and it will help you learn more about the endocrine system and reading scientific articles. | libretexts | 2025-03-17T22:26:54.276827 | 2018-10-18T16:30:14 | {
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"book_url": "https://commons.libretexts.org/book/med-9530",
"title": "9.3: Examples of Internal Balance",
"author": "Sanja Hinić-Frlog"
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/01%3A_Some_Basic_Algebra | 1: Some Basic Algebra Last updated Save as PDF Page ID 125025 1.1: Constants, Variables, and Expressions | libretexts | 2025-03-17T22:27:19.971671 | 2023-04-03T20:57:40 | {
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https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/01%3A_Some_Basic_Algebra/1.01%3A_Constants_Variables_and_Expressions | 1.1: Constants, Variables, and Expressions
Constants and variables, at least one of these objects appear in every mathematical expression one can imagine. Let’s get a sense of just what they are.
A VARIABLE is a quantity that has a capacity for change in a particular context.
A CONSTANT is a quantity that has no capacity for change in a particular context.
Let’s put both in the context of hiring a programmer to write a program that performs some particular task. Suppose there are three programmers, A , B , and C , we are considering.
Programmer A charges a flat fee of $25,000 for writing the program.
Programmer A’s fee is constant.
In this context, the fee has no capacity for change. The fee is $25,000, no more, no less. Outside this context, maybe writing a less complicated
Programmer B charges $100 hour for writing the program.
Programmer B's fee is variable.
In this context, the fee has the capacity for change. The total fee varies with the amount of time taken to write the program.
Programmer C charges a flat fee of $15,000 and $100/whole hour (1, 2, 3, …, 50) for writing the program.
Programmer C’s fee is variable since the total fee varies since it has the capacity for change. The fee varies with the amount of time taken to write the program. Programmer C's fee structure comprises both a constant, the flat fee of $15,000, and a variable, the $100/ whole hour. But because it contains a variable, the entire quantity is variable.
A bit more detail
It is convenient to think of a variable as a container that can hold different objects at different times. In the programmer example, we might let the letter \(x\) represent the number of hours Programmer C takes to write the program. The number of hours can vary from, say, 1 to 50. Think of \(x\) as a container into which could be placed the numbers 1, 2, 3, and so on up to and including 50.
- If C takes only 1 hour to write the program, think of the number 1 being placed into the container named letter \(x\) . Then C ’s total fee would be \(\$ 15,000+1 \times \$ 100=\$ 15,100\)
- If C takes 2 hours to write the program, think of the number 2 being placed into the container named letter \(x\) . Then C ’s total fee would be \(\$ 15,000+2 \times \$ 100=\$ 15,200\)
- If C takes 50 hours to write the program, think of the number 50 being placed into the container named letter \(x\) . Then C ’s total fee would be \(\$ 15,000+50 \times \$ 100=\$ 20,000\)
Mathematical Expressions
For example, using the letter \(x\) to represent the number of hours C takes to write the program, we could express Programmer C ’s total fee with the expression
\(15,000+100 x\)
The variable \(x\) (the container named \(x\) ) can hold, one at a time, any of the fifty numbers 1, 2, 3, … 50. See this expression as a set of computing instructions that take an input value, one of the numbers 1, 2, 3, … 50, and converts it to a single output value.
Using Teechnology
We can use technology to evaluate expressions.
Go to www.wolframalpha.com
To evaluate \(15,000+100 x\) at \(x = 14\), use the “evaluate” command. Enter evaluate \(15000 + 100x, x = 14\) in the entry field. Wolframalpha tells you what it thinks you entered, then tells you its answer. In this case, \(15000+100 x, x=14\) .
To evaluate \(15,000+100x\) at \(x = 14\) through 18, use the “table” command. Enter table \(15000 + 100x, x = 14\).. 18 in the entry field. Wolframalpha tells you what it thinks you entered, then tells you its answer. In this case it shows you a table with answers for \(15000+100x, x = 14..18\) .
Try These
Suppose a subscription to a photograph service costs $50 year and that each downloaded photograph costs $2.
- Which of the two quantities is the variable quantity?
- Which of the two quantities is the constant?
- Write the expression that produces the annual cost of subscribing and downloading \(x\) number of photographs.
- What is the annual cost of subscribing and downloading 20 photographs?
- Answer
-
- The variable quantity is the download cost.
- The constant is the fixed service cost.
- Annual cost = 50 + 2 \(x\) , where \(x\) represents the number of downloaded photographs.
- Annual cost = 50 + 2 \(\bullet\) 20 = 50 + 40 = 90 The annual cost of downloading 20 photos is $90
What is the minimum number of cookies a person must eat to be happy? What is the minimum number of cookies beyond that number must eat to feel sick? These numbers are likely different for all of us. Let the variable \(x\) represent the minimum number of cookies someone must eat to be happy, and the variable \(y\) be the minimum number that makes that person sick.
- How many variable quantities are in this problem?
- Are there any constants in this problem?
- Answer
-
- There are 2 variable quantities in this problem.
- There are no constants in this problem | libretexts | 2025-03-17T22:27:20.045795 | 2023-04-03T20:57:42 | {
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https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/02%3A_Vectors_In_Two_Dimensions | 2: Vectors In Two Dimensions Last updated Save as PDF Page ID 125027 2.1: Vectors 2.2: Addition, Subtraction, and Scalar Multiplication of Vectors 2.3: Magnitude, Direction, and Components of a Vector 2.4: The Dot Product of Two Vectors, the Length of a Vector, and the Angle Between Two Vectors 2.5: Parallel and Perpendicular Vectors, The Unit Vector 2.6: The Vector Projection of One Vector onto Another | libretexts | 2025-03-17T22:27:20.124002 | 2023-04-03T20:57:44 | {
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https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/02%3A_Vectors_In_Two_Dimensions/2.01%3A_Vectors | 2.1: Vectors
Vectors are fundamental objects in applied mathematics; they efficiently convey information about a mathematical or physical object. Let’s get a sense of what they are.
A VECTOR is a representation of an object that has both direction and magnitude. By direction, we mean the place toward which something faces, and by magnitude, we mean the size of something.
A vector can be depicted visually by an arrow, with an initial point called the tail and a terminal point called the head. The length of the arrow represents the vector’s magnitude.
An example of a vector is a car’s velocity. Velocity is a vector since it has both magnitude (speed) and direction. A car might be moving west at 60 mph. Other examples of vectors are displacement, acceleration, and force.
The temperature of some medium is not a vector since it has only magnitude. But if the medium is being heated, its temperature is increasing and has a direction; it is going upward. The increase or decrease in temperature is a vector.
Vectors in Standard Position
Components of a Vector
For example, the vector \(\vec{v}\) in the diagram can be broken into two components,
- its horizontal, or \(x\)-component, and
- its vertical, or \(y\)-component.
The vector \(\vec{v}\) in component form is expressed using angle brackets as \(\vec{v}=\langle 3,6\rangle\), where
- the first component, 3 is the length and direction of its \(x\)-component, and
- the second component, 6 is the length and direction of its \(y\)-component.
The vector \(\overrightarrow{u}\) in the picture below has
FIRST COMPONENT = (terminal \(x\) -value) – (initial \(x\) -value) = \(2-7=-5\) , and
SECOND COMPONENT = (terminal \(y\) -value) – (initial \(y\) -value) = \(4-6=-2\) ,
so that \(\overrightarrow{u}=\left\langle -5,\left.-2\right\rangle \right.\) .
we get <-5,-2>." src="/@api/deki/files/97342/clipboard_eaa414d12e539b2aa26d4cc28d0dc560e.png">
Row and Column forms of a Vector
Vectors are represented by a single column matrix or a single row matrix. The vectors \(\overrightarrow{v}=\left\langle 3,\left.6\right\rangle \right.\) , and
\(\overrightarrow{u}=\left\langle -5,\left.-2\right\rangle \right.\) above, can be represented by the 2x1 row matrix and the 1x2 column matrix, respectively as
\[\vec{v}=\left[\begin{array}{ll}
3 & 6
\end{array}\right] \text { and } \vec{u}=\left[\begin{array}{l}
-5 \\
-2
\end{array}\right] \nonumber \]
Equal Vectors
Two vectors are EQUAL if they have the same direction and magnitude. They may start and end at different positions, but their representing arrows will be parallel.
In the diagram vectors \(\vec{a}\) and \(\vec{b}\) are equal but appear in different locations in the x-yplane.
Try These
Express the vectors \(\vec{v}\) and \(\vec{u}\) in component form.
- Answer
-
\(\overrightarrow{v}=\left\langle -3,\left.2\right\rangle \right.\) and \(\overrightarrow{u}=\left\langle 13,\left.3\right\rangle \right.\)
Explain why the two vectors are equal.
- Answer
-
Two vectors are equal because they have the same direction and magnitude. | libretexts | 2025-03-17T22:27:20.194794 | 2023-04-03T20:57:46 | {
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https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/02%3A_Vectors_In_Two_Dimensions/2.02%3A_Addition_Subtraction_and_Scalar_Multiplication_of_Vectors | 2.2: Addition, Subtraction, and Scalar Multiplication of Vectors
Addition and Subtraction of Vectors
To add or subtract two vectors, add, or subtract their corresponding components.
To ADD the vectors \(\vec{u}\) and \(\vec{v}\)
Solution
To ADD the vectors \(\vec{u}\) and \(\vec{v}\) , begin by writing each in component form.
\(\vec{u}=\langle-3,-8\rangle \text { and } \vec{v}=\langle 6,3\rangle\)
ADD their corresponding components
\(\vec{u}+\vec{v}=\langle-3+6,-8+3\rangle=\langle 3,-5\rangle\)
So, \(\vec{u}+\vec{v}=\langle 3,-5\rangle\)
Now, graph this sum.
- Start at the origin.
- Since the horizontal component is 3, move 3 units to the right.
- Since the vertical component is −5 , move 5 units downward.
The addition of two vectors \(\vec{u}\) and \(\vec{v}\) can be demonstrated by placing the tail of one vector at the head of the other. Then connect the tail of \(\vec{u}\) to the head of \(\vec{v}\) .
To SUBTRACT the vector \(\vec{u}\) from the vector \(\vec{v}\)
Solution
To SUBTRACT the vector \(\vec{u}\) from the vector \(\vec{v}\) , begin by writing each in component form.
\(\vec{u}=\langle-3,-8\rangle\) and \(\vec{v}=\langle 6,3\rangle\)
SUBTRACT the components of \(\vec{u}\) from the corresponding components of \(\vec{v}\)
\(\vec{v}-\vec{u}=\langle 6-(-3), 3-(-8)\rangle=\langle 6+3,3+8\rangle=\langle 9,11\rangle\)
So, \(\vec{v}-\vec{u}=\langle 9,11\rangle\)
Now, graph this sum.
- Start at the origin.
- Since the horizontal component is 9, move 9 units to the right.
- Since the vertical component is 11, move 11 units upward.
Scalars
In contrast to a vector, and having both direction and magnitude, a SCALAR is a physical quantity defined by only its magnitude.
Examples are speed, time, distance, density, and temperature. They are represented by real numbers (both positive and negative), and they can be operated on using the regular laws of algebra.
The term scalar derives from this usage: a scalar is that which scales, resizes a vector .
Scalar multiplication is the multiplication of a vector by a real number (a scalar).
Suppose we let the letter \(k\) represent a real number and \(\vec{v}\) be the vector \(\left\langle x\right.,\left.y\right\rangle .\) Then, the scalar multiple of the vector \(\vec{v}\) is
\(k \vec{v}=\langle k x, k y\rangle\)
To multiply a vector by a scalar (a constant), multiply each of its components by the constant.
- Suppose \(\overrightarrow{u}=\left\langle -3,\left.-8\right\rangle \right.\) and \(k=3\) . Then \[k\overrightarrow{u}=3\overrightarrow{u}=3\left\langle -3,\left.-8\right\rangle \right.=\left\langle 3(-3),\left.3(-8)\right\rangle \right.=\left\langle -9,\left.-24\right\rangle \right. \nonumber \]
- Suppose \(\overrightarrow{v}=\left\langle 6,\left.3\right\rangle \right.\) and \(k=\frac{-1}{3}\) . Then \[k \vec{u}=\frac{-1}{3} \vec{u}=\frac{-1}{3}\langle 6,3\rangle=\left\langle\frac{-1}{3}(6), \frac{-1}{3}(3)\right\rangle=\langle-2,-1\rangle \nonumber \]
- Suppose \(\overrightarrow{u}=\left[ \begin{array}{c} -2 \\ 6 \end{array} \right]\) and \(\overrightarrow{v}=\left[ \begin{array}{c} 5 \\ 3 \end{array} \right]\) . Then \[3\overrightarrow{u}+4\overrightarrow{v}=3\left[ \begin{array}{c} -2 \\ 6 \end{array} \right]+4\left[ \begin{array}{c} 5 \\ 3 \end{array} \right]=\left[ \begin{array}{c} -6 \\ 18 \end{array} \right]+\left[ \begin{array}{c} 20 \\ 12 \end{array} \right]=\left[ \begin{array}{c} 14 \\ 30 \end{array} \right] \nonumber \]
Using Technology
We can use technology to add and subtract vectors and to multiply a vector by a scalar.
Go to www.wolframalpha.com.
For the vectors \(\overrightarrow{u}=\left[ \begin{array}{c} -2 \\ 6 \end{array} \right]\) and \(\overrightarrow{v}=\left[ \begin{array}{c} 5 \\ 3 \end{array} \right]\) , use WolframAlpha to find \(3\overrightarrow{u}+4\overrightarrow{v}\) . Enter evaluate 3 \(\mathrm{<}\) -2, 6 \(\mathrm{>}\) + 4 \(\mathrm{<}\) 5, 3 \(\mathrm{>}\) in the entry field. Wolframalpha tells you what it thinks you entered, then tells you its answer. In this case, \(<14,\ 30>\) .
multiplied by 3 and <5,3> multiplied by 4. The result is the vector <14,30>." src="/@api/deki/files/97350/clipboard_eaf3e6c7b171f693605d899d7edbd53b1.png">
Try These
Find the sum of the two vectors \(\vec{u}=\langle-5,2\rangle\) and \(\vec{v}=\langle 10,-1\rangle\).
- Answer
-
\(\overrightarrow{u}\ +\ \overrightarrow{v}=\left\langle -5,\left.1\right\rangle \right.\)
Subtract the vector \(\vec{u}=\langle-5,2\rangle\) from the vector \(\vec{v}=\langle 10,-1\rangle\)
- Answer
-
\(\overrightarrow{v}-\ \overrightarrow{u}=\left\langle 15,\left.-3\right\rangle \right.\)
Suppose \(\vec{u}=\langle-5,2\rangle, \vec{v}=\langle 1,6\rangle\), and \(\vec{w}=\langle 4,-3\rangle\). Perform the operation \(2 \vec{u}-4 \vec{v}+3 \vec{w}\).
- Answer
-
\(\langle -29,-5\rangle\) | libretexts | 2025-03-17T22:27:20.356683 | 2023-04-03T20:57:48 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/02%3A_Vectors_In_Two_Dimensions/2.02%3A_Addition_Subtraction_and_Scalar_Multiplication_of_Vectors",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "2.2: Addition, Subtraction, and Scalar Multiplication of Vectors",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/02%3A_Vectors_In_Two_Dimensions/2.03%3A_Magnitude_Direction_and_Components_of_a_Vector | 2.3: Magnitude, Direction, and Components of a Vector
The Magnitude of a Vector
It is productive to represent the horizontal and vertical components of a vector \(\vec{v}\) as \(v_x\) and \(v_y\) , respectively.
The magnitude (the length) of a vector \(\vec{v}=\left\langle v_x, v_y\right\rangle\) is \[\|\vec{v}\|=\sqrt{v_x^2+v_y^2} \nonumber \]
Magnitude of vector \(\vec{v}=\left\langle 5,\left.-8\right\rangle \right.\):
Solution
The vector \(\vec{v}=\left\langle 5,\left.-8\right\rangle \right.\) has magnitude:
\(\left\|\overrightarrow{v}\right\|=\sqrt{{v_x}^2+{v_y}^2}\) = \(\sqrt{5^2+{\left(-8\right)}^2}=\ \sqrt{25+64}\) = \(\sqrt{89}\)
Interpret this as the length of the vector \(\overrightarrow{v}=\left\langle 5,\left.-8\right\rangle \right.\) is \(\sqrt{89}\) units.
The Direction of a Vector
The direction of a vector \(\vec{v}\) is the angle the vector makes with the positive x-axis.
It is typically represented with the uppercase Greek letter theta \(\theta\) . We use some trigonometry to determine the angle \(\theta\) .
\(\tan \theta=\dfrac{y}{x}\) or \(\theta=\tan ^{-1} \dfrac{y}{x}\)
The angle \(\theta \) is always between \(0^{\circ}\) and \(360^{\circ}\).
To approximate the direction of the vector \(\vec{v}=\langle 5,8\rangle\), use \(\theta=\tan ^{-1} \frac{y}{x}\), with \(x=5\) and \(y=8\).
Solution
\(\begin{aligned}
\vec{v} & =\langle 5,8\rangle \\
\theta & =\tan ^{-1} \frac{y}{x} \\
\theta & =\tan ^{-1} \frac{8}{5}
\end{aligned}\)
Using a calculator, we get
\(\begin{aligned}
& \theta=57.99^{\circ} \\
& \theta=58^{\circ}
\end{aligned}\)
To approximate the direction of the vector \(\vec{v}=\langle 5,-8\rangle\), use \(\theta=\tan ^{-1} \frac{y}{x}\), with \(x=5\) and \(y=-8\).
Solution
\(\begin{gathered}
\vec{v}=\langle 5,-8\rangle \\
\theta=\tan ^{-1} \frac{y}{x} \\
\theta=\tan ^{-1} \frac{-8}{5}
\end{gathered}\)
Using a calculator, we get
\(\theta=-57.99^{\circ}\)
Vertical component is in Quadrant IV and \(\theta\) must be in the interval \(\left[0,\left.360\right)\right.\) , therefore we calculate \(\theta\) by
\[\theta =360{}^\circ -57.99{}^\circ =302.005{}^\circ\nonumber \] \[\theta =302{}^\circ \nonumber \]
The Components of a Vector
The lengths of the \(x\) - and \(y\) - components of a vector \(\left\|\vec{v}\right\|=\sqrt{{v_x}^2+{v_y}^2}\) in two dimensions can be found using trigonometric ratios.
\(\vec{v}_x=\|\vec{v}\|_{\cos \theta}\)and \(\vec{v}_y=\|\vec{v}\|_{\sin \theta}\)
\(\vec{v}_x\) is the horizontal component of \(\vec{v}\) and \(\vec{v}_y\) is the vertical component.
The angle \(\theta\) is always between \(0^{\circ}\) and \(360^{\circ}\).
Suppose the magnitude of a vector \(\vec{v}\boldsymbol{=}\left\langle v_x,\left.v_y\right\rangle \right.\) is 20 units, and that \(\vec{v}\) makes a 60 \(\mathrm{{}^\circ}\) angle with the horizontal. Then, the components of \(\vec{v}\) are
\[\begin{aligned}
\vec{v}_x & =\|\vec{v}\| \cos \theta \\
= & 20 \cos 60^{\circ} \\
& =20 \cdot \frac{1}{2} \\
& =10
\end{aligned} \nonumber \]
and
\[\begin{aligned}
\vec{v}_y & =\|\vec{v}\|_{\sin } \theta \\
= & 20 \sin 60^{\circ} \\
& =20 \cdot \frac{\sqrt{3}}{2} \\
& =10 \sqrt{3}
\end{aligned} \nonumber \]
So, we could write \(\vec{v}\boldsymbol{=}\left\langle v_x,\left.v_y\right\rangle \right.\) as \(\vec{v}\boldsymbol{=}\left\langle 10,\left.10\sqrt{3}\right\rangle \right.\)
Using Technology
We can use technology to determine the magnitude of a vector.
Go to www.wolframalpha.com.
To find the magnitude of the vector \(\vec{v}\boldsymbol{=}\left\langle 2,\left.4\right\rangle ,\right.\) enter magnitude of the vector \(\mathrm{<}\) 2,4 \(\mathrm{>}\) in the entry field. Wolframalpha tells you what it thinks you entered, then tells you its answer. In this case, \(\left\|\vec{v}\right\|=2\sqrt{5}\) .
. The result is 2 times the square root of 5." src="/@api/deki/files/97361/clipboard_e757f835523c600e69c47a6190429dff2.png">
To find the direction of the vector \(\vec{v}\boldsymbol{=}\left\langle 5,\left.8\right\rangle ,\right.\) enter direction of the vector \(\mathrm{<}\) 5,8 \(\mathrm{>}\) in the entry field. Wolframalpha answers \(57.9946{}^\circ \approx 58{}^\circ\) .
. The result is 9.43398 radians (r) and 57.9946 degrees from the x-axis (theta)." src="/@api/deki/files/97362/clipboard_e0ddb516d9c06e26d0792e439e307323d.png">
Try These
Find the magnitude of the vector \(\vec{v}=\langle 3,-4\rangle\).
- Answer
-
\(\left\|\overrightarrow{v}\right\|=5\)
Find the magnitude of the vector \(\vec{v}=\langle -3,-3\rangle\).
- Answer
-
\(\left\|\overrightarrow{v}\right\|=3\sqrt{2}\)
Find the components of the vector \(\vec{v}\) if the magnitude of \(\vec{v}\) is 6 and it makes a \(30^{\circ}\) angle with the horizontal.
- Answer
-
\({\overrightarrow{v}}_x=\ 3\sqrt{3}\) and \({\overrightarrow{v}}_y\mathrm{=3}\)
Approximate the direction of the vector \(\vec{v}=\langle 3,10\rangle\).
- Answer
-
\(\theta \approx 73.3008{}^\circ\) | libretexts | 2025-03-17T22:27:20.439875 | 2023-04-03T20:57:49 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/02%3A_Vectors_In_Two_Dimensions/2.03%3A_Magnitude_Direction_and_Components_of_a_Vector",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "2.3: Magnitude, Direction, and Components of a Vector",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/02%3A_Vectors_In_Two_Dimensions/2.04%3A_The_Dot_Product_of_Two_Vectors_the_Length_of_a_Vector_and_the_Angle_Between_Two_Vectors | 2.4: The Dot Product of Two Vectors, the Length of a Vector, and the Angle Between Two Vectors
The Dot Product of Two Vectors
The length of a vector or the angle between two vectors \(\vec{u}\boldsymbol{=}\left\langle u_x,\left.u_y\right\rangle \right.\) and \(\vec{v}\boldsymbol{=}\left\langle v_x,\left.v_y\right\rangle \right.\) can be found using the dot product.
The dot product of vectors \(\vec{u}=\left\langle u_x, u_y\right\rangle\) and \(\vec{v}=\left\langle v_x, v_y\right\rangle\) is a scalar (real number) and is defined to be \[\vec{u} \cdot \vec{v}=u_x v_x+u_y v_y \nonumber \]
Since \(u_x,\ u_y,\ v_x\) and \(v_y\) are real numbers, you can see that the dot product is itself a real number and not a vector.
To compute the dot product of the vectors \(\vec{u}\boldsymbol{=}\left\langle 5,\left.2\right\rangle \right.\) and \(\vec{v}\boldsymbol{=}\left\langle 3,\left.4\right\rangle \right.\)
Solution
We compute: \[\overrightarrow{u}\bullet \overrightarrow{v}\boldsymbol{=}5\bullet 3\mathrm{+\ }2\bullet 4=15+8=23 \nonumber \]
Since the dot product is a scalar, it follows the properties of real numbers.
- \(\vec{u} \cdot \vec{v}=\vec{v} \cdot \vec{u}\)the dot product is commutative
- \(\vec{u} \cdot(\vec{v}+\vec{w})=\vec{u} \cdot \vec{v}+\vec{u} \cdot \vec{w}\) the dot product distributes over vector addition
- \(\vec{u} \cdot \overrightarrow{0}=0\),the dot product with the zero vector \(\overrightarrow{0}\), is the scalar 0.
- \(\vec{u} \cdot \vec{u}=\|\vec{u}\|^2\)
Compute the dot product \(\overrightarrow{u}\bullet \left(\overrightarrow{v}+\overrightarrow{w}\right)=\overrightarrow{u}\bullet \overrightarrow{v}+\overrightarrow{u}\bullet \overrightarrow{w}\) , where \(\overrightarrow{u}=\left\langle 5,\left.-2\right\rangle \right.\) , \(\overrightarrow{v}=\left\langle 6,\left.4\right\rangle \right.\) , and \(\overrightarrow{w}=\left\langle -3,\left.7\right\rangle \right.\) .
Solution
\[\begin{align} \vec{u} \cdot(\vec{v}+\vec{w})&=\vec{u} \cdot \vec{v}+\vec{u} \cdot \vec{w} \nonumber\\
\vec{u} \cdot(\vec{v}+\vec{w})&=\langle 5,-2\rangle \cdot\langle 6,4\rangle+\langle 5,-2\rangle \cdot\langle-3,7\rangle \nonumber \\
&=(5 \cdot 6+(-2) \cdot 4)+(5 \cdot(-3)+(-2) \cdot 7) \nonumber \\
&=30-8-15-14 \nonumber \\
&=-7 \nonumber
\end{align} \nonumber \]
The Length of a Vector
The length (magnitude) of a vector you know is given by \(\left\|\overrightarrow{v}\right\|=\sqrt{{v_x}^2+{v_y}^2}\) . The length can also be found using the dot product. If we dot a vector \(\overrightarrow{v}=\left\langle v_x,\left.v_y\right\rangle \right.\) with itself, we get
\[\overrightarrow{v}\bullet \overrightarrow{v}=\left\langle v_x,\left.v_y\right\rangle \right.\bullet \left\langle v_x,\left.v_y\right\rangle \right. \nonumber \] \[\overrightarrow{v}\bullet \overrightarrow{v}=v_x\bullet v_x+v_y\bullet v_y \nonumber \] \[\overrightarrow{v}\bullet \overrightarrow{v}= {v_x}^2+\ {v_y}^2 \nonumber \]
By Vector Property 4, \(\overrightarrow{v}\bullet \overrightarrow{v}=\ {\left\|\overrightarrow{v}\right\|}^2\) . This gives \({\left\|\overrightarrow{v}\right\|}^2={v_x}^2+{v_y}^2\) .
Taking the square root of each side produces
\[\sqrt{{\|\overrightarrow{v}\|}^2}=\sqrt{{v_x}^2+{v_y}^2} \nonumber \]
\[\|\overrightarrow{v}\|=\sqrt{{v_x}^2+{v_y}^2} \nonumber \]
Which is the length of the vector \(\overrightarrow{v}\) .
The dot product of a vector \(\vec{v}=\left\langle v_x, v_y\right\rangle\) with itself gives the length of the vector.
\[\begin{equation}
\|\vec{v}\|=\sqrt{v_x^2+v_y^2}
\end{equation} \nonumber \]
Use the dot product to find the length of the vector \(\overrightarrow{v}=\left[ \begin{array}{c} 2 \\ 6 \end{array} \right]\) .
Solution
In this case, \(v_x=2\) and \(v_y=6.\)
Using \(\|\overrightarrow{v}\|=\sqrt{{v_x}^2+{v_y}^2}\) , we get \[\|\overrightarrow{v}\|=\sqrt{2^2+6^2} \nonumber \] \[\|\overrightarrow{v}\|=\sqrt{40} \nonumber \] \[\|\overrightarrow{v}\|=\sqrt{4\bullet 10} \nonumber \] \[\|\overrightarrow{v}\|=\sqrt{4}\bullet \sqrt{10} \nonumber \] \[\|\overrightarrow{v}\|=2\sqrt{10} \nonumber \]
The length of the vector \(\overrightarrow{v}=\left[ \begin{array}{c} 2 \\ 6 \end{array} \right]\) is \(2\sqrt{10}\) units.
The Angle between two Vectors
The dot product and elementary trigonometry can be used to find the angle \(\theta\) between two vectors.
If \(\theta\) is the smallest nonnegative angle between two non-zero vectors \(\vec{u}\) and \(\vec{v}\) then
\[\begin{equation}
\cos \theta=\frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \cdot\|\vec{v}\|} \; \text { or } \; \theta=\cos ^{-1} \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \cdot\|\vec{v}\|}
\end{equation} \nonumber \]
\[\begin{equation}
\text { where } \; 0 \leq \theta \leq 2 \pi \; \text { and } \; \|\vec{u}\|=\sqrt{u_x^2+u_y{ }^2} \; \text { and } \; \|\vec{v}\|=\sqrt{v_x^2+v_y{ }^2}
\end{equation} \nonumber \]
Find the angle between the vectors \(\overrightarrow{u}=\left\langle 5,\left.-3\right\rangle \right.\) and \(\overrightarrow{v}=\left\langle 2,\left.4\right\rangle \right.\) .
Solution
Using \(\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{\overrightarrow{u}\bullet \overrightarrow{v}}{\|\overrightarrow{u}\|\bullet \|\overrightarrow{v}\|}\) , we get
\[\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{\left\langle 5,\left.-3\right\rangle \right.\bullet \left\langle 2,\left.4\right\rangle \right.}{\sqrt{5^2+{\left(-3\right)}^2}\bullet \sqrt{2^2+4^2}} \nonumber \]
\[\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{5\bullet 2+\left(-3\right)\bullet 4}{\sqrt{25+9}\bullet \sqrt{4+16}} \nonumber \]
\[\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{-2}{\sqrt{34}\bullet \sqrt{20}} \nonumber \]
\[\theta =94.4 \nonumber \]
We conclude that the angle between these two vectors is close to 94.4 \(\mathrm{{}^\circ}\) .
Using Technology
We can use technology to find the angle \(\theta\) between two vectors.
Go to www.wolframalpha.com.
To find the angle between the vectors \(\overrightarrow{u}=\left\langle 5,\left.-3\right\rangle \right.\) and \(\overrightarrow{v}=\left\langle 2,\left.4\right\rangle \right.\) , enter angle between the vectors \(\mathrm{<}\) 5, -3 \(\mathrm{>}\) and \(\mathrm{<}\) 2, 4 \(\mathrm{>}\) in the entry field. Wolframalpha tells you what it thinks you entered, then tells you its answer. In this case, \(\theta =94.4\) , rounded to one decimal place.
Try These
Find the dot product of the vectors \(\vec{u}=\langle-2,3\rangle \) and \(\vec{v}=\langle 5,-1\rangle \).
- Answer
-
\(\overrightarrow{u}\bullet \overrightarrow{v}=-13\)
Find the dot product of the vectors \(\vec{u}=\langle-4,6\rangle \) and \(\vec{v}=\langle 3,2\rangle \).
- Answer
-
\(\overrightarrow{u}\bullet \overrightarrow{v}=0\)
Find the length of the vector \(\vec{u}=\langle 4,-7\rangle \).
- Answer
-
\(\sqrt{65}\)
Find the length of the vectors \(\vec{v}=\langle 0,5\rangle \).
- Answer
-
\(5\)
Find the angle between the vectors \(\vec{u}=\langle-2,3\rangle \) and \(\vec{v}=\langle 5,-1\rangle \).
- Answer
-
\(135{}^\circ\)
Find the angle between the vectors \(\vec{u}=\langle -4,6\rangle \) and \(\vec{v}=\langle 3,2\rangle \).
- Answer
-
\(90{}^\circ\) | libretexts | 2025-03-17T22:27:20.528030 | 2023-04-03T20:57:51 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/02%3A_Vectors_In_Two_Dimensions/2.04%3A_The_Dot_Product_of_Two_Vectors_the_Length_of_a_Vector_and_the_Angle_Between_Two_Vectors",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "2.4: The Dot Product of Two Vectors, the Length of a Vector, and the Angle Between Two Vectors",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/02%3A_Vectors_In_Two_Dimensions/2.05%3A_Parallel_and_Perpendicular_Vectors_The_Unit_Vector | 2.5: Parallel and Perpendicular Vectors, The Unit Vector
Parallel and Orthogonal Vectors
Two vectors \(\vec{u}=\left\langle u_x, u_y\right\rangle\) and \(\vec{v}=\left\langle v_x, v_y\right\rangle\) are parallel if the angle between them is \(0^{\circ}\) or \(180^{\circ}\).
Also, two vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle u_x,\left.u_y\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle v_x,\left.v_y\right\rangle \right.\) are parallel to each other if the vector \(\overrightarrow{u}\) is some multiple of the vector \(\overrightarrow{v}\) . That is, they will be parallel if the vector \(\overrightarrow{u}=c\overrightarrow{v}\) , for some real number \(c\) . That is, \(\overrightarrow{u}\) is some multiple of \(\overrightarrow{v}\) .
Two vectors \(\vec{u}=\left\langle u_x, u_y\right\rangle\) and \(\vec{v}=\left\langle v_x, v_y\right\rangle\) are orthogonal (perpendicular to each other) if the angle between them is \(90^{\circ}\) or \(270^{\circ}\).
Use this shortcut: Two vectors are perpendicular to each other if their dot product is 0.
The two vectors \(\overrightarrow{u}\boldsymbol{=}\left.\boldsymbol{\langle }2,-3\right\rangle\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle -8,\left.12\right\rangle \right.\) are parallel to each other since the angle between them is \(180{}^\circ\) .
Solution
\(\begin{aligned}
& \theta=\cos ^{-1} \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \cdot\|\vec{v}\|} \\
& \theta=\cos ^{-1} \frac{\langle 2,-3\rangle \cdot\langle-8,12\rangle}{\sqrt{2^2+(-3)^2} \cdot \sqrt{(-8)^2+12^2}} \\
& \theta=\cos ^{-1} \frac{2 \cdot(-8)+(-3) \cdot 12}{\sqrt{4+9} \cdot \sqrt{64+144}} \\
& \theta=\cos ^{-1} \frac{-52}{\sqrt{13} \cdot \sqrt{208}} \\
& \theta=180^{\circ}
\end{aligned}\)
To show that the two vectors \(\overrightarrow{u}\boldsymbol{=}\left.\boldsymbol{\langle }5,10\right\rangle\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 6,\left.-3\right\rangle \right.\) are orthogonal (perpendicular to each other), we just need to show that their dot product is 0.
Solution
\[\left.\boldsymbol{\langle }5,10\right\rangle \boldsymbol{\bullet }\left\langle 6,\left.-3\right\rangle \right.\boldsymbol{=}5\bullet 6+10\bullet \left(-3\right)=30-30=0 \nonumber \]
The Unit Vector
A unit vector is a vector of length 1.
A unit vector in the same direction as the vector \(\overrightarrow{v}\) is often denoted with a “hat” on it as in \(\hat{v}\) . We call this vector “v hat.”
The unit vector \(\hat{v}\)corresponding to the vector \(\vec{v}\)is defined to be \[\hat{v}=\frac{\vec{v}}{\|\vec{v}\|} \nonumber \]
The unit vector corresponding to the vector \(\overrightarrow{v}\boldsymbol{=}\left\langle -8,\left.12\right\rangle \right.\).
Solution
\(\begin{gathered}
\hat{v}=\frac{\vec{v}}{\|\vec{v}\|} \\
\hat{v}=\frac{\langle-8,12\rangle}{\sqrt{(-8)^2+(12)^2}} \\
\hat{v}=\frac{\langle-8,12\rangle}{\sqrt{64+144}} \\
\hat{v}=\frac{\langle-8,12\rangle}{\sqrt{208}} \\
\hat{v}=\left\langle\frac{-8}{\sqrt{208}}, \frac{12}{\sqrt{208}}\right\rangle
\end{gathered}\)
Using Technology
We can use technology to find the angle \(\theta\) between two vectors.
Go to www.wolframalpha.com.
To show that the vectors \(\overrightarrow{u}\boldsymbol{=}\left.\boldsymbol{\langle }2,-3\right\rangle\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle -8,\left.12\right\rangle \right.\) are parallel, enter angle between the vectors \(\mathrm{<}\) 2, -3 \(\mathrm{>}\) and \(\mathrm{<}\) -8, 12 \(\mathrm{>}\) in the entry field. Wolframalpha tells you what it thinks you entered, then tells you its answer. In this case, \(\theta =180{}^\circ\) , indicating the two vectors are parallel.
and
Try These
Determine if the vectors \(\vec{u}=\langle 2,1\rangle\) and \(\vec{v}=\langle 3,-6\rangle\) are parallel to each other, perpendicular to each other, or neither parallel nor perpendicular to each other.
- Answer
-
Parallel
Determine if the vectors \(\vec{u}=\langle 2,16\rangle\) and \(\vec{v}=\left\langle\frac{1}{2}, 4\right\rangle\) are parallel to each other, perpendicular to each other, or neither parallel nor perpendicular to each other.
- Answer
-
Perpendicular
Determine if the vectors \(\vec{u}=\langle 7,6\rangle\) and \(\vec{v}=\langle 2,-1\rangle\) are parallel to each other, perpendicular to each other, or neither parallel nor perpendicular to each other.
- Answer
-
Neither parallel nor perpendicular
Find the unit vector corresponding to the vector \(\vec{v}=\langle 2,-1\rangle\).
- Answer
-
\(\hat{v}\boldsymbol{=}\left\langle \frac{2}{\sqrt{5}},\left.\frac{-1}{\sqrt{5}}\right\rangle \right.\) | libretexts | 2025-03-17T22:27:20.606232 | 2023-04-03T20:57:53 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/02%3A_Vectors_In_Two_Dimensions/2.05%3A_Parallel_and_Perpendicular_Vectors_The_Unit_Vector",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "2.5: Parallel and Perpendicular Vectors, The Unit Vector",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/02%3A_Vectors_In_Two_Dimensions/2.06%3A_The_Vector_Projection_of_One_Vector_onto_Another | 2.6: The Vector Projection of One Vector onto Another
Projection
Let’s project vector \(\overrightarrow{u}\boldsymbol{=}\left\langle u_x,\left.u_y\right\rangle \right.\) onto the vector \(\overrightarrow{v}\boldsymbol{=}\left\langle v_x,\left.v_y\right\rangle \right.\) .
To do so, imagine a light bulb above \(\overrightarrow{u}\) shining perpendicular onto \(\overrightarrow{v}\) .
The light from the bulb will cast a shadow of \(\overrightarrow{u}\) onto \(\overrightarrow{v}\) , and it is this shadow that we are looking for. The shadow is the projection of \(\overrightarrow{u}\) onto \(\overrightarrow{v}\) .
The red vector is the projection of \(\overrightarrow{u}\) onto \(\overrightarrow{v}\) . The notation commonly used to represent the projection of \(\overrightarrow{u}\) onto \(\overrightarrow{v}\) is \({\mathrm{proj}}_{\overrightarrow{\mathrm{v}}}\overrightarrow{u}\) .
Vector parallel to \(\overrightarrow{v}\) with magnitude \(\frac{\overrightarrow{u}\bullet \overrightarrow{v}}{\left\|\overrightarrow{v}\right\|}\) in the direction of \(\overrightarrow{v}\) is called projection of \(\overrightarrow{u}\) onto \(\overrightarrow{v}\) .
The formula for \({\mathrm{proj}}_{\overrightarrow{\mathrm{v}}}\overrightarrow{u}\) is
\[\operatorname{proj}_{\overrightarrow{\mathrm{v}}} \vec{u}=\frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2} \vec{v} \nonumber \]
To find the projection of \(\vec{u}=\langle 4,3\rangle\) onto \(\vec{v}=\langle 2,8\rangle\), we need to compute both the dot product of \(\overrightarrow{u}\) and \(\overrightarrow{v}\) , and the magnitude of \(\overrightarrow{v}\) , then apply the formula.
Solution
\(\begin{gathered}
\operatorname{proj}_{\overrightarrow{\mathrm{v}}} \vec{u}=\frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2} \vec{v} \\
\operatorname{proj}_{\vec{v}} \vec{u}=\frac{\langle 4,3\rangle \cdot\langle 2,8\rangle}{\|\langle 2,8\rangle\|^2}\langle 2,8\rangle \\
\operatorname{proj}_{\overrightarrow{\mathrm{v}}} \vec{u}=\frac{4 \cdot 2+3 \cdot 8}{\left(\sqrt{2^2+8^2}\right)^2}\langle 2,8\rangle \\
\operatorname{proj}_{\vec{v}} \vec{u}=\frac{32}{(\sqrt{4+64})^2}\langle 2,8\rangle \\
\operatorname{proj}_{\vec{v}} \vec{u}=\frac{32}{68}\langle 2,8\rangle \\
\operatorname{proj}_{\vec{v}} \vec{u}=\frac{8}{17}\langle 2,8\rangle \\
\operatorname{proj}_{\vec{v}} \vec{u}=\left\langle\frac{16}{17}, \frac{64}{17}\right\rangle
\end{gathered}\)
Using Technology
We can use technology to determine the projection of one vector onto another.
Go to www.wolframalpha.com.
To find the projection of \(\overrightarrow{u}=\left\langle 4,\left.3\right\rangle \right.\) onto \(\vec{v}=\langle 2,8\rangle\), use the “projection” command. In the entry field enter projection of \(\mathrm{<}\) 4, 3 \(\mathrm{>}\) onto \(\mathrm{<}\) 2, 8 \(\mathrm{>}\) .
\\
Wolfram alpha tells you what it thinks you entered, then tells you its answer. In this case, \(\left\langle \frac{16}{17},\left.\frac{64}{17}\right\rangle \right.\) .
As an applied example, suppose a video game has a ball moving near a wall.
Solution
We take the origin at the bottom-left-most corner of the screen. The wall is at a 30 \(\mathrm{{}^\circ}\) angle to the horizontal, and at a point in time, the ball is at position \(\overrightarrow{v}\boldsymbol{=\ }\) \(\mathrm{\langle }\) 4, \(\ \left.7\right\rangle\) . To find the perpendicular distance from the ball to the wall, we use the projection formula to project the vector \(\overrightarrow{v}\boldsymbol{=\ }\) \(\mathrm{\langle }\) 4, \(\ \left.7\right\rangle\) onto the wall.
We begin by decomposing \(\overrightarrow{v}\) into two vectors \({\overrightarrow{v}}_1\) and \({\overrightarrow{v}}_2\) so that \(\overrightarrow{v}=\ {\overrightarrow{v}}_1+{\overrightarrow{v}}_2\) and \({\overrightarrow{v}}_1\) lies along the wall .
The length (magnitude) of the vector \(\overrightarrow{v}\) is then the distance from the ball to the wall.
The vector \({\overrightarrow{v}}_1\) is the projection of \(\overrightarrow{v}\) onto the wall. We can get \({\overrightarrow{v}}_1\) by scaling (multiplying) a unit vector \(\overrightarrow{w}\) that lies along the wall and, thus, along with \({\overrightarrow{v}}_1\) .
Since \(\overrightarrow{w}\) lies at a 30 \(\mathrm{{}^\circ}\) angle to the horizontal, \(\overrightarrow{w}\boldsymbol{=}\left.\boldsymbol{\langle }\mathrm{cos}30{}^\circ ,\ \ \mathrm{sin}30{}^\circ \right\rangle \boldsymbol{=}\left.\boldsymbol{\langle }\mathrm{0.866},\ \mathrm{0.5}\right\rangle\) , using the projection formula, we get the projection of \(\overrightarrow{v}\) that lies along the wall.
\(\begin{gathered}
\vec{v}_1=\operatorname{proj}_{\vec{w}} \vec{v}=\frac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|^2} \vec{w} \\
\vec{v}_1 \frac{\langle 4,7\rangle \cdot\langle 0.866,0.5\rangle}{\left\|\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\right\|^2}\langle 0.866,0.5\rangle =\frac{4 \cdot(0.866)+7 \cdot(0.5)}{\left(\sqrt{(0.866)^2+(.5)^2}\right)^2}\langle 0.866,0.5\rangle \\
\vec{v}_1=\frac{6.964}{(\sqrt{1})^2}\langle 0.866,0.5\rangle\\
\vec{v}_1=(6.964)\langle 0.866,0.5\rangle \\
\vec{v}_1=\langle 6.031,3.482\rangle
\end{gathered}\)
Since that \(\vec{v}=\vec{v}_{\mathbf{1}}+\vec{v}_{\mathbf{2}}\), subtraction get us
\(\begin{gathered}
\vec{v}_2=\vec{v}-\vec{v}_1 \\
\vec{v}_2=\langle 4,7\rangle-\langle 6.031,3.482\rangle \\
\vec{v}_2=\langle 4-6.031,7-3.482\rangle \\
\vec{v}_2=\langle-2.031,3.518\rangle
\end{gathered}\)
To get the magnitude of \(\vec{v}_2\), we use
\(\begin{gathered}
\left\|\vec{v}_2\right\|=\sqrt{v_x^2+v_y^2} \\
\left\|\overrightarrow{\boldsymbol{v}}_2\right\|=\sqrt{(-2.031)^2+3.518^2} \\
\left\|\vec{v}_2\right\|=\sqrt{4.125+12.376} \\
\left\|\overrightarrow{\vec{v}_2}\right\|=4.062
\end{gathered}\)
Try These
Find the projection of the vector \(\vec{v}=\langle 3,5\rangle\) onto the vector \(\vec{u}=\langle 6,2\rangle\).
- Answer
-
\(\left\langle \frac{21}{5}\right.,\ \left.\frac{7}{5}\right\rangle\)
Find \(\operatorname{proj}_{\vec{v}} \vec{u}\), with \(\vec{u}=\langle-2,5\rangle\) and \(\vec{v}=\langle 6,-5\rangle\).
- Answer
-
\(\left\langle \frac{-222}{61}\right.,\ \left.\frac{185}{61}\right\rangle\) | libretexts | 2025-03-17T22:27:20.682285 | 2023-04-03T20:57:54 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/02%3A_Vectors_In_Two_Dimensions/2.06%3A_The_Vector_Projection_of_One_Vector_onto_Another",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "2.6: The Vector Projection of One Vector onto Another",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions | 3: Vectors in Three Dimensions Last updated Save as PDF Page ID 125034 3.1: Three Dimensional Vectors 3.2: Magnitude and Direction Cosines of a Vector 3.3: Arithmetic on Vectors in 3-Dimensional Space 3.4: The Unit Vector in 3-Dimensions and Vectors in Standard Position 3.5: The Dot Product, Length of a Vector, and the Angle between Two Vectors in Three Dimensions 3.6: The Cross Product- Algebra 3.7: The Cross Product- Geometry | libretexts | 2025-03-17T22:27:20.760469 | 2023-04-03T20:57:56 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "3: Vectors in Three Dimensions",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions/3.01%3A_Three_Dimensional_Vectors | 3.1: Three Dimensional Vectors
3-Dimensional Space
To this point, we have been working with vectors in 2-dimenisional space. Now, we will expand our discussion to 3-dimensional space.
The 2-dimensional coordinate system is built around a set of two axes that intersect at right angles and one particular point called the origin. Points in the plane are described by ordered pairs \((x, y)\) and vectors in standard position by \(\langle x, y\rangle\).
The 3-dimensional coordinate system is built around a set of three axes that intersect at right angles and one particular point again called the origin. Points in the plane are described by ordered triples \((x, y, z)\) and vectors in standard position by \(\langle x, y, z\rangle\).
The Distance between two Points in 2 & 3-Dimensional Space
In two-dimensional space , the distance \(d\) between two points say \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) is given by the distance formula
\[d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \nonumber \]
In three-dimensional space , the distance \(d\) between two points say \(P(x_1,y_1,z_1)\) and \(Q(x_2,y_2,z_2)\) is given by the distance formula
\[d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2} \nonumber \]
The distance between the two points \(P(2,\ 2,\ 5)\) and \(Q(5,\ 6,\ 2)\) is
Solution
\(\begin{gathered}
d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2} \\
d=\sqrt{(5-2)^2+(6-2)^2+(2-5)^2} \\
d=\sqrt{(3)^2+(4)^2+(-3)^2} \\
d=\sqrt{9+16+9} \\
d=\sqrt{34} \approx 5.8 \text { units }
\end{gathered}\)
The distance between the two points \(P(2,\ 2,\ 5)\) and \(Q(5,\ 6,\ 2)\) is \(\sqrt{34}\) \(\approx 5.8\\) units.
Using Technology
We can use technology to find the distance between points.
Go to www.wolframalpha.com.
To find the distance between the two points \((-3,\ 5)\) and \((-7,\ 4)\) enter distance \((-3,\ 5)\) and \((-7,\ 4)\) in the entry field. Wolframalpha tells you what it thinks you entered, then tells you its answer. In this case, \(\sqrt{17}\ \approx 4.1\) 2311.
The Equation of a Circle and a Sphere
We can use the distance formulas to get equations of circles and spheres.
The center-radius form of a circle with center at the point \(C(h,k)\) and radius \(r\) is
\[(x-h)^2+(y-k)^2=r^2 \nonumber \]
The center-radius form of a sphere with center at the point \(C(h,k,j)\) and radius \(r\) is
\[(x-h)^2+(y-k)^2+(z-j)^2=r^2 \nonumber \]
To write the equation of a circle that has the point \(C\left(4,\ 7\right)\) as its center and radius 8, we use the center-radius form \({\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\) with \(h=4\) , \(k=7\) , and \(r=8\) .
Solution
\(\begin{aligned}
& (x-h)^2+(y-k)^2=r^2 \\
& (x-4)^2+(y-7)^2=8^2 \\
& (x-4)^2+(y-7)^2=64
\end{aligned}\)
To write the equation of a sphere that has the point \(C\left(4,\ 7,\ 1\right)\) as its center and radius 8, we use the center-radius form \({\left(x-h\right)}^2+{\left(y-k\right)}^2+\ {\left(z-j\right)}^2=r^2\) with \(h=4\) , \(k=7\) , \(j=1\) , and \(r=8\) .
Solution
\(\begin{aligned}
& (x-h)^2+(y-k)^2+(z-j)^2=r^2 \\
& (x-4)^2+(y-7)^2+(z-1)^2=8^2 \\
& (x-4)^2+(y-7)^2+(z-1)^2=64
\end{aligned}\)
Try These
Find the distance between the two points (2,4) and (-3,6). Round to one decimal place.
- Answer
-
\(\sqrt{29}\ \approx 5.4\) units
Find the distance between the two points (-3,5,-6) and (7,-4,2). Round to one decimal place.
- Answer
-
\(7\sqrt{5}\ \approx 15.6\) units
Write the equation of a circle that has the point C(2,9) as its center and radius 1.
- Answer
-
\({\left(x-2\right)}^2+{\left(y-9\right)}^2=1\)
Write the equation of a sphere that has the point C(-2,5,-7) as its center and radius 4.
- Answer
-
\({\left(x+2\right)}^2+{\left(y-5\right)}^2+\ {\left(z+7\right)}^2=16\) | libretexts | 2025-03-17T22:27:20.840750 | 2023-04-03T20:57:58 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions/3.01%3A_Three_Dimensional_Vectors",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "3.1: Three Dimensional Vectors",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions/3.02%3A_Magnitude_and_Direction_Cosines_of_a_Vector | 3.2: Magnitude and Direction Cosines of a Vector
The Magnitude of a Vector
You likely recall that the magnitude (the length) of a vector \(\overrightarrow{v}\boldsymbol{=}\boldsymbol{\langle }v_x\boldsymbol{,\ }v_{\boldsymbol{y}}\boldsymbol{\rangle }\) in 2-dimensions is
. The blue vector v and v_x have their starting points connected together. Vector v and v_y are connected together by their terminal points. The terminal point of v_x is connected to the starting point of v_y." src="/@api/deki/files/97413/clipboard_e54c0403f113cb233f89c9f459d41eeaa.png">
\(\|\vec{v}\|=\sqrt{v_x^2+v_y^2}\)
The magnitude of vector \(\vec{v}=\langle 2,4\rangle\)
Solution
\(\begin{aligned}
& \|\vec{v}\|=\sqrt{v_x{ }^2+v_y^2} \\
& \sqrt{2^2+4^2}=\sqrt{4+16}=\sqrt{20}=2 \sqrt{5} \\
& \|\vec{v}\|=2 \sqrt{5}
\end{aligned}\)
Interpret this as the length of the vector \(\vec{v}=\langle 2,4\rangle \) is \(2 \sqrt{5}\) units.
The formula for the length of the vector \(\vec{v}=\left\langle v_x, v_y, v_z\right\rangle\) in 3-dimensions is
\(\|\vec{v}\|=\sqrt{v_x^2+v_y^2+v_z^2}\)
The magnitude of vector \(\vec{v}=\langle 2,4,-6\rangle\)
Solution
\(\begin{gathered}
\|\vec{v}\|=\sqrt{v_x^2+v_y^2+v_z^2} \\
\|\vec{v}\|={\sqrt{2^2+4^2+(-6)^2}}^{\|\vec{v}\|}=\sqrt{4+16+36} \\
\|\vec{v}\|=\sqrt{56} \\
\|\vec{v}\|=2 \sqrt{14}
\end{gathered}\)
Interpret this as the length of the vector \(\vec{v}=\langle 2,4,-6\rangle \text { is } 2 \sqrt{14}\) units.
The Direction Cosines of Vectors in 2- and 3-Dimensions
The direction cosines of a vector \(\overrightarrow{v}=\langle v_x,\ v_y\rangle\) or \(\overrightarrow{v}=\langle v_x,\ v_y,v_z\rangle\) are the cosines of the angles the vector forms with the coordinate axes.
The direction cosines are important as they uniquely determine the direction of the vector.
Direction cosines are found by dividing each component of the vector by the magnitude (length) of the vector.
\(\cos \alpha=\frac{v_x}{\|\vec{v}\|}, \quad \cos \beta=\frac{v_y}{\|\vec{v}\|}\)
\(\cos \alpha=\frac{v_x}{\|\vec{v}\|^{\prime}} \quad \cos \beta=\frac{v_y}{\|\vec{v}\|^{\prime}} \quad \cos \theta=\frac{v_z}{\|\vec{v}\|^{\prime}}\)
Find the direction cosines of the vector \(\overrightarrow{v}=\left\langle 4,\left.5,\ 2\right\rangle \right.\) .
Solution
First, find the magnitude of the vector \(\overrightarrow{v}=\left\langle 4,\left.5,\ 2\right\rangle \right.\)
\(\left\|\overrightarrow{v}\right\|=\sqrt{{v_x}^2+{v_y}^2+{v_z}^2}=\ \sqrt{4^2+5^2+2^2}=\ \sqrt{16+25+4}=\sqrt{45}=\sqrt{9\times 5}=3\sqrt{5} \)
Get the direction cosines by dividing each component, 4, 5, and 2, by this magnitude.
\(\begin{aligned}
& \cos \alpha=\frac{v_x}{\|\vec{v}\|}=\frac{4}{3 \sqrt{5}} \approx 0.596 \\
& \cos \beta=\frac{v_y}{\|\vec{v}\|}=\frac{5}{3 \sqrt{5}} \approx 0.745 \\
& \cos \theta=\frac{v_z}{\|\vec{v}\|}=\frac{2}{3 \sqrt{5}} \approx 0.298
\end{aligned}\)
Find the vector \(\overrightarrow{v}\) that has magnitude 32 and direction cosines \(\mathrm{cos}\alpha =5/8\) and \(\ \mathrm{cos}\beta =-3/8\) .
Solution
Since \(\cos \alpha=\frac{v_x}{\|\vec{v}\|}\) and \(\cos \beta=\frac{v_y}{\|\vec{v}\|}\),
\(v_x=\|\vec{v}\| \cdot \cos \alpha=32 \cdot \frac{5}{8}=20\), and
\(v_y=\|\vec{v}\| \cdot \cos \beta=32 \cdot \frac{-3}{8}=-12\).
So, \(\vec{v}=\langle 20,-12\rangle\).
Using Technology
We can use technology to determine the magnitude of a vector.
Go to www.wolframalpha.com.
To find the magnitude of the vector \(\overrightarrow{v}\boldsymbol{=}\left\langle 2,\left.4,\ -6\right\rangle ,\right.\) enter magnitude of \(\mathrm{<}\) 2, 4, -6 \(\mathrm{>}\) in the entry field. Wolframalpha tells you what it thinks you entered, then tells you its answer. In this case, \(\left\|\overrightarrow{v}\right\|=2\sqrt{14}\) .
. The answer is 2 multiplied by the square root of 14." src="/@api/deki/files/97417/clipboard_ed73083bf6ed52a173f4dff61da5c0bce.png">
To find the direction cosines of the vector \(\overrightarrow{v}\boldsymbol{=}\left\langle 4,\left.5,\ 2\right\rangle ,\right.\) enter evaluate 4/(magnitude of \(\mathrm{<}\) 4,5,2 \(\mathrm{>}\) ), 5/(magnitude of \(\mathrm{<}\) 4,5,2 \(\mathrm{>}\) ), 2/(magnitude of \(\mathrm{<}\) 4,5,2 \(\mathrm{>}\) ) in the entry field. WolframAlpha answers \(\{0.596285,\ 0.745356,\ 0.298142\}\) .
We can use WolframAlpha to approximate a vector give its magnitude and direction cosines.
, 5 divided by the magnitude of <4,5,2>, and 2 divided by the magnitude of <4,5,2>. The answer is 4 over 3 times the square root of 5, square root of 5 divided by 3, 2 over start denominator three times the square root of 5 end denominator." src="/@api/deki/files/97418/clipboard_efcdfc03181976fbe8aeb9da126077dd4.png">
Try These
Find the magnitude of the vector \(\vec{v}=\langle-3,4,-2\rangle\).
- Answer
-
\(\left\|\overrightarrow{v}\right\|=\sqrt{29}\)
Find the magnitude of the vector \(\vec{v}=\langle 1,-1\rangle\).
- Answer
-
\(\left\|\overrightarrow{v}\right\|=\sqrt{2}\)
Find the cosines of the vector \(\vec{v}=\langle 3,-1,2\rangle\). Round to three decimal places.
- Answer
-
\(\mathrm{\{}\) 0.802, -0.267, 0.535 \(\mathrm{\}}\)
Approximate the vector \(\vec{v}\) that has magnitude 24 and direction cosines \(\cos \alpha=-3 / 4, \cos \beta=-1 / 4, \cos \theta=7 / 8\).
- Answer
-
\(<-18,\ -6,\ 21>\) | libretexts | 2025-03-17T22:27:20.920514 | 2023-04-03T20:58:00 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions/3.02%3A_Magnitude_and_Direction_Cosines_of_a_Vector",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "3.2: Magnitude and Direction Cosines of a Vector",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions/3.03%3A_Arithmetic_on_Vectors_in_3-Dimensional_Space | 3.3: Arithmetic on Vectors in 3-Dimensional Space
Addition and Subtraction of Vectors
To add or subtract two vectors, add or subtract their corresponding components.
Add the vectors \(\overrightarrow{u}=\left\langle 2,\ 5,\ \left.4\right\rangle \right.\) and \(\overrightarrow{v}=\left\langle 4,\left.2,1\right\rangle \right.,\)
Solution
To add the vectors \(\overrightarrow{u}=\left\langle 2,\ 5,\ \left.4\right\rangle \right.\) and \(\overrightarrow{v}=\left\langle 4,\left.2,1\right\rangle \right.,\) add their corresponding components.
\[\overrightarrow{u}\ +\ \overrightarrow{v}=\left\langle 2+4,\left.5+2,\ 4+1\right\rangle \right.=\ \left\langle 6,\left.7,\ 5\right\rangle \right. \nonumber \] So, \(\overrightarrow{u}\ +\ \overrightarrow{v}=\left\langle 6,\left.7,\ 5\right\rangle \right.\)
Now, graph this sum. Start at the origin.
Since the \(x-\) component is 6, move 6 units in the \(x-\) direction.
Since the \(y-\) component is 7, move 7 units in the \(y-\) direction.
Since the \(z-\) component is 5, move 5 units upward.
Subtract the vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle 2,\ 5,\ \left.4\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 4,\left.2,\ 1\right\rangle \right.\)
Solution
To subtract the vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle 2,\ 5,\ \left.4\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 4,\left.2,\ 1\right\rangle \right.\) subtract their corresponding components.
\[\overrightarrow{u}-\ \overrightarrow{v}\boldsymbol{=}\left\langle 2-4,\left.5-2,\ 4-1\right\rangle \right.\boldsymbol{=\ }\left\langle -2,\left.3,\ 3\right\rangle \right. \nonumber \]
So, \(\overrightarrow{u}-\ \overrightarrow{v}\boldsymbol{=}\left\langle -2,\left.3,\ 3\right\rangle \right.\)
Scalar Multiplication
Scalar multiplication is the multiplication of a vector by a real number (a scalar).
Suppose we let the letter \(k\) represent a real number and \(\overrightarrow{v}\) be the vector \(\left\langle x\right.,\left.y,z\right\rangle .\boldsymbol{\ }\) Then, the scalar multiple of the vector \(\overrightarrow{v}\) is
\[k \vec{v}=\langle k x, k y, k z\rangle \nonumber \]
Suppose \(\overrightarrow{u}=\left\langle -3,\left.-8,\ 5\right\rangle \right.\) and \(k=3\) .
Solution
Then \(k\overrightarrow{u}=\) 3 \(\overrightarrow{u}=3\left\langle -3,\left.-8,\ 5\right\rangle \right.=\ \left\langle 3(-3),\left.3\left(-8\right),3\eqref{GrindEQ__5_}\right\rangle \right.=\left\langle -9,\left.-24,\ 15\right\rangle \right.\)
Suppose \(\overrightarrow{v}=\left\langle 6,\left.3,-12\right\rangle \right.\) and \(k=\frac{-1}{3}\) .
Solution
Then \(k\overrightarrow{u}=\) \(\frac{-1}{3}\overrightarrow{u}=\frac{-1}{3}\left\langle 6,\left.3,\ -12\right\rangle \right.=\ \left\langle \frac{-1}{3}\left(6\right),\left.\frac{-1}{3}\left(3\right),\frac{-1}{3}(-12)\right\rangle \right.=\left\langle -2,\left.-1,\ 4\right\rangle \right.\)
Suppose \(\overrightarrow{u}=\left[ \begin{array}{c} \begin{array}{c} -2 \\ 6 \end{array} \\ 0 \end{array} \right]\) , \(\overrightarrow{v}=\left[ \begin{array}{c} 1 \\ 2 \\ -8 \end{array} \right]\) , and \(\overrightarrow{w}=\left[ \begin{array}{c} -3 \\ -1 \\ 2 \end{array} \right]\) . Find \(3\overrightarrow{u}+4\overrightarrow{v}-2\overrightarrow{w}.\)
Solution
Then \(3\overrightarrow{u}+4\overrightarrow{v}-2\overrightarrow{w}=3\left[ \begin{array}{c} \begin{array}{c} -2 \\ 6 \end{array} \\ 0 \end{array} \right]+4\left[ \begin{array}{c} 1 \\ 2 \\ -8 \end{array} \right]-2\left[ \begin{array}{c} -3 \\ -1 \\ 2 \end{array} \right]=\left[ \begin{array}{c} \begin{array}{c} -6 \\ 18 \end{array} \\ 0 \end{array} \right]+\left[ \begin{array}{c} 4 \\ 8 \\ -32 \end{array} \right]+\left[ \begin{array}{c} 6 \\ 2 \\ -4 \end{array} \right]=\left[ \begin{array}{c} 4 \\ 28 \\ -36 \end{array} \right] \nonumber \)
Using Technology
We can use technology to determine the value of adding or subtracting vectors.
Go to www.wolframalpha.com.
Suppose \(\overrightarrow{u}=\left[ \begin{array}{c} \begin{array}{c} -2 \\ 6 \end{array} \\ 0 \end{array} \right]\) , \(\overrightarrow{v}=\left[ \begin{array}{c} 1 \\ 2 \\ -8 \end{array} \right]\) , and \(\overrightarrow{w}=\left[ \begin{array}{c} -3 \\ -1 \\ 2 \end{array} \right]\) . Use WolframAlpha to find \(3\overrightarrow{u}+4\overrightarrow{v}-2\overrightarrow{w}.\) In the entry field enter evaluate \(3 *[-2,6,0]+4 *[1,2,-8]-2 *[-3,-1,2]\).
plus 4 times <1,2, negative 8> minus 2 times
WolframAlpha answers \((4,\ 28,\ -36)\) which is WolframAlpha’s notation for \(\left[ \begin{array}{c} 4 \\ 28 \\ -36 \end{array} \right]\) .
Try These
Add the vectors \(\vec{u}=\langle-3,4,6\rangle\) and \(\vec{v}=\langle 8,7,-5\rangle\).
- Answer
-
\(\overrightarrow{u}\ +\ \overrightarrow{v}=\left\langle 5,\left.11,\ 1\right\rangle \right.\)
Subtract the vector \(\vec{v}=\langle 8,7,-5\rangle\) from the vector \(\vec{u}=\langle-3,4,6\rangle\).
- Answer
-
\(\overrightarrow{u}-\ \overrightarrow{v}=\left\langle -11,\left.-3,\ 11\right\rangle \right.\)
Given the three vectors, \(\vec{u}=\langle 2,4,-5\rangle, \vec{v}=\langle-3,4,-8\rangle\), and \(\vec{w}=\langle 0,1,2\rangle\), find \(2 \vec{u}+3 \vec{v}-4 \vec{w}\).
- Answer
-
\(2\overrightarrow{u}+3\overrightarrow{v}-4\overrightarrow{w}\ =\left\langle -5,\left.16,\ -42\right\rangle \right.\)
Suppose \(\vec{u}=\left[\begin{array}{c}
3 \\
4 \\
-2
\end{array}\right], \vec{v}=\left[\begin{array}{c}
-1 \\
6 \\
4
\end{array}\right]\), and \(\vec{w}=\left[\begin{array}{l}
0 \\
5 \\
2
\end{array}\right]\), find \(4 \vec{u}-4 \vec{v}-\vec{w}\)
- Answer
-
\(4\overrightarrow{u}-4\overrightarrow{v}-\overrightarrow{w}=\left\langle 16,\left.-13,\ -26\right\rangle \right.\) | libretexts | 2025-03-17T22:27:21.002527 | 2023-04-03T20:58:02 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions/3.03%3A_Arithmetic_on_Vectors_in_3-Dimensional_Space",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "3.3: Arithmetic on Vectors in 3-Dimensional Space",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions/3.04%3A_The_Unit_Vector_in_3-Dimensions_and_Vectors_in_Standard_Position | 3.4: The Unit Vector in 3-Dimensions and Vectors in Standard Position
The Unit Vector in 3-Dimensions
The unit vector, as you will likely remember, in 2-dimensions is a vector of length 1. A unit vector in the same direction as the vector \(\overrightarrow{v}\) is often denoted with a “hat” on it as in \(\hat{v}\) . We call this vector “v hat.”
The unit vector \(\hat{v}\) corresponding to the vector \(\hat{v}\) is defined to be \[\hat{v}=\frac{\vec{v}}{\|\vec{v}\|} \nonumber \]
Unit vector corresponding to the vector \(\overrightarrow{v}\boldsymbol{=}\left\langle -8,\left.12\right\rangle \right.\)
Solution
The unit vector corresponding to the vector \(\overrightarrow{v}\boldsymbol{=}\left\langle -8,\left.12\right\rangle \right.\) is
\[\hat{v}=\frac{\overrightarrow{v}}{\left\|\overrightarrow{v}\right\|}\nonumber \]
\[\begin{gathered}
\hat{v}=\frac{\langle-8,12\rangle}{\sqrt{(-8)^2+(12)^2}} \\
\hat{v}=\frac{\langle-8,12\rangle}{\sqrt{64+144}} \\
\hat{v}=\frac{\langle-8,12\rangle}{\sqrt{208}} \\
\hat{v}=\left\langle\frac{-8}{\sqrt{208}}, \frac{12}{\sqrt{208}}\right\rangle
\end{gathered} \nonumber \]
A unit vector in 3-dimensions and in the same direction as the vector \(\overrightarrow{v}\) is defined in the same way as the unit vector in 2-dimensions.
The unit vector \(\hat{v}\) corresponding to the vector \(\overrightarrow{v}\) is defined to be \(\hat{v}=\frac{\overrightarrow{v}}{\left\|\overrightarrow{v}\right\|}\) , where \(\overrightarrow{v}\boldsymbol{=}\left\langle x,\left.y,\ z\right\rangle \right..\)
For example, the unit vector corresponding to the vector \(\overrightarrow{v}=\left\langle 5,\left.-3,\ 4\right\rangle \right.\) is
\[\hat{v}=\frac{\overrightarrow{v}}{\left\|\overrightarrow{v}\right\|} \nonumber \]
\[\hat{v}=\frac{\left\langle 5,\left.-3,\ 4\right\rangle \right.}{\sqrt{5^2+{\left(-3\right)}^2+4^2}}\nonumber \] \[\hat{v}=\frac{\left\langle 5,\left.-3,\ 4\right\rangle \right.}{\sqrt{25+9+16}}\nonumber \] \[\hat{v}=\frac{\left\langle 5,\left.-3,\ 4\right\rangle \right.}{\sqrt{50}}\nonumber \] \[\hat{v}=\frac{\left\langle 5,\left.-3,\ 4\right\rangle \right.}{5\sqrt{2}}\nonumber \] \[\hat{v}=\left\langle \frac{5}{5\sqrt{2}},\left.\frac{-3}{5\sqrt{2}},\frac{4}{5\sqrt{2}}\right\rangle \right. \nonumber \]
Vectors in Standard Position
A vector with its initial point at the origin in a Cartesian coordinate system is said to be in standard position . A common notation for a unit vector in standard position uses the lowercase letters i, j, and k is to represent the unit vector in
the \(x\) -direction with the vector \(\hat{i}\) , where \(\hat{i}=\left\langle 1,\left.0,\ 0\right\rangle \right.\) , and
the \(y\) -direction with the vector \(\hat{j}\) , where \(\hat{j}=\left\langle 0,\left.1,\ 0\right\rangle \right.\) , and
the \(z\) -direction with the vector \(\hat{k}\) , where \(\hat{k}=\left\langle 0,\left.0,\ 1\right\rangle \right.\) .
The figure shows these three unit vectors.
Any vector can be expressed as a combination of these three unit vectors.
The vector \(\overrightarrow{v}=\left\langle 5,\ \left.4,\ 7\right\rangle \right.\) can be expressed as \(\overrightarrow{v}=5\hat{i}+4\hat{j}+7\hat{k}\) .
Solution
Now, the unit-vector in the direction of \(\overrightarrow{v}=5\hat{i}+4\hat{j}+7\hat{k}\) is
\[\hat{v}=\frac{\overrightarrow{v}}{\left\|\overrightarrow{v}\right\|} \nonumber \] \[\hat{v}=\frac{\left\langle 5,\left.4,\ 7\right\rangle \right.}{\sqrt{5^2+4^2+7^2}}\nonumber \] \[\hat{v}=\frac{\left\langle 5,\left.4,\ 7\right\rangle \right.}{\sqrt{25+16+49}}\nonumber \] \[\hat{v}=\frac{\left\langle 5,\left.-3,\ 4\right\rangle \right.}{\sqrt{90}}\nonumber \] \[\hat{v}=\frac{\left\langle 5,\left.4,\ 7\right\rangle \right.}{\sqrt{9\bullet 10}}\nonumber \] \[\hat{v}=\frac{\left\langle 5,\left.4,\ 7\right\rangle \right.}{3\sqrt{10}}\nonumber \] \[\hat{v}=\left\langle \frac{5}{3\sqrt{10}},\left.\frac{4}{3\sqrt{10}},\frac{7}{3\sqrt{10}}\right\rangle \right.\nonumber \] \[\overrightarrow{v}=\frac{5}{3\sqrt{10}}\hat{i}+\frac{4}{3\sqrt{10}}\hat{j}+\frac{7}{3\sqrt{10}}\hat{k}\nonumber \]
Normalizing a Vector
Normalizing a vector is a common practice in mathematics and it also has practical applications in computer graphics. Normalizing a vector \(\overrightarrow{v}\) is the process of identifying the unit vector of a vector \(\overrightarrow{v}\) .
Using Technology
We can use technology to find the unit vector in the direction of the given vector.
Go to www.wolframalpha.com.
Use WolframAlpha to find the unit vector in the direction of \(\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.4,\ 3\right\rangle \right.\) . Enter normalize \(\mathrm{<}\) 5,4,3 \(\mathrm{>}\) in the entry field and WolframAlpha gives you an answer.
. The result is 1 over the square root of 2, 2 times the square root of 2 over 5, and 3 over the 5 times the square root of 2." src="/@api/deki/files/97444/clipboard_e93161cd99c52ed22a96bec8b04b475d0.png">
Translate WolframAlpha’s answer to \(\frac{1}{\sqrt{2}}\hat{i}+\frac{2\sqrt{2}}{5}\hat{j}+\frac{3}{5\sqrt{2}}\hat{k}\) .
Try These
Write the unit vector that corresponds to \(\vec{v}=\langle 2,-3,4\rangle\).
- Answer
-
\(\frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k}\)
Write the unit vector that corresponds to \(\vec{v}=\langle 1,-1,1\rangle\).
- Answer
-
\(\frac{1}{\sqrt{3}}\hat{i}-\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}\)
Write the unit vector that corresponds to \(\vec{v}-\vec{u}=\langle 6,7,2\rangle-\langle 2,7,6\rangle\).
- Answer
-
\(\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{k}\)
Normalize the vector \(\vec{v}=\langle 4,3,2\rangle\).
- Answer
-
\(\frac{4}{\sqrt{29}}\hat{i}+\frac{3}{\sqrt{29}}\hat{j}+\frac{2}{\sqrt{29}}\hat{k}\) | libretexts | 2025-03-17T22:27:21.079272 | 2023-04-03T20:58:05 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions/3.04%3A_The_Unit_Vector_in_3-Dimensions_and_Vectors_in_Standard_Position",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "3.4: The Unit Vector in 3-Dimensions and Vectors in Standard Position",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions/3.05%3A_The_Dot_Product_Length_of_a_Vector_and_the_Angle_between_Two_Vectors_in_Three_Dimensions | 3.5: The Dot Product, Length of a Vector, and the Angle between Two Vectors in Three Dimensions
The Dot Product of two Vectors
The dot product of two vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle u_x,\left.u_y\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle v_x,\left.v_y\right\rangle \right.\) in two dimensions is nicely extended to three dimensions.
The dot product of vectors \(\vec{u}=\left\langle u_x, u_y, u_z\right\rangle\) and \(\vec{v}=\left\langle v_x, v_y, v_z\right\rangle\) is a scalar (real number) and is defined to be \[\vec{u} \cdot \vec{v}=u_x v_x+u_y v_y+u_z v_z \nonumber \]
Since \(u_x,\ u_y,\ {u_z,\ v}_x\) , \(v_y\) , and \(v_z\) are real numbers, you can see that the dot product is itself a real number and not a vector.
Compute the dot product of the vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.2,\ 4\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 3,\left.4,-7\right\rangle \right.\)
Solution
To compute the dot product of the vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.2,\ 4\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 3,\left.4,-7\right\rangle \right.\) , we compute
\[\overrightarrow{u}\bullet \overrightarrow{v}\boldsymbol{=}5\bullet 3\mathrm{+\ }2\bullet 4+4\bullet \left(-7\right)=15+8-28=-5 \nonumber \]
Since the dot product is a scalar, it follows the properties of real numbers.
- \(\vec{u} \cdot \vec{v}=\vec{v} \cdot \vec{u}\) the dot product is commutative
- \(\vec{u} \cdot(\vec{v}+\vec{w})=\vec{u} \cdot \vec{v}+\vec{u} \cdot \vec{w}\), the dot product distributes over vector addition
- \(\vec{u} \cdot \overrightarrow{0}=0\), the dot product with the zero vector \(\overrightarrow{0}\), is the scalar 0.
- \(\vec{u} \cdot \vec{u}=\|\vec{u}\|^2\)
Compute the dot product \(\overrightarrow{u}\bullet \left(\overrightarrow{v}+\overrightarrow{w}\right)=\overrightarrow{u}\bullet \overrightarrow{v}+\overrightarrow{u}\bullet \overrightarrow{w}\) , where \(\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.-2,\ -3\right\rangle \right.\) , \(\overrightarrow{v}\boldsymbol{=}\left\langle 6,\left.4,\ 1\right\rangle \right.\) , and \(\overrightarrow{w}\boldsymbol{=}\left\langle -3,\left.7,\ -2\right\rangle \right.\) ,
Solution
\[\begin{aligned}
\vec{u} \cdot(\vec{v}+\vec{w})&=\langle 5,-2,-3\rangle \cdot\langle 6,4,1\rangle+\langle 5,-2,-3\rangle \cdot\langle-3,7,-2\rangle\\&=(5 \cdot 6+(-2) \cdot 4+(-3) \cdot 1)+(5 \cdot(-3)+(-2) \cdot 7)+(-3) \cdot(-2))\\ &=30-8-3-15-14+6 \\
&=-4
\end{aligned} \nonumber \]
The Length of a Vector in Three Dimensions
The length (magnitude) of a vector in two dimensions is nicely extended to three dimensions.
The dot product of a vector 𝑣\(\vec{v}=\left\langle v_x, v_y\right\rangle\) with itself gives the length of the vector. \[\|\vec{v}\|=\sqrt{v_x^2+v_y^2} \nonumber \]
You can see that the length of the vector is the square root of the sum of the squares of each of the vector’s components. The same is true for the length of a vector in three dimensions.
The dot product of a vector \(\vec{v}=\left\langle v_x, v_y, v_z\right\rangle\) with itself gives the length of the vector. \[\|\vec{v}\|=\sqrt{v_x^2+v_y^2+v_z^2} \nonumber \]
Use the dot product to find the length of the vector \(\overrightarrow{v}=\left[ \begin{array}{c} 4 \\ 2 \\ 6 \end{array} \right]\) .
Solution
In this case, \(v_x=4\) , \(v_y=2\) , and \(v_z=6\)
Using \(\|\overrightarrow{v}\|=\sqrt{{v_x}^2+{v_y}^2+\ {v_z}^2}\) , we get
\[\begin{gathered}
\|\vec{v}\|=\sqrt{4^2+2^2+6^2} \\
\|\vec{v}\|=\sqrt{56} \\
\|\vec{v}\|=\sqrt{4 \cdot 14} \\
\|\vec{v}\|=\sqrt{4} \cdot \sqrt{14} \\
\|\vec{v}\|=2 \sqrt{14}
\end{gathered} \nonumber \]
The length of the vector \(\overrightarrow{v}=\left[ \begin{array}{c} 4 \\ 2 \\ 6 \end{array} \right]\) is \(2\sqrt{14}\) units.
The Angle between two Vectors
The formula for the angle between two vectors in two dimensions is nicely extended to three dimensions.
If \(\theta\) is the smallest nonnegative angle between two non-zero vectors \(\vec{u}\) and \(\vec{v}\), then
\[\cos \theta=\frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \cdot\|\vec{v}\|} \text { or } \theta=\cos ^{-1} \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \cdot\|\vec{v}\|} \nonumber \]
\[\text { where } 0 \leq \theta \leq 2 \pi \text { and }\|\vec{u}\|=\sqrt{u_x{ }^2+u_y{ }^2+u_z{ }^2} \text { and }\|\vec{v}\|=\sqrt{v_x{ }^2+v_y{ }^2+v_z{ }^2} \nonumber \]
Find the angle between the vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.-3,\ -1\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 2,\left.4,\ -5\right\rangle \right.\) .
Solution
Using \(\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{\overrightarrow{u}\bullet \overrightarrow{v}}{\|\overrightarrow{u}\|\bullet \|\overrightarrow{v}\|}\) , we get
\[\begin{gathered}
\theta=\cos ^{-1} \frac{\langle 5,-3,-1\rangle \cdot\langle 2,4,-5\rangle}{\sqrt{5^2+(-3)^2+(-1)^2} \cdot \sqrt{2^2+4^2+(-5)^2}} \\
\theta=\cos ^{-1} \frac{5 \cdot 2+(-3) \cdot 4+(-1) \cdot(-5)}{\sqrt{25+9+1} \cdot \sqrt{4+16+25}} \\
\theta=\cos ^{-1} \frac{3}{\sqrt{35} \cdot \sqrt{45}} \\
\theta=85.66
\end{gathered} \nonumber \]
We conclude that the angle between these two vectors is close to 85.7 \(\mathrm{{}^\circ}\) rounded to one decimal place.
Using Technology
We can use technology to find the magnitude of the vector and the angle \(\theta\) between two vectors.
Go to www.wolframalpha.com.
To find the magnitude (length) of the vector \(\overrightarrow{v}\boldsymbol{=}\left\langle 4,\left.2,\ 4\right\rangle ,\right.\) enter magnitude of \(\mathrm{<}\) 4, 2, 4 \(\mathrm{>}\) in the entry field. Wolframalpha tells you what it thinks you entered, then tells you its answer. In this case, \(\left\|\overrightarrow{v}\right\|=6\) .
. The result is 6." src="/@api/deki/files/97445/clipboard_e5b9b6fa9319843040ab3d357fd7e33ef.png">
To find the angle between the vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.-3\boldsymbol{,\ }-1\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 2,\left.4,\ -5\right\rangle \right.\) , enter angle between vectors \(\mathrm{<}\) 5, -3, -1 \(\mathrm{>}\) and \(\mathrm{<}\) 2, 4, -5 \(\mathrm{>}\) in the entry field. Wolframalpha tells you what it thinks you entered, then tells you its answer. In this case, \(\theta =85.7\mathrm{{}^\circ }\) , rounded to one decimal place.
and <2,4, negative 5>. The result is 85.6647 degrees." src="/@api/deki/files/97446/clipboard_ecd5ac2462c48e45c3c2106a9f3324bd9.png">
Try These
Find the dot product of the vectors \(\vec{u}=\langle-2,3,-9\rangle\) and \(\vec{v}=\langle 5,-1,2\rangle\)
- Answer
-
\(\overrightarrow{u}\bullet \overrightarrow{v}=-31\)
Find the dot product of the vectors \(\vec{u}=\langle 6,2,-1\rangle\) and \(\vec{v}=\langle 2,-7,-2\rangle\).
- Answer
-
\(\overrightarrow{u}\bullet \overrightarrow{v}=0\)
Find the length of the vector \(\vec{u}=\langle 4,-7,-6\rangle\).
- Answer
-
\(\sqrt{101}\)
Find the length of the vector \(\vec{v}=\langle 0,5,0\rangle\)
- Answer
-
\(5\)
Find the angle between the vectors \(\vec{u}=\langle 3,4,5\rangle\) and \(\vec{v}=\langle-3,-1,8\rangle\).
- Answer
-
\(63.6{}^\circ\)
Find the angle between the vectors \(\vec{u}=\langle 1,-2,1\rangle\) and \(\vec{v}=\langle 3,5,7\rangle\)
- Answer
-
\(90{}^\circ\) | libretexts | 2025-03-17T22:27:21.169743 | 2023-04-03T20:58:08 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions/3.05%3A_The_Dot_Product_Length_of_a_Vector_and_the_Angle_between_Two_Vectors_in_Three_Dimensions",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "3.5: The Dot Product, Length of a Vector, and the Angle between Two Vectors in Three Dimensions",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions/3.06%3A_The_Cross_Product-_Algebra | 3.6: The Cross Product- Algebra
The Cross Product of two Vectors
A vector that is perpendicular to both vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle u_x,\left.u_y\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle v_x,\left.v_y\right\rangle \right.\) , can be found using the cross product. The cross product requires that both vectors be in three-dimensional space.
The cross product of vectors \(\vec{u}=\left\langle u_x, u_y, u_z\right\rangle\) and \(\vec{v}=\left\langle v_x, v_y, v_z\right\rangle\) is a vector and is defined to be \[\vec{u} \times \vec{v}=\left\langle u_y v_z-u_z v_y, u_z v_x-u_x v_z, u_x v_y-u_y v_z\right\rangle \nonumber \]
This formula is challenging to remember. A nice device to help you remember both this formula and the dot product formula is to visualize them in a 3x3 square of components. The square shows how vectors can interact with one another.
For the cross product,
The \(x\) -component has a product that involves no \(x\) -components: \({\ u}_yv_z-\ u_zv_y\)
The \(y\) -component has a product that involves no \(y\) -components: \({\ u}_zv_x-u_xv_z\)
The \(z\) -component has a product that involves no \(z\) -components: \({{\ u}_xv}_y\mathrm{-}\mathrm{\ }u_yv_z\)
Each component is a difference of two diagonal products.
To produce the 𝑥𝑥-component, (top right) - (bottom left) = y*z – z*y
To produce the 𝑦𝑦-component, (bottom left) - (top right) = z*x – x*z
To produce the 𝑧𝑧-component, (top right) – (bottom left) = x*y – y*x
The DOT product is the interaction between two vectors having similar components:
\[x \cdot x, \quad y \cdot y, \quad z \cdot z \nonumber \]
The dot product measures similarity since it combines only interactions of matching components.
The CROSS product is the interaction between two vectors having different components:
\[x \cdot y, x \cdot z, y \cdot x, y \cdot z, z \cdot x, z \cdot y \nonumber \]
The cross product measures cross interactions since it combines interactions of different components.
Find the cross product of the vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.2,\ 4\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 3,\left.4,\ -7\right\rangle \right.\) .
Solution
To produce the \(x\) -component, (top right) - (bottom left) = 2*(-7) – 4*4 = -30
To produce the \(y\) -component, (bottom left) - (top right) = 4*3 – 5*(-7) = 47
To produce the \(z\) -component, (top right) – (bottom left) = 5*4 – 2*3 = 14
\[\overrightarrow{u}\times \overrightarrow{v}\boldsymbol{=}\left\langle -30,\ 47,\ \left.14\right\rangle \right. \nonumber \]
*Be careful with the computation. It goes (bottom left) – (top right) whilethe first and last go (top right) – (bottom left).
Using Technology
We can use technology to find the cross product between two vectors.
Go to www.wolframalpha.com
To find the cross product of the vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.2,\ 4\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 3,\left.4,\ -7\right\rangle \right.\boldsymbol{,\ }\) use either the “cross” or the x command. Wolframalpha tells you what it thinks you entered, then it tells you its answer. In this case, \(\overrightarrow{u}\times \overrightarrow{v}\boldsymbol{=}\left\langle -30,\ 47,\ \left.14\right\rangle \right.\) .
and <3,4,negative 7>, or <5,2,4> x <3,4,negative 7>. The result is the vector (negative 30,47,14)." src="/@api/deki/files/97451/clipboard_e9640f1af100c5e766bbe45d473b19fb1.png">
The Right-Hand Rule
You can see that the cross product of the two vectors \(\overrightarrow{u}\) and \(\overrightarrow{v}\) , is itself a vector. But where is this vector \(\overrightarrow{u}\times \overrightarrow{v}?\) The cross product of two vectors is a vector that is perpendicular to the plane formed by the two vectors. What about the two perpendicular directions? Does this perpendicular vector lie above or below the plane formed by the two vectors? We use the right-hand rule .
Hold your hand as shown in the picture, your index and middle fingers extended. Your thumb points in the direction of the cross product.
Since the dot product is a scalar, it follows the properties of real numbers.
- \(\vec{u} \times \vec{v}=-\vec{v} \times \vec{u}\), , the cross product is anti-commutative
- \(\vec{u} \times(\vec{v}+\vec{w})=\vec{u} \times \vec{v}+\vec{u} \times \vec{w}\), the cross product distributes over vector addition
- \(k(\vec{u} \times \vec{v})=(k \vec{u}) \times \vec{v}=\vec{u} \times(k \vec{u})\)
- \(\vec{u} \times \overrightarrow{0}=\overrightarrow{0}\), the cross product with the zero vector \(\vec{0}\), is the zero vector \(\vec{0}\)
Using Technology
For example, use WolframAlpha to compute both the cross product \(\overrightarrow{u}\times \overrightarrow{v}\) and \(\overrightarrow{v}\times \overrightarrow{u}\) , with
\(\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.-2,\ -3\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 6,\left.4,\ 1\right\rangle \right.\) , to show that one is the opposite of the other.
x <6,4,1>. The result is <10,negative 23,32>." src="/@api/deki/files/97453/clipboard_e067e85cbf91bc75737263e604bf16bbe.png">
x <5,negative 2,negative 3>. The result is
Notice that \(\left\langle 10,\left.-23,-32\right\rangle \right.=\ -\left\langle -10,\left.23,\ -32\right\rangle \right.\) , verifying property 1.
Try These
Find the cross product of the vectors \(\vec{u}=\langle 4,-2,1\rangle\) and \(\vec{v}=\langle 5,-1,3\rangle\).
- Answer
-
\(\overrightarrow{u}\times \overrightarrow{v}\boldsymbol{=\ }\left\langle -5,\left.-7,\ 6\right\rangle \right.\)
Find the cross product of the vectors \(\vec{u}=\langle-2,3,-9\rangle\) and \(\vec{v}=\langle-8,12,-36\rangle\).
- Answer
-
\(\overrightarrow{u}\times \overrightarrow{v}=\overrightarrow{0}\)
Find \(\vec{u} \times \vec{v} \cdot \vec{w}\), where \(\vec{u}=\langle-2,5,3\rangle\), \(\vec{v}=\langle 4,4,-2\rangle\), and \(\vec{w}=\langle 2,6,-5\rangle\).
- Answer
-
144
* Note that the cross product must be computed first since if it is not, we would be crossing a vector with a scalar. | libretexts | 2025-03-17T22:27:21.249417 | 2023-04-03T20:58:12 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions/3.06%3A_The_Cross_Product-_Algebra",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "3.6: The Cross Product- Algebra",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/03%3A_Vectors_in_Three_Dimensions/3.07%3A_The_Cross_Product-_Geometry | 3.7: The Cross Product- Geometry
The Cross Product of two Vectors and the Right-Hand Rule
The cross product of the two vectors \(\overrightarrow{u}\) and \(\overrightarrow{v}\) , is itself a vector. Where is this vector \(\overrightarrow{u}\times \overrightarrow{v}\boldsymbol{?}\) The cross product of two vectors is a vector perpendicular to the plane formed by the two vectors. What if there are two perpendicular directions? Does this perpendicular vector lie above or below the plane formed by the two vectors? Let’s use the right-hand rule .
The Geometry of the Cross Product
If \(\theta\) is the angle between the two vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle u_x,\left.u_y,\ u_z\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle v_x,\left.v_y,\ v_z\right\rangle \right.\) , then the length (magnitude) of the cross product \(\overrightarrow{u}\times \overrightarrow{v}\) is
\[\|\vec{u} \times \vec{v}\|=\|\vec{u}\|\|\vec{v}\| \sin \theta \nonumber \]
\[\|\vec{u}\|=\sqrt{u_x^2+u_y{ }^2+u_z^2} \text { and }\|\vec{v}\|=\sqrt{v_x^2+v_y^2+v_z^2} \nonumber \]
Length of the vector \(\overrightarrow{u}\times \overrightarrow{v}\) , where \(\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.2,\ 4\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 3,\left.4,\ -7\right\rangle \right.\)
Solution
The length of the vector \(\overrightarrow{u}\times \overrightarrow{v}\) , where \(\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.2,\ 4\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 3,\left.4,\ -7\right\rangle \right.\) is
\[\left\|\overrightarrow{u}\times \overrightarrow{v}\right\|=\ \left\|\overrightarrow{u}\right\|\left\|\overrightarrow{v}\right\|\mathrm{sin}\theta \nonumber \]
\[\sqrt{{u_x}^2+{u_y}^2+\ {u_z}^2}\bullet \sqrt{{v_x}^2+{v_y}^2+\ {v_z}^2}\bullet \mathrm{sin}\theta \nonumber \]
We now need to get \(\mathrm{sin}\theta\) . We’ll use the formula for the angle between two vectors.
\[\begin{gathered}
\theta=\cos ^{-1} \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \cdot\|\vec{v}\|} \\
\theta=\cos ^{-1} \frac{\langle 5,2,4\rangle \cdot\langle 3,4,-7\rangle}{\sqrt{5^2+2^2+4^2} \cdot \sqrt{3^2+4^2+(-7)^2}} \\
\theta=\cos ^{-1} \frac{5 \cdot 3+2 \cdot 4+4 \cdot(-7)}{\sqrt{45} \cdot \sqrt{74}} \\
\theta=\cos ^{-1} \frac{-5}{\sqrt{45} \cdot \sqrt{74}} \\
\theta=94.97^{\circ}
\end{gathered} \nonumber \]
Now we compute \(\left\|\overrightarrow{u}\times \overrightarrow{v}\right\|=\sqrt{45}\bullet \sqrt{74}{\mathrm{sin} \left(94.97{}^\circ \right)\ }=57.49\) units.
Using Technology
We can use technology to find the magnitude of the cross product of two vectors.
Go to www.wolframalpha.com.
To find the length of the cross product of the vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.2,\ 4\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 3,\left.4,\ -7\right\rangle \right.\) enter magnitude of \(\mathrm{<}\) 5, 2, 4 \(\mathrm{>}\) cross \(\mathrm{<}\) 3, 4, -7 \(\mathrm{>}\) in the entry field. Wolframalpha tells you what it thinks you entered and its answer. In this case it shows you result of \(\sqrt{3305}\) . Click on the approximate form button to get the result in decimal form as \(57.49\) .
x <3,4, negative 7>. The result is 57.4891." src="/@api/deki/files/97458/clipboard_e13da1df53b00e241350a56220a04e76b.png">
Area of a Parallelogram
Geometrically, \(\left\|\overrightarrow{u}\times \overrightarrow{v}\right\|\) produces the area of a parallelogram determined by \(\overrightarrow{u}\) and \(\overrightarrow{v}\) .
\[\text { Area }=\|\vec{u} \times \vec{v}\|=\|\vec{u}\|\|\vec{v}\| \sin \theta \nonumber \]
The area of the parallelogram determined by the vectors \(\vec{u}=\langle 5,2,4\rangle\) and \(\vec{v}=\langle 3,4,-7\rangle\) from the above example is 57.49 square units.
The Cross Product of Perpendicular and Parallel Vectors
If vectors \(\vec{u}\) and \(\vec{v}\) are perpendicular to each other, then the angle between them is 90° and sin(90°) = 1, so that \[\vec{u} \times \vec{v}=\|\vec{u}\|\|\vec{v}\| \nonumber \]
If vectors \(\vec{u}\) and \(\vec{v}\) are parallel to each other, then the angle between them is \(0^{\circ}\) and \(\sin \left(0^{\circ}\right)=0\).
It makes sense then to define the cross product of parallel vectors to be the zero vector, \(\overrightarrow{0}\) . Also, if at least one of the vectors \(\overrightarrow{u}\) and \(\overrightarrow{v}\) is the zero vector \(\overrightarrow{0}\) , then the cross product \(\overrightarrow{u}\times \overrightarrow{v}\) is defined to be the zero vector. We can say that if the cross product of two vectors is zero, then the vectors are parallel to each other. Also, if two vectors are parallel to each other, then their cross product is zero. We combine these statements together in an if-and-only-if statement.
Nonzero vectors \(\vec{u}\) and \(\vec{v}\) are parallel to each other if and only if \(\vec{u} \times \vec{v}=0\).
- \(\vec{u} \times \vec{v}=-\vec{v} \times \vec{u}\), the cross product is anti-commutative
- \(k(\vec{u} \times \vec{v})=k \vec{u} \times \vec{v}=\vec{u} \times k \vec{v}\), multiplication by a scalar
- \(\vec{u} \times(\vec{v}+\vec{w})=\vec{u} \times \vec{v}+\vec{u} \times \vec{w}\), the cross product distributes over vector addition
- \(\vec{u} \times \overrightarrow{0}=\overrightarrow{0}\), the cross product with the zero vector \(\vec{0}\), is the zero vector \(\vec{0}\)
Use WolframAlpha to verify that \[\overrightarrow{u}\times \left(\overrightarrow{v}+\overrightarrow{w}\right)=\overrightarrow{u}\times \overrightarrow{v}+\overrightarrow{u}\times \overrightarrow{w} \nonumber \]
where \(\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.-2,\ -3\right\rangle \right.\) , \(\overrightarrow{v}\boldsymbol{=}\left\langle 6,\left.4,\ 1\right\rangle \right.\) , and \(\overrightarrow{w}\boldsymbol{=}\left\langle -3,\left.7,\ 2\right\rangle \right.\) .
Solution
Use WA to first compute \(\overrightarrow{u}\times \left(\overrightarrow{v}+\overrightarrow{w}\right)\boldsymbol{=}\overrightarrow{u}\) and then \(\overrightarrow{u}\times \overrightarrow{v}+\overrightarrow{u}\times \overrightarrow{w}\) . Determine if the results do or do not match. We can do this in one step by entering
\(\mathrm{<}\) 5,-2,-3 \(\mathrm{>}\) x ( \(\mathrm{<}\) 6,4,1 \(\mathrm{>}\) + \(\mathrm{<}\) -3,7,2 \(\mathrm{>}\) ) = \(\mathrm{<}\) 5,-2,-3 \(\mathrm{>}\) x \(\mathrm{<}\) 6,4,1 \(\mathrm{>}\) + \(\mathrm{<}\) 5,-2,-3 \(\mathrm{>}\) x \(\mathrm{<}\) -3,7,2 \(\mathrm{>}\)
If the statement on the left of the = equals the statement on the right, WA responds with True.
If the statement on the left of the \(\mathrm{\neq}\) equals the statement on the right, WA responds with False.
In this case, we get a True response and have verified the truth of the statement.
x (<6,4,1>+
Try These
Find the cross product of the vectors \(\vec{u}=\langle-4,3,5\rangle\) and \(\vec{v}=\langle 5,-1,2\rangle\).
- Answer
-
\(\overrightarrow{u}\times \overrightarrow{v}\boldsymbol{=\ }\left\langle 11,\left.33,-11\right\rangle \right.\)
Find the cross product of the vectors \(\vec{u}=\langle-2,3,-9\rangle\) and \(\vec{v}=\langle 6,-9,27\rangle\).
- Answer
-
\(\overrightarrow{u}\times \overrightarrow{v}=\left.\boldsymbol{\langle }0,\ 0,\ 0\right\rangle = \overrightarrow{0}\)
Find the length of the vector formed by the cross product of the vectors \(\vec{u}=\langle 3,-5,4\rangle\) and \(\vec{v}=\langle 2,-4,1\rangle\).
- Answer
-
\(5\sqrt{6}\) = 12.2 units
Find the angle between the vectors \(\vec{u}=\langle 4,-7,-6\rangle\) and \(\vec{v}=\langle 5,-1,2\rangle\).
- Answer
-
\(\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{\overrightarrow{u}\bullet \overrightarrow{v}}{‖\overrightarrow{u}‖∙‖\overrightarrow{v}‖}\) \(\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{\left\langle 4,\left.-7,6\right\rangle \right.\boldsymbol{\bullet }\left\langle 5,\left.-1,\ 2\right\rangle \right.}{\sqrt{4^2+{(-7)}^2+\ 6^2}\bullet \sqrt{5^2+{(-1)}^2+\ 2^2}}\) \(\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{4∙5+(-7)∙(-1)+6∙2}{\sqrt{91}\bullet \sqrt{30}}\) \(\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{25}{\sqrt{101}\bullet \sqrt{30}}\) \(\theta =\ \mathrm{74.19}\mathrm{{}^\circ }\)
Determine if the vectors \(\vec{u}=\langle 3,-2,1\rangle\) and \(\vec{v}=\langle 0,2,4\rangle\) are perpendicular or parallel to each other.
- Answer
-
Perpendicular since \(\overrightarrow{u}\bullet \overrightarrow{v}\boldsymbol{=}0\)
Find the area of the parallelogram and the triangle formed by the vectors \(\vec{u}=\langle 1,-2,-4\rangle\) and \(\vec{v}=\langle 4,3,-5\rangle\).
- Answer
-
Parallelogram is \(26.94\) square units. Triangle is (½ of \(26.94)\) = \(13.47\) square units. | libretexts | 2025-03-17T22:27:21.339075 | 2023-04-03T20:58:13 | {
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"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "3.7: The Cross Product- Geometry",
"author": ""
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https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices | 4: Matrices Last updated Save as PDF Page ID 125042 4.1: Matrices 4.2: Addition, Subtraction, Scalar Multiplication, and Products of Row and Column Matrices 4.3: Matrix Multiplication 4.4: Rotation Matrices in 2-Dimensions 4.5: Finding the Angle of Rotation Between Two Rotated Vectors in 2-Dimensions 4.6: Rotation Matrices in 3-Dimensions | libretexts | 2025-03-17T22:27:21.419138 | 2023-04-03T20:58:15 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "4: Matrices",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices/4.01%3A_Matrices | 4.1: Matrices
Matrix
A matrix is a rectangular array of objects, often numbers.
The rectangular array of numbers \(\left[ \begin{array}{c} \begin{array}{cc} 5 & -2 \\ 0 & 4 \\ -6 & 3 \end{array} \\ \ \end{array} \right]\) is a matrix having 3 rows and 2 columns.
Dimension of a Matrix
Matrix having \(m\) number of rows and \(n\) number of columns has dimension (size) \(m \times n\) (pronounced as "m by n") and is called an \(m \times n\) matrix.
The matrix in Example \(3\times 2\) matrix since it is composed of 3 rows and 2 columns. When specifying the dimension of a matrix, the number of rows is stated first and the number of columns second.
is aElements of a Matrix
It is common to use an uppercase letter of the alphabet to name a matrix and the corresponding lowercase letter to name an element (entry or member) of the matrix. Subscripts are attached to the lowercase letter to specify its position in the matrix.
The first number in subscript indicates the row in which the element resides and the second number the column.
The subscript numbers appear adjacent to each other and typically without a comma separating them.
We could name the matrix of Example . We specify the element \(-2\) in row 1, column 2, with the notation \(a_{12}\) . The lowercase \(a\) is used to indicate that the element is from matrix \(A\) and the subscripts indicate we are observing the entry in row 1, column 2. The subscript is not the number 12, but rather the two individual numbers, 1 and 2.
In general, the notation \(a_{i j}\) denotes the entry in row \(i\) and column \(j\).
Some other elements of \(A\) are
\(a_{11}=5,\) the number in row 1, column 1
\(a_{31}=-6,\) the number in row 3, column 1
\(a_{22}=4,\) the number in row 2, column 2
In general, an \(m\mathrm{\times }n\) matrix has the form \(\left[ \begin{array}{c} \begin{array}{c} \begin{array}{cc} a_{\mathrm{11}} & a_{\mathrm{12}} \end{array} \mathrm{\ \ \ \ }\mathrm{\cdots }\mathrm{\ \ \ }a_{\mathrm{1}n} \\ \begin{array}{cc} a_{\mathrm{21}} & a_{\mathrm{22}} \end{array} \mathrm{\ \ \ \ }\mathrm{\cdots }\mathrm{\ \ \ }a_{\mathrm{2}n} \\ \begin{array}{cc} a_{\mathrm{31}} & a_{\mathrm{32}} \end{array} \mathrm{\ \ \ \ }\mathrm{\cdots }\mathrm{\ \ \ }a_{\mathrm{3}n} \end{array} \\ \mathrm{\vdots } \\ \begin{array}{cc} a_{m\mathrm{1}} & a_{m\mathrm{2}} \end{array} \mathrm{\ \ \ \ }\mathrm{\cdots }\mathrm{\ \ \ }a_{mn} \end{array} \right]\) . For some number \(m\) , the element \(a_{m\mathrm{2}}\) is the number in row \(m\) , column 2.
Your turn:
In the matrix \(B\mathrm{=}\left[ \begin{array}{c} \begin{array}{ccc} 0 & \mathrm{-}\mathrm{4} & \mathrm{2} \\ \mathrm{1} & \mathrm{-}\mathrm{1} & \mathrm{5} \\ \mathrm{-}\mathrm{3} & \mathrm{3} & \mathrm{8} \end{array} \\ \mathrm{\ } \end{array} \right],\)
- Specify the size of \(B\) .
- Find the value of \(b_{11}.\)
- Find the value of \(b_{13}.\)
- Find the value of \(b_{32}.\)
ANS: (a) \(3\times 3\) , (b) 0, (c) 2, (d) 3
Equal Matrices
Two matrices \(A\) and \(B\) are said to be equal, written as \(A=B\), if they are the same size and all the corresponding entries are equal.
In matrix notation, for all \(i\) and \(j\) , \(\mathrm{\ }A\mathrm{=}B\) if \(a_{ij}\mathrm{=}b_{ij}.\) The notation \(a_{ij}\) names the element in row \(i\) and column \(j\) of matrix \(A\) . Similarly, the notation \(b_{ij}\) names the element in row \(i\) and column \(j\) of matrix \(B\) . The notation \(a_{ij}\mathrm{=}b_{ij}\) indicates that the element in row \(i\) and column \(j\) of matrix \(A\) is the same as the element in row \(i\) and column \(j\) of matrix \(B.\)
Square Matrices
A matrix is called square if it has the same number of columns as rows.
The \(2\times 2\ \) matrices \(A\) and \(B\) are both equal and square.
\(A=\left[ \begin{array}{cc} 3 & 4 \\ 2 & 1 \end{array} \right]\) and \(B=\left[ \begin{array}{cc} 3 & 4 \\ 2 & 1 \end{array} \right]\)
Main Diagonal of a Square Matrix
Consider a square matrix, say \[A=\left[\begin{array}{ccc}
2 & 1 & 0 \\
3 & 4 & -5 \\
-1 & 6 & 7
\end{array}\right] \nonumber \]
Imagine a line passing from the top left element to the bottom right element as in the picture.
\[\left.A=\mid \begin{array}{ccc}
8 & 1 & 0 \\
3 & 4 & -5 \\
-1 & 6 & 7
\end{array}\right] \nonumber \]
This diagonal set of elements from the top left element to the bottom right is called the main diagonal of the matrix.
Diagonal and Non-Diagonal Elements of a Matrix
The elements lying on the main diagonal of matrix \(A\) are called the diagonal elements of matrix \(A\) . The elements 2, 4, and 7 are the diagonal elements of matrix \(A\) . The elements lying off the main diagonal of matrix \(A\) are called the non-diagonal or off-diagonal elements of matrix \(A\) . The elements 1, 0, 3, -5, -1, and 6 are the non-diagonal elements of matrix \(A\) .
The Identity Matrix
An Identity matrix is a square matrix that has only 1’s on its main diagonal and 0’s everywhere else.
A matrix in which every diagonal element is 1 and every non-diagonal element is 0 is an identity matrix. Identity matrices are typically named with the uppercase letter \(I.\) It is not uncommon to write the size of the matrix as a subscript on the \(I.\)
The square matrix \(I=\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]\) is the \(3\times 3\) . We could write \(I_{3\times 3}\) to indicate the \(3\times 3\) identity matrix.
The Zero Matrix
The zero matrix is a matrix, in which every element is 0.
Zero matrices are commonly named with a 0.
The matrix \(0=\left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{array} \right]\) is a zero matrix.
The Transpose of a Matrix
Consider some \(m\times n\) matrix \(A.\ \) For example, suppose \(A\) is the \(2\times 3\) matrix \(A=\left[ \begin{array}{ccc} 2 & 3 & 4 \\ 5 & 6 & 7 \end{array} \right]\) .Form a new matrix, call it \(A\) -transpose and denote it by \(A^T\) , by making
- The first row of \(A\) the first column of \(A^T\) ,
- The second row of \(A\) the second column of \(A^T\) .
Then \(A^T=\left[ \begin{array}{cc} 2 & 5 \\ 3 & 6 \\ 4 & 7 \end{array} \right]\) is the transpose of \(A=\ \left[ \begin{array}{ccc} 2 & 3 & 4 \\ 5 & 6 & 7 \end{array} \right]\) .
The rows of a matrix are the columns of its transpose. If the matrix \(A\) is size \(m \times n\), then dimension of \(A^T\) is \(n \times m\).
Row Matrices and Column Matrices
A row matrix is a matrix with only one row and any number of columns.
The matrix \(R=\left[ \begin{array}{ccc} 3 & 4 & 5 \end{array} \right]\) is a row matrix with 3 columns. It is a \(1\times 3\) matrix.
A column matrix is a matrix with only one column and any number of rows.
The matrix \(C=\left[ \begin{array}{c} 6 \\ 7 \end{array} \right]\) is a column matrix with 2 rows. It is a \(2\times 1\) matrix.
Vectors as Matrices
When we first described vectors, we expressed them using the bracket notation. For example, we could write a vector as \(\left\langle 2,\ 4,\ 6\right\rangle\) . We can just as easily describe this vector using a row matrix \(\left[ \begin{array}{ccc} 2 & 4 & 6 \end{array} \right]\ \ \) or column matrix \(\left[ \begin{array}{c} 2 \\ 4 \\ 6 \end{array} \right].\)
Try These
Specify the dimension of each matrix.
-
\(S=\left[\begin{array}{ccc}
0 & 2 & 5 \\
-6 & -3 & 2 \\
1 & 9 & 2 \\
8 & -1 & 4
\end{array}\right]\) -
\(T=\left[\begin{array}{lll}
5 & 6 & -3 \\
0 & 0 & -3
\end{array}\right]\) -
\(Q=\left[\begin{array}{lll}
1 & 0 & -1
\end{array}\right]\)
- Answer
-
- \(4\times 3\)
- \(2\times 3\)
- \(1\times 3\)
True or False: The transpose of a square matrix is also a square matrix.
- Answer
-
True
In the matrix \(S=\left[\begin{array}{ccc}
0 & 2 & 5 \\
-6 & -3 & 2 \\
1 & 9 & 2 \\
8 & -1 & 4
\end{array}\right]\)
- Find the value of \(S_{13}\).
- Find the value of \(S_{23}\).
- Find the value of \(S_{31}\).
- Find the value of \(S_{43}\).
- Answer
-
- 5
- 2
- 1
- 4
Construct and name the transpose of \(S=\left[\begin{array}{ccc}
0 & 2 & 5 \\
-6 & -3 & 2 \\
1 & 9 & 2 \\
8 & -1 & 4
\end{array}\right]\)
- Answer
-
\(S^T=\left[ \begin{array}{ccc} 0 & -6 & 1 \\ 2 & -3 & 9 \\ 5 & 2 & 2 \end{array} \ \ \ \ \begin{array}{c} 8 \\ -1 \\ 4 \end{array} \right]\)
Construct \(I_{4 \times 4}\).
- Answer
Construct the transpose of \(I_{3 \times 3}\).
- Answer
-
\(I^T_{3\times 3}=I_{3\times 3}\)
Write the column matrix \(\left[\begin{array}{l}
4 \\
3 \\
2
\end{array}\right]\) using vector bracket notation, < >.
- Answer
-
\(\mathrm{<}\) 4,3,2 \(\mathrm{>}\)
Construct a 2 × 2 matrix in which the diagonal elements are 5 and 6 and the non-diagonal elements are 0 and 2.
- Answer
-
\(\left[ \begin{array}{cc} 5 & 0 \\ 2 & 6 \end{array} \right]\) or \(\left[ \begin{array}{cc} 5 & 2 \\ 0 & 6 \end{array} \right]\) | libretexts | 2025-03-17T22:27:21.606338 | 2023-04-03T20:58:17 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices/4.01%3A_Matrices",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "4.1: Matrices",
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} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices/4.02%3A_Addition_Subtraction_Scalar_Multiplication_and_Products_of_Row_and_Column_Matrices | 4.2: Addition, Subtraction, Scalar Multiplication, and Products of Row and Column Matrices
Addition and Subtraction of Matrices
Let \(A\) and \(B\ \) be \(m\times n\) matrices. Then the sum, \(A+B\) , is the new matrix formed by adding corresponding entries together. The difference, \(A-B\) , is the new matrix formed by subtracting each entry in matrix \(B\) from its corresponding entry in matrix \(A\) .
To add or subtract two or more matrices together, they all must be of the same size. That is, they all must have the same number of rows and the same numbers of columns. To add them together, add the corresponding elements together. To subtract one from the other, subtract corresponding elements from each other
If the addition and subtraction is defined (if it is possible), perform each operation.
\[A=\left[\begin{array}{cc}
2 & 5 \\
-1 & 4 \\
6 & 0
\end{array}\right], B=\left[\begin{array}{ll}
3 & 1 \\
0 & 4 \\
2 & 7
\end{array}\right], C=\left[\begin{array}{cc}
9 & -4 \\
2 & 6
\end{array}\right] \nonumber \]
Solution
\[A+B=\left[\begin{array}{cc}
2 & 5 \\
-1 & 4 \\
6 & 0
\end{array}\right]+\left[\begin{array}{ll}
3 & 1 \\
0 & 4 \\
2 & 7
\end{array}\right]=\left[\begin{array}{cc}
2+3 & 5+1 \\
-1+0 & 4+4 \\
6+2 & 0+7
\end{array}\right]=\left[\begin{array}{cc}
5 & 6 \\
-1 & 8 \\
8 & 7
\end{array}\right] \nonumber \]
\[A-B=\left[\begin{array}{cc}
2 & 5 \\
-1 & 4 \\
6 & 0
\end{array}\right]-\left[\begin{array}{ll}
3 & 1 \\
0 & 4 \\
2 & 7
\end{array}\right]=\left[\begin{array}{cc}
2-3 & 5-1 \\
-1-0 & 4-4 \\
6-2 & 0-7
\end{array}\right]=\left[\begin{array}{cc}
-1 & 4 \\
-1 & 0 \\
4 & -7
\end{array}\right] \nonumber \]
\(A+C\) is not defined as they are different sizes. Matrix \(A\) is a \(3\times 2\) matrix whereas matrix \(B\) is a \(2\times 2\) matrix.
Your Turn : Compute \(B\ -A.\)
Scalar Multiplication
You might recall that a scalar is a physical quantity that is defined by only its magnitude and that some examples are speed, time, distance, density, and temperature. They are represented by real numbers (both positive and negative), and they can be operated on using the regular laws of algebra.
To multiply a matrix by a scalar, multiply every element of the matrix by the scalar.
Symbolically, \(k\bullet \left[ \begin{array}{c} \begin{array}{cc} a_{11} & a_{12} \end{array} \ \ \ \cdots \ \ \ a_{1n} \\ \begin{array}{cc} a_{21} & a_{22}\ \ \ \cdots \ \ \ a_{2n} \end{array} \\ \begin{array}{cc} \vdots & \vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots \end{array} \\ \begin{array}{cc} a_{m1} & a_{m2} \end{array} \ \ \ \cdots \ \ \ a_{mn} \end{array} \right]=\ \left[ \begin{array}{c} \begin{array}{cc} {k\bullet a}_{11} & {k\bullet a}_{12} \end{array} \ \ \ \cdots \ \ \ {k\bullet a}_{1n} \\ \begin{array}{cc} {k\bullet a}_{21} & {k\bullet a}_{22}\ \ \ \cdots \ \ \ {k\bullet a}_{2n} \end{array} \\ \begin{array}{cc} \vdots & \vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots \end{array} \\ \begin{array}{cc} {k\bullet a}_{m1} & k\bullet a_{m2} \end{array} \ \ \ \cdots \ \ \ {k\bullet a}_{mn} \end{array} \right]\)
Perform scalar multiplication.
Solution
\[6 \cdot\left[\begin{array}{ccc}
4 & 1 & -3 \\
0 & 3 & 5
\end{array}\right]=\left[\begin{array}{ccc}
6 \cdot 4 & 6 \cdot 1 & 6 \cdot(-3) \\
6 \cdot 0 & 6 \cdot 3 & 6 \cdot 5
\end{array}\right]=\left[\begin{array}{ccc}
24 & 6 & -18 \\
0 & 18 & 30
\end{array}\right] \nonumber \]
Your Turn : Multiply \(\mathrm{7\ }\mathrm{\bullet }\left[ \begin{array}{cc} \mathrm{4} & \mathrm{2} \\ --\mathrm{1} & \mathrm{3} \end{array} \right]\) .
Multiplication with Row and Column Matrices
Suppose we have two matrices, \(A\) and \(B,\) where \(\mathrm{\ }A\) is a \(\mathrm{1\times }n\) matrix and \(B\) is an \(n\mathrm{\times 1}\) matrix. That is, \(A\) has one row and \(n\) columns and \(B\) has \(n\) rows and only 1 column.
\[A=\left[\begin{array}{llll}
a_1 & a_2 & \cdots & a_n
\end{array}\right] \text { and } B=\left[\begin{array}{c}
b_1 \\
b_2 \\
\vdots \\
b_n
\end{array}\right] \nonumber \]
The product \(A\bullet B\) is the new matrix obtained by multiplying together the corresponding elements of each matrix then adding those sums together.
\[A \cdot B=\left[\begin{array}{llll}
a_1 & a_2 & \cdots & a_n
\end{array}\right] \cdot\left[\begin{array}{c}
b_1 \\
b_2 \\
\vdots \\
b_n
\end{array}\right] \nonumber \]
\[A\bullet B=\left[a_1\bullet b_1+a_2\bullet b_2+\cdots +a_n\bullet b_n\right] \nonumber \]
This product is the sum (addition) of the
first entry in \(A\) times the first entry in \(B\)
second entry in \(A\) times the second entry in \(B\)
\(\vdots \)
last entry in \(A\) times the last entry in \(B\)
Multiply the two matrices \(A=\left[ \begin{array}{ccc} 2 & 4 & 5 \end{array} \right]\ \) and \(B=\left[ \begin{array}{c} 1 \\ 4 \\ 3 \end{array} \right]\)
Solution
Suppose \(A=\left[ \begin{array}{ccc} 2 & 4 & 5 \end{array} \right]\ \) and \(B=\left[ \begin{array}{c} 1 \\ 4 \\ 3 \end{array} \right]\) . Then,
\[A \cdot B=\left[\begin{array}{lll}
2 & 4 & 5
\end{array}\right] \cdot\left[\begin{array}{l}
1 \\
4 \\
3
\end{array}\right] \nonumber \]
\[\begin{aligned}
& A \cdot B=[2 \cdot 1+4 \cdot 4+5 \cdot 3] \\
& A \cdot B=[2+16+15] \\
& A \cdot B=[33]
\end{aligned} \nonumber \]
Notice the dimensions of the two matrices. The number of rows of \(B,\) is 3 which is equal to the number of columns of \(A,\) which is also 3. The product is a \(1\times 1\) matrix whose dimension is the (number of rows of \(A)\ \times\) (number of columns of \(B\) ).
Your Turn : Suppose \(A\mathrm{=}\left[ \begin{array}{cc} \mathrm{3} & \mathrm{1} \end{array} \mathrm{\ } \begin{array}{cc} \mathrm{\ \ \ }-\mathrm{2} & \mathrm{5} \end{array} \right]\) and \(B\mathrm{=}\left[ \begin{array}{c} \begin{array}{c} \mathrm{1} \\ -\mathrm{6} \end{array} \\ 0 \\ \mathrm{4} \end{array} \right]\) . Show that \(A\mathrm{\bullet }B\mathrm{=[17].}\)
Motivation for the Process of Multiplication with Row and Column Matrices
This process of multiplication may not seem intuitive; however, we can motivate it with an example. You probably know, or at least believe, that the revenue \(R\ \) realized by selling \(n\) number of units of some product for \(p\) dollars per unit is given by \(R=np\) . Revenue equals (the number of units sold) times the (price of each unit).
Suppose your business sells three sizes of boxes, small-sized boxes, medium-sized boxes, and large-sized boxes. Small boxes sell for $3 each, medium boxes for $5 each, and large boxes for $7 each. What would your total revenue be if you sold 20 small-sized boxes, 30 medium-sized boxes, and 40 large-sized boxes?
Solution
Using \(R=n\mathrm{\bullet }p\) , your revenue from the sale of the
small boxes is \(R=20\bullet \$3=\$60\)
medium boxes is \(R=30\bullet \$5=\$150\)
large boxes is \(R=40\bullet \$7=\$280\)
The total revenue is the sum of these three products, \[20\bullet \$3+30\bullet \$5+ 40\bullet \$7=\$60+\$150+\$280=\$490. \nonumber \]
We can compute the total revenue using two matrices and matrix multiplication.
Let the first matrix be the row matrix of the number of boxes sold \(N\) ,
\[N=\left[ \begin{array}{ccc} 20 & 30 & 40 \end{array} \right] \nonumber \]
and the second matrix be the column matrix of the price per boxes sold \(P\) .
\[P=\left[ \begin{array}{c} \$3 \\ \$5 \\ \$7 \end{array} \right] \nonumber \]
The total revenue is the matrix product
\[R=N\mathrm{\bullet }P \nonumber \] \[R\mathrm{=}\left[ \begin{array}{ccc} \mathrm{20} & \mathrm{30} & \mathrm{40} \end{array} \right]\mathrm{\bullet }\left[ \begin{array}{c} \mathrm{\$3} \\ \mathrm{\$5} \\ \mathrm{\$7} \end{array} \right] \nonumber \]
\[\mathrm{=}\left[\mathrm{20}\mathrm{\bullet }\mathrm{\$3+30}\mathrm{\bullet }\mathrm{\$5+40}\mathrm{\bullet }\mathrm{\$7}\right] \nonumber \]
\[\mathrm{=}\left[\mathrm{\$490}\right] \nonumber \]
Important Observation – See this
Notice that the result of a row and column matrix multiplication is a matrix with exactly one entry. That entry is the sum of a collection of products. In Example 4, the result of the row and column matrix multiplication is a matrix with exactly one entry, $490. The $490 is the sum of the products \(\mathrm{20}\mathrm{\bullet }\mathrm{\$3,}\) \(\mathrm{30}\mathrm{\bullet }\mathrm{\$5}\) , and \(\mathrm{40}\mathrm{\bullet }\mathrm{\$7.}\) Don’t let the phrase the sum of a collection of products befuddle you. It means it is the addition (the sum) of a collection of multiplications (products). This idea will be helpful in the next section when we discuss multiplication of matrices of larger dimensions.
Dimension Matters
Notice the dimensions of the two matrices \(N\) and \(P\) from Example 4. The number of rows of \(P\) is 3, which is equal to the number columns of \(N,\) which is also 3. The product is a \(1\times 1\) matrix whose dimension is (the number of rows of \(N)\ \times\) (the number of columns of \(P\) ).
To multiply a row matrix \(A\) and column matrix \(B\) together, it must be that the (number of rows of \(B\)) = (number of columns of \(A\))
Symbolically, if \(A\) has \(n\) number of columns, \(B\) must have \(n\) number of rows
Suppose \(A=\left[ \begin{array}{ccc} 2 & 4 & 5 \end{array} \right]\ \) and \(B=\left[ \begin{array}{c} \begin{array}{c} 1 \\ 4 \\ 3 \end{array} \\ 2 \end{array} \right]\) .
Solution
This multiplication will not work, it is not defined. Matrix \(B\) has 4 rows, but \(A\) has only 3 columns.
\(A\bullet B\) = \(\left[ \begin{array}{ccc} 2 & 4 & 5 \end{array} \right]\) \(\bullet \left[ \begin{array}{c} \begin{array}{c} 1 \\ 4 \\ 3 \end{array} \\ 2 \end{array} \right]=\left[2\bullet 1+4\bullet 4+5\bullet 3+Now\ what?\right]\)
Try these
Using these six matrices, perform each operation if it is defined (if it is possible).
\[A=\left[\begin{array}{cc}
1 & 2 \\
-1 & 3 \\
0 & 4
\end{array}\right] \quad B=\left[\begin{array}{cc}
3 & 2 \\
1 & 0 \\
-1 & -2
\end{array}\right] \quad C=\left[\begin{array}{lll}
2 & 1 & 3 \\
4 & 2 & 1
\end{array}\right] \quad D=\left[\begin{array}{ll}
4 & 9
\end{array}\right] \quad E=\left[\begin{array}{l}
5 \\
2
\end{array}\right] \quad F=[-2] \nonumber \]
\(A+B\)
- Answer
-
\(\left[ \begin{array}{cc} 4 & 4 \\ 0 & 3 \\ -1 & 2 \end{array} \right]\)
\(B-A\)
- Answer
-
\(\left[ \begin{array}{cc} -2 & 0 \\ 2 & -3 \\ -1 & -6 \end{array} \right]\)
\(D+E\)
- Answer
-
Not possible
\(D \cdot E\)
- Answer
-
\(\left[38\right]\)
\(E \cdot D\)
- Answer
-
\(\left[38\right]\)
\(-3 \cdot\left[\begin{array}{cc}
-4 & 0 \\
2 & -1
\end{array}\right]\)
- Answer
-
\(\left[ \begin{array}{cc} 12 & 0 \\ -6 & 3 \end{array} \right]\)
\(A+C\)
- Answer
-
Not defined
\(3 \cdot(D \cdot E)\)
- Answer
-
\(\left[114\right]\)
\((D \cdot E) \cdot F\)
- Answer
-
\(\left[-76\right]\)
\(F^2\)
- Answer
-
\(\left[4\right]\) | libretexts | 2025-03-17T22:27:21.718226 | 2023-04-03T20:58:19 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices/4.02%3A_Addition_Subtraction_Scalar_Multiplication_and_Products_of_Row_and_Column_Matrices",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "4.2: Addition, Subtraction, Scalar Multiplication, and Products of Row and Column Matrices",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices/4.03%3A_Matrix_Multiplication | 4.3: Matrix Multiplication
Compatible Matrices
We are going to multiply together two matrices, one of size \(m\times n\) , and one of size \(n\times p\) . The multiplication will be possible, and the product exists because the sizes make them compatible with each other.
Notice the number of columns of the leftmost matrix is equal to the number of rows of the rightmost matrix.
For the product, \(A \cdot B\), of two matrices to exist it must be that (the number of columns of matrix \(A\) ) = (the number of rows of matrix \(B\)) Matrices for which this is true are said to be compatible with each other.
Matrices as Collections of Row and Column Matrices
It is productive to think of a matrix as a collection of individual row matrices and column matrices.For example, we can think of the matrix \(A=\left[ \begin{array}{cc} 3 & 1 \\ --4 & 2 \\ 0 & 5 \end{array} \right]\) as being composed of
- the three row matrices, \(\left[ \begin{array}{cc} 3 & 1 \end{array} \right],\ \ \left[ \begin{array}{cc} --4 & 2 \end{array} \right],\) and \(\left[ \begin{array}{cc} 0 & 5 \end{array} \right],\ \) and
- the two column matrices \(\left[ \begin{array}{c} 3 \\ --4 \\ 0 \end{array} \right]\) and \(\left[ \begin{array}{c} 1 \\ 2 \\ 5 \end{array} \right]\) .
(If you need a review of row and column matrices, see Section 4.2)
Multiplication of Two Matrices
To multiply two compatible matrices \(A\) and \(B\) together, multiply every row matrix of \(A\) through every column matrix of \(B\).
Suppose the size of matrix \(A\) is \(3\times 4\) and the size of matrix \(B\) is \(4\times 5\) . The matrices are compatible with each other and the size of the product is \(3\times 5\)
Some of the entries of the product \(A\bullet B\) are
\(a_{11}\) : The entry in row 1, column 1, is the result of multiplying the
1st row of matrix \(A\) through the 1st column of matrix \(B\) .
\(a_{12}\) : The entry in row 1, column 2, is the result of multiplying the
1st row of matrix \(A\) through the 2nd column of matrix \(B\) .
\(a_{24}\) : The entry in row 2, column 4, is the result of multiplying the
2nd row of matrix \(A\) through the 4th column of matrix \(B\) .
\(a_{35}\) : The entry in row 3, column \(5\) , is the result of multiplying the
3rd row of matrix \(A\) through the 5th column of matrix \(B\) .
\(a_{33}\) : The entry in row 3, column 3, is the result of multiplying the
3rd row of matrix \(A\) through the 3rd column of matrix \(B\) .
Do you see the general rule for producing any particular entry?
To get the entry in row \(i\) and column \(\ j\) , \(a_{ij},\ \) multiply
the \(ith\) row of matrix \(A\) through the \(jth\) column of matrix \(B.\)
Compute the product of the matrices \(A=\left[ \begin{array}{cc} 3 & 1 \\ --4 & 2 \\ 0 & 5 \end{array} \right]\) and \(B=\left[ \begin{array}{cc} 3 & 2 \\ 4 & 1 \end{array} \right]\) .
First note that the two matrices are compatible
Solution
\[A \cdot B=\left[\begin{array}{cc}
3 & 1 \\
-4 & 2 \\
0 & 5
\end{array}\right] \cdot\left[\begin{array}{ll}
3 & 2 \\
4 & 1
\end{array}\right] \nonumber \]
The product is the \(3\times 2\) matrix of the form \(\left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{array} \right]\)
Since we are multiplying 3 rows through 2 columns, there will be 6 entries. The six entries of \(A\bullet B\) are
\(a_{11}=\ \) the 1st row of \(A\) times the 1st column of \(B\)
\[=\left[\begin{array}{ll}
3 & 1
\end{array}\right] \cdot\left[\begin{array}{l}
3 \\
4
\end{array}\right]=[3 \cdot 3+1 \cdot 4]=[13] \nonumber \]
\(a_{12}=\ \) the 1st row of \(A\) times the 2nd column of \(B\)
\[=\left[\begin{array}{ll}
3 & 1
\end{array}\right] \cdot\left[\begin{array}{l}
2 \\
1
\end{array}\right]=[3 \cdot 2+1 \cdot 1]=[7] \nonumber \]
\(a_{21}=\ \) the 2nd row of \(A\) times the 1st column of \(B\)
\[=\left[\begin{array}{ll}
-4 & 2
\end{array}\right] \cdot\left[\begin{array}{l}
3 \\
4
\end{array}\right]=[-4 \cdot 3+2 \cdot 4]=[-4] \nonumber \]
\(a_{22}=\ \) the 2nd row of \(A\) times the 2nd column of \(B\)
\[=\left[\begin{array}{ll}
-4 & 2
\end{array}\right] \cdot\left[\begin{array}{l}
2 \\
1
\end{array}\right]=[-4 \cdot 2+2 \cdot 1]=[-6] \nonumber \]
\(a_{31}=\ \) the 3rd row of \(A\) times the 1st column of \(B\)
\[=\left[\begin{array}{ll}
0 & 5
\end{array}\right] \cdot\left[\begin{array}{l}
3 \\
4
\end{array}\right]=[0 \cdot 3+5 \cdot 4]=[20] \nonumber \]
\(a_{32}=\) the 3rd row of \(A\) times the 2nd column of \(B\)
\[=\left[\begin{array}{ll}
0 & 5
\end{array}\right] \cdot\left[\begin{array}{l}
2 \\
1
\end{array}\right]=[0 \cdot 2+5 \cdot 1]=[5] \nonumber \]
So, \(A \cdot B=\left[\begin{array}{cc}
13 & 7 \\
-4 & -6 \\
20 & 5
\end{array}\right]\)
Your Turn : Show that the product of the matrices \(A\mathrm{=}\left[ \begin{array}{cc} \mathrm{2} & \mathrm{3} \\ \mathrm{4} & \mathrm{1} \end{array} \right]\) and \(B\mathrm{=}\left[ \begin{array}{ccc} \mathrm{2} & \mathrm{3} & 0 \\ \mathrm{1} & \mathrm{2} & \mathrm{4} \end{array} \right]\) is \(\left[ \begin{array}{ccc} \mathrm{7} & \mathrm{12} & \mathrm{12} \\ \mathrm{9} & \mathrm{14} & \mathrm{4} \end{array} \right]\) .
Using Technology
You can see that multiplying matrices together involves a lot of arithmetic and can be cumbersome. We can use technology to help us through the process.
Go to www.wolframalpha.com.
To find the product of the two matrices of above Your Turn Example, enter [[2,3], [4,1]] * [[2,3,0], [1,2,4]] in the entry field. WolframAlpha sees a matrix as a collection of row matrices.
Both entries and rows are separated by commas and WA does not see spaces.
Wolframalpha tells you what it thinks you entered, then tells you its answer \(\left[ \begin{array}{ccc} 7 & 12 & 12 \\ 9 & 14 & 4 \end{array} \right]\) .
Try these
\[A=\left[\begin{array}{cc}
1 & 2 \\
-1 & 3 \\
0 & 4
\end{array}\right] \quad B=\left[\begin{array}{cc}
3 & 2 \\
1 & 0 \\
-1 & -2
\end{array}\right] \quad C=\left[\begin{array}{lll}
2 & 1 & 3 \\
4 & 2 & 1
\end{array}\right] \quad D=\left[\begin{array}{cc}
-2 & 6 \\
4 & 1
\end{array}\right] \quad E=\left[\begin{array}{l}
5 \\
2
\end{array}\right] \quad F=\left[\begin{array}{l}
2 \\
1 \\
5
\end{array}\right] \nonumber \]
\(A \cdot C\)
- Answer
-
\(\left[ \begin{array}{ccc} 10 & 5 & 5 \\ 10 & 5 & 0 \\ 16 & 8 & 4 \end{array} \right]\)
\(C \cdot A\)
- Answer
-
\(\left[ \begin{array}{cc} 1 & 19 \\ 2 & 18 \end{array} \right]\)
Compare your answers to question 1 and 2. If you got them right, would you say that matrix multiplication is or is not commutative?
- Answer
-
Is not commutative
\(D \cdot C\)
- Answer
-
\(\left[ \begin{array}{ccc} 20 & 10 & 0 \\ 12 & 6 & 13 \end{array} \right]\)
\(C \cdot F\)
- Answer
-
\(\left[ \begin{array}{c} 20 \\ 15 \end{array} \right]\)
\(A \cdot E\)
- Answer
-
\(\left[ \begin{array}{c} 9 \\ 1 \\ 8 \end{array} \right]\)
\(D^2\)
- Answer
-
\(\left[ \begin{array}{cc} 28 & -6 \\ -4 & 25 \end{array} \right]\)
\(D \cdot\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
- Answer
-
\(D\)
\(B \cdot D\)
- Answer
-
\(\left[ \begin{array}{cc} 2 & 20 \\ -2 & 6 \\ -6 & -8 \end{array} \right]\)
\(B \cdot D \cdot C\)
- Answer
-
\(\left[ \begin{array}{ccc} 84 & 42 & 26 \\ 20 & 10 & 0 \\ -44 & -22 & -26 \end{array} \right]\)
\(D \cdot B\)
- Answer
-
Not defined | libretexts | 2025-03-17T22:27:21.818133 | 2023-04-03T20:58:21 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices/4.03%3A_Matrix_Multiplication",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "4.3: Matrix Multiplication",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices/4.04%3A_Rotation_Matrices_in_2-Dimensions | 4.4: Rotation Matrices in 2-Dimensions
The Rotation Matrix
To this point, we worked with vectors and with matrices. Now, we will put them together to see how to use a matrix multiplication to rotate a vector in the counterclockwise direction through some angle \(\theta\) in 2-dimensions.
Our plan is to rotate the vector
\(v=\left[ \begin{array}{c} x \\ y \end{array} \right]\)
counterclockwise through some angle
\(\theta\)
to the new position given by the vector \(v^{\prime}=\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]\). To do so, we use the rotation matrix, a matrix that rotates points in the
\(xy\)
-plane counterclockwise through an angle
\(\theta\)
relative to the
\(x\)
-axis.
\[\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right] \nonumber \]
The Rotation Process
To get the coordinates of the new vector \(\left[{ \begin{array}{c} x \\ y^{\mathrm{'}} \end{array} }^{\mathrm{'}}\right]\mathrm{,\ }\) perform the matrix multiplication
\[\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right] \nonumber \]
Find the vector \(\left[{ \begin{array}{c} x \\ y^{\mathrm{'}} \end{array} }^{\mathrm{'}}\right]\mathrm{\ }\) that results when the vector \(\left[ \begin{array}{c} x \\ y \end{array} \right]\mathrm{=}\left[ \begin{array}{c} \mathrm{1} \\ --\mathrm{1} \end{array} \right]\) is rotated 90 \(\mathrm{{}^\circ}\) counterclockwise.
Solution
Using the rotation formula \(\left[{ \begin{array}{c} x \\ y^{\mathrm{'}} \end{array} }^{\mathrm{'}}\right]\mathrm{=}\left[ \begin{array}{cc} \mathrm{cos}\theta & \mathrm{-}\mathrm{sin}\theta \\ \mathrm{sin}\theta & \mathrm{cos}\theta \end{array} \right]\left[ \begin{array}{c} x \\ y \end{array} \right]\) with \(\left[ \begin{array}{c} x \\ y \end{array} \right]\mathrm{=}\left[ \begin{array}{c} \mathrm{1} \\ --\mathrm{1} \end{array} \right]\)
and \(\theta \mathrm{=90{}^\circ ,}\) we get
\[\begin{aligned}
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{cc}
\cos 90^{\circ} & -\sin 90^{\circ} \\
\sin 90^{\circ} & \cos 90^{\circ}
\end{array}\right]\left[\begin{array}{c}
1 \\
-1
\end{array}\right]} \\
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\left[\begin{array}{c}
1 \\
-1
\end{array}\right]=\left[\begin{array}{c}
0 \cdot 1+(-1) \cdot(-1) \\
1 \cdot 1+0 \cdot(-1)
\end{array}\right]} \\
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{l}
1 \\
1
\end{array}\right]}
\end{aligned} \nonumber \]
When rotated counterclockwise 90°, the vector \(\left[\begin{array}{c}
1 \\
-1
\end{array}\right] \) becomes \(\left[\begin{array}{c}
1 \\
1
\end{array}\right]\)
Find the vector
\(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]\)
that results when the vector \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
2 \\
6
\end{array}\right]\) is rotated 60° counterclockwise.
Solution
Using the rotation formula \(\left[\begin{array}{c}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) with \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
2 \\
6
\end{array}\right]\) and
\(\theta =60\mathrm{{}^\circ },\)
we get
\[\begin{aligned}
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{cc}
\cos 60^{\circ} & -\sin 60^{\circ} \\
\sin 60^{\circ} & \cos 60^{\circ}
\end{array}\right]\left[\begin{array}{l}
2 \\
6
\end{array}\right]} \\
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
1 / 2 & -\sqrt{3} / 2 \\
\sqrt{3} / 2 & 1 / 2
\end{array}\right]\left[\begin{array}{l}
2 \\
6
\end{array}\right]=\left[\begin{array}{c}
1 / 2 \cdot 2+(-\sqrt{3} / 2) \cdot 6 \\
\sqrt{3} / 2 \cdot 2+1 / 2 \cdot 6
\end{array}\right]} \\
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{l}
1-3 \sqrt{3} \\
3+\sqrt{3}
\end{array}\right]}
\end{aligned} \nonumber \]
When rotated counterclockwise 60
\(\mathrm{{}^\circ}\)
, the vector \(\left[\begin{array}{l}
2 \\
6
\end{array}\right]\) becomes \(\left[\begin{array}{l}
1-3 \sqrt{3} \\
3+\sqrt{3}
\end{array}\right]\).
Using Technology
We can use technology to help us find the rotation. WolframAlpha evaluates the trig functions for us.
Go to www.wolframalpha.com.
We can check the above problem from Example 2 by using WolframAlpha. Find the vector \(\left[\begin{array}{c}
x^{\prime} \\
y^{\prime}
\end{array}\right]\) that results when the vector \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
2 \\
6
\end{array}\right]\) is rotated 60° counterclockwise. To find rotation of the vector enter evaluate \([[\cos (60),-\sin (60)],[\sin (60), \cos (60)]] *[2,6]\) into the entry field.
When rotated counterclockwise 60 \(\mathrm{{}^\circ}\) , the vector \(\left[ \begin{array}{c} 2 \\ 6 \end{array} \right]\) becomes \(\left[ \begin{array}{c} 1--3\sqrt{3} \\ 3+\sqrt{3} \end{array} \right]\) .
Try these
Find the vector \(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]\) that results when \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
1 \\
1
\end{array}\right]\) is rotated 90° counterclockwise.
- Answer
-
\(\left[ \begin{array}{c} -1 \\ 1 \end{array} \right]\)
Find the vector \(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]\) that results when \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
1 \\
1
\end{array}\right]\) is rotated 180° counterclockwise.
- Answer
-
\(\left[ \begin{array}{c} -1 \\ -1 \end{array} \right]\)
Find the vector \(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]\) that results when \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
1 \\
1
\end{array}\right]\) is rotated 270° counterclockwise.
- Answer
-
\(\left[ \begin{array}{c} 1 \\ -1 \end{array} \right]\)
Find the vector \(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]\) that results when \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
0 \\
1
\end{array}\right]\) is rotated 90° counterclockwise.
- Answer
-
\(\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]\)
Find the vector \(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]\) that results when \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
1 \\
1
\end{array}\right]\) is rotated 45° counterclockwise.
- Answer
-
\(\left[ \begin{array}{c} 0 \\ \sqrt{2} \end{array} \right]\)
Find the vector \(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]\) that results when \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
-1 \\
1
\end{array}\right]\) is rotated 45° counterclockwise.
- Answer
-
\(\left[ \begin{array}{c} -\sqrt{2} \\ 0 \end{array} \right]\)
Find the vector \(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]\) that results when \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
– 2.20205 \\
4.48898
\end{array}\right]\) is rotated -63° counterclockwise.
- Answer
-
\(\left[ \begin{array}{c} 3 \\ 4 \end{array} \right]\)
Find the vector \(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]\) that results when \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
-3 \\
-3
\end{array}\right]\) is rotated -90° counterclockwise.
- Answer
-
\(\left[ \begin{array}{c} -3 \\ 3 \end{array} \right]\)
Approximate, to five decimal places, the coordinates of the vector \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
-1 \\
1
\end{array}\right]\) when it is rotated counterclockwise 30°.
- Answer
-
\(\left[ \begin{array}{c} -1.36603 \\ 0.36603 \end{array} \right]\) | libretexts | 2025-03-17T22:27:21.917169 | 2023-04-03T20:58:22 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices/4.04%3A_Rotation_Matrices_in_2-Dimensions",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "4.4: Rotation Matrices in 2-Dimensions",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices/4.05%3A_Finding_the_Angle_of_Rotation_Between_Two_Rotated_Vectors_in_2-Dimensions | 4.5: Finding the Angle of Rotation Between Two Rotated Vectors in 2-Dimensions
Given the Rotated Vector, Find the Angle of Rotation
Suppose we did not know the angle \(\theta\) of rotation. We can get it by working backwards and solving a system of equations. The rotation formula
\[\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right] \nonumber \]
produces the system of equations
\[\left\{\begin{array}{c}
x^{\prime}=x \cdot \cos \theta+y \cdot(-\sin \theta) \\
y^{\prime}=x \cdot \sin \theta+y \cdot \cos \theta
\end{array}\right. \nonumber \]
In
Example 4.4.1 of Section 4.4
, we found that when the vector \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
1 \\
-1
\end{array}\right]\) was rotated counterclockwise by 90°, it became the vector
\(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{l}
1 \\
1
\end{array}\right]\)
. We got this rotated vector by applying the rotation formula \(\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\).
\[\begin{gathered}
{\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]} \\
{\left[\begin{array}{l}
1 \\
1
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{c}
1 \\
-1
\end{array}\right]} \\
{\left[\begin{array}{l}
1 \\
1
\end{array}\right]=\left[\begin{array}{c}
1 \cdot \cos \theta+(-1) \cdot(-\sin \theta) \\
1 \cdot \sin \theta+(-1) \cdot \cos \theta
\end{array}\right]}
\end{gathered} \nonumber \]
Since two vectors are equal only if their corresponding components are equal, we have the system of two equations
\[\left\{\begin{array}{c}
1=1 \cdot \cos \theta+(-1) \cdot(-\sin \theta) \\
1=1 \cdot \sin \theta+(-1) \cdot \cos \theta
\end{array}\right. \nonumber \]
Using Technology
We can use WolframAlpha to help us solve above system for the angle of rotation, \(\theta .\)
Go to www.wolframalpha.com.
Since we want to rotate only one time around the coordinate system, we want to instruct WA to give us solutions only where the angle \(\theta\) is between 0 and 2 \(\pi\) .
Using the English letter \(x\) in place of the Greek letter \(\theta\), enter
Solve \(1=1^* \cos (x)+(-1)^*(-\sin (x)), 1=1^* \sin (x)+(-1)^* \cos (x), 0<=x<=2^* \text { pi }\) in the entry field.
WA shows the angle of rotation is \(\theta =\frac{\pi }{2}\) , which is 90°. We conclude that the angle of rotation is 90°.
In
Example 4.4.2 of Section 4.4
, we found that when the vector
\(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
2 \\
6
\end{array}\right]\)
was rotated counterclockwise by 60°, it became the vector \(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{c}
1-3 \sqrt{3} \\
3+\sqrt{3}
\end{array}\right]\). We got this rotated vector by applying the rotation formula \(\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\).
\[\begin{aligned}
& {\left[\begin{array}{c}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]} \\
& {\left[\begin{array}{c}
1-3 \sqrt{3} \\
3+\sqrt{3}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{l}
2 \\
6
\end{array}\right]} \\
& {\left[\begin{array}{c}
1-3 \sqrt{3} \\
3+\sqrt{3}
\end{array}\right]=\left[\begin{array}{c}
2 \cdot \cos \theta+6 \cdot(-\sin \theta) \\
2 \cdot \sin \theta+6 \cdot \cos \theta
\end{array}\right]}
\end{aligned} \nonumber \]
Since two vectors are equal only if their corresponding components are equal, we have the system of two equations
\[\left\{\begin{array}{c}
1-3 \sqrt{3}=2 \cdot \cos \theta+6 \cdot(-\sin \theta) \\
3+\sqrt{3}=2 \cdot \sin \theta+6 \cdot \cos \theta
\end{array}\right. \nonumber \]
We will use WolframAlpha to help us solve this system for the angle of rotation, \(\theta\).
Using the English letter x in place of the Greek letter \(\theta ,\) enter
Solve \(1-3 \operatorname{sqrt}(3)=2^* \cos (x)+6^*(-\sin (x)), 3+\operatorname{sqrt}(3)=2^* \sin (x)+6^* \cos (x), 0<=x<=2^* \text { pi }\)
in the entry field. Separate the two equations with a comma.
WA shows the angle of rotation is \(\theta =\frac{\pi }{3}\) , which is 60°. We conclude that the angle of rotation is 60°.
Try these
Find the angle \(\theta\) through which the vector \(\left[\begin{array}{l}
3 \\
3
\end{array}\right]\) is rotated to become \(\left[\begin{array}{c}
0 \\
3 \sqrt{2}
\end{array}\right]\).
- Answer
-
\(\theta =\frac{\pi }{4}=45{}^\circ\)
Find the angle \(\theta\) through which the vector \(\left[\begin{array}{l}
2 \\
-2
\end{array}\right]\) is rotated to become \(\left[\begin{array}{c}
1+\sqrt{3} \\
-1+\sqrt{3}
\end{array}\right]\).
- Answer
-
\(\theta =\frac{\pi }{3}=60{}^\circ\)
Find the angle \(\theta\) through which the vector \(\left[\begin{array}{l}
2 \\
0
\end{array}\right]\) is rotated to become \(\left[\begin{array}{c}
0 \\
-2
\end{array}\right]\).
- Answer
-
\(\theta =\frac{3\pi }{2}=270{}^\circ\)
Find the angle \(\theta\) through which the vector \(\left[\begin{array}{l}
-2 \\
-2
\end{array}\right]\) is rotated to become \(\left[\begin{array}{l}
-1+\sqrt{3} \\
-1-\sqrt{3}
\end{array}\right]\).
- Answer
-
\(\theta =\frac{\pi }{3}=60{}^\circ\)
Find the angle \(\theta\) through which the vector \(\left[\begin{array}{l}
1 \\
1
\end{array}\right]\) is rotated to become \(\left[\begin{array}{c}
\sqrt{2} \\
0
\end{array}\right]\).
- Answer
-
\(\theta =\frac{7\pi }{4}=315{}^\circ = -45{}^\circ\) | libretexts | 2025-03-17T22:27:21.998865 | 2023-04-03T20:58:24 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices/4.05%3A_Finding_the_Angle_of_Rotation_Between_Two_Rotated_Vectors_in_2-Dimensions",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "4.5: Finding the Angle of Rotation Between Two Rotated Vectors in 2-Dimensions",
"author": ""
} |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices/4.06%3A_Rotation_Matrices_in_3-Dimensions | 4.6: Rotation Matrices in 3-Dimensions
The Three Basic Rotations
A basic rotation of a vector in 3-dimensions is a rotation around one of the coordinate axes. We can rotate a vector counterclockwise through an angle \(\theta\) around the \(x\)–axis, the \(y\)–axis, or the \(z\)–axis.
To get a counterclockwise view, imagine looking at an axis straight on toward the origin.
Our plan is to rotate the vector
\(\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\)
counterclockwise around one of the axes through some angle
\(\theta\)
to the new position given by the vector \(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]\). To do so, we will use one of the three rotation matrices.
The Rotation Matrices
The rotation matrices for \(x\) , \(y\) , and \(z\) axes are, respectively,
\[\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \theta & -\sin \theta \\
0 & \sin \theta & \cos \theta
\end{array}\right] \quad\left[\begin{array}{ccc}
\cos \theta & 0 & \sin \theta \\
0 & 1 & 0 \\
-\sin \theta & 0 & \cos \theta
\end{array}\right] \quad\left[\begin{array}{ccc}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{array}\right] \nonumber \]
The Rotation Process
\(x\)–axis
To rotate the vector
\(\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\)
counterclockwise through an angle
\(\theta\)
around the
\(x\)
–axis to a new position
\(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]\),
perform the matrix multiplication,
\[\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \theta & -\sin \theta \\
0 & \sin \theta & \cos \theta
\end{array}\right] \cdot\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] \nonumber \]
\(y\)–axis
To rotate the vector
\(\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\)
counterclockwise through an angle
\(\theta\)
around the
\(y\)
–axis to a new position
\(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]\),
perform the matrix multiplication,
\[\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]=\left[\begin{array}{ccc}
\cos \theta & 0 & \sin \theta \\
0 & 1 & 0 \\
-\sin \theta & 0 & \cos \theta
\end{array}\right] \cdot\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] \nonumber \]
\(z\)–axis
To rotate the vector
\(\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\)
counterclockwise through an angle
\(\theta\)
around the
\(z\)
–axis to a new position
\(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]\),
perform the matrix multiplication,
\[\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]=\left[\begin{array}{ccc}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{array}\right] \cdot\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] \nonumber \]
Find the vector \(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]\)
that results when the vector
\(\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]=\left[ \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right]\)
is rotated 90° counterclockwise around
\(x\)-
axis.
Solution
Using the rotation formula \(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \theta & -\sin \theta \\
0 & \sin \theta & \cos \theta
\end{array}\right] \cdot\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) with
\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
and
\(\theta =90\mathrm{{}^\circ },\)
we get
\[\begin{aligned}
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \theta & -\sin \theta \\
0 & \sin \theta & \cos \theta
\end{array}\right] \cdot\left[\begin{array}{c}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos 90^{\circ} & -\sin 90^{\circ} \\
0 & \sin 90^{\circ} & \cos 90^{\circ}
\end{array}\right] \cdot\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]} \\
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & 0
\end{array}\right] \cdot\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]=\left[\begin{array}{c}
1 \cdot 1+0 \cdot 2+0 \cdot 3 \\
0 \cdot 1+0 \cdot 2+(-1) \cdot 3 \\
0 \cdot 1+1 \cdot 2+0 \cdot 3
\end{array}\right]} \\
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]=\left[\begin{array}{c}
1 \\
-3 \\
2
\end{array}\right]} \\
&
\end{aligned} \nonumber \]
When rotated counterclockwise 90° around the
\(x\)
–axis, the vector
\(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
becomes \(\left[\begin{array}{l}
1 \\
-3 \\
2
\end{array}\right]\).
Using Technology
We can use technology to help us find the rotation. WolframAlpha evaluates the trig functions for us.
Go to www.wolframalpha.com
In Example
Now we will use WolframAlpha to rotate vector
\(\left[ \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right]\)
90° around the
\(y\)-
axis. We use the
\(y\)-
axis rotation matrix
\(\left[\begin{array}{ccc}
\cos \theta & 0 & \sin \theta \\
0 & 1 & 0 \\
-\sin \theta & 0 & \cos \theta
\end{array}\right]\).
To perform the rotation, enter Evaluate \([[\cos (90), 0, \sin (90)],[0,1,0],[-\sin (90), 0, \cos (90)]] *[1,2,3]\) into the entry field.
Both entries and rows are separated by commas as WA does not see spaces.
Wolframalpha tells you what it thinks you entered, then tells you its answer.
When rotated counterclockwise 90 \(\mathrm{{}^\circ}\) around the \(y\) –axis, the vector \(\left[ \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right]\) becomes \(\left[ \begin{array}{c} 3 \\ 2 \\ -1 \end{array} \right]\) .
Find the vector
\(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]\)
that results when the vector
\(\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]=\left[ \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right]\)
is rotated 45° counterclockwise around the
\(z\)-
axis.
Solution
Since we are rotating the vector around the \(z\) –axis, we use the \(z\)- axis rotation matrix
\[\left[\begin{array}{ccc}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{array}\right] \nonumber \]
Using WolframAlpha with \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\) and
\(\theta =45\mathrm{{}^\circ },\)
we get
When rotated counterclockwise 45° around the
\(z\)
–axis, the vector \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\) becomes \(\left[\begin{array}{c}
-1 / \sqrt{2} \\
3 / \sqrt{2} \\
3
\end{array}\right]\).
Try these
Find the vector \(\left[\begin{array}{l}
x^{\prime} \\
y^{\prime} \\
z^{\prime}
\end{array}\right]\) that results when the given vector \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) is rotated the given angle \(\theta\) counterclockwise around the given axis.
\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]\) through 90° around the \(x\)-axis.
- Answer
-
\(\left[ \begin{array}{c} 1 \\ -1 \\ 1 \end{array} \right]\)
\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
0 \\
1
\end{array}\right]\) through 45° around the \(z\)-axis.
- Answer
-
\(\left[ \begin{array}{c} \sqrt{2} \\ 0 \\ 1 \end{array} \right]\)
\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
3 \\
4 \\
5
\end{array}\right]\) through 30° around the \(y\)-axis
- Answer
-
\(\left[ \begin{array}{c} 3 \\ -5 \\ 4 \end{array} \right]\) | libretexts | 2025-03-17T22:27:22.094549 | 2023-04-03T20:58:26 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/04%3A_Matrices/4.06%3A_Rotation_Matrices_in_3-Dimensions",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "4.6: Rotation Matrices in 3-Dimensions",
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https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/05%3A_Some_Basic_Trigonometry | 5: Some Basic Trigonometry Last updated Save as PDF Page ID 125049 5.1: The Basic Trigonometric Functions 5.2: Circular Trigonometry 5.3: Graphs of the Sine Function 5.4: Graphs of the Cosine Function 5.5: Amplitude and Period of the Sine and Cosine Functions | libretexts | 2025-03-17T22:27:22.172901 | 2023-04-03T20:58:28 | {
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"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/05%3A_Some_Basic_Trigonometry",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "5: Some Basic Trigonometry",
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https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/05%3A_Some_Basic_Trigonometry/5.01%3A_The_Basic_Trigonometric_Functions | 5.1: The Basic Trigonometric Functions
Right Triangle Trigonometry
There are six trigonometric functions associated with right triangles. Since our focus is on the mathematics of games, we will concentrate on only three of them, the sine function, the cosine function, and the tangent function.
The sine function is useful for producing the vertical motion of an object and the cosine function for producing the horizontal motion.
The figures just below show right triangles with angle \(\theta\) , and sides opposite angle \(\theta\), adjacent to angle \(\theta\) , and the hypotenuse of the triangle \(.\)
The angle \(\theta\) has two measures associated with it:
- Its degree measure, which we can label \(\theta {}^\circ\) , and
- Its trigonometric measure.
A trigonometric measure of an angle is a ratio (quotient) of two of the sides of the triangle.
We will discuss all three of these ratios, the sine, the cosine, and the tangent of an angle.
The Sine of an Angle
In words: In a right triangle, the sine of angle \(\theta\) is the ratio of the length of the side opposite \(\theta\) to the length of the hypotenuse. We abbreviate the phrase “the sine of angle \(\theta\) ” with \(\sin \theta\) .
\[\sin \theta=\frac{\text { opposite }}{\text { hypotenuse }} \nonumber \]
\[\sin \theta=\frac{y}{r} \nonumber \]
The Cosine of an Angle
In words: In a right triangle, the cosine of angle \(\theta\) is the ratio of the length of the side adjacent to \(\theta\) to the length of the hypotenuse. We abbreviate the phrase “the cosine of angle \(\theta\) ” with \(\cos \theta\) .
\[\cos \theta=\frac{\text { adjacent }}{\text { hypotenuse }} \nonumber \]
\[\cos \theta=\frac{x}{r} \nonumber \]
The Tangent of an Angle
In words: In a right triangle, the tangent of angle \(\theta\) is the ratio of the length of the side opposite \(\theta\) to the length of the side adjacent to \(\theta\) . We abbreviate the phrase “the tangent of angle \(\theta\) ” with \(\tan \theta\) .
\[\tan \theta=\frac{\text { opposite }}{\text { adjacent }} \nonumber \]
\[\tan \theta=\frac{y}{x} \nonumber \]
Find \(\sin \theta \) , \(\cos \theta \) and \(\tan \theta\) for the 3-4-5 triangle.
Solution
\(\begin{aligned}
& \sin \theta=\frac{\text { opposite }}{\text { hypotenuse }}=\frac{3}{5}=0.6 \\
& \cos \theta=\frac{\text { adjacent }}{\text { hypotenuse }}=\frac{4}{5}=0.8 \\
& \tan \theta=\frac{\text { opposite }}{\text { adjacent }}=\frac{3}{4}=0.75
\end{aligned}\)
Find \(\sin \theta \) , \(\cos \theta \), and \(\tan \theta\) for the triangle.
Solution
\(\begin{aligned}
&\sin \theta=\frac{\text { opposite }}{\text { hypotenuse }}=\frac{1}{\sqrt{2}} \approx 0.7071\\
&\begin{aligned}
& \cos \theta=\frac{\text { adjacent }}{\text { hypotenuse }}=\frac{1}{\sqrt{2}} \approx 0.7071 \\
& \tan \theta=\frac{\text { opposite }}{\text { adjacent }}=\frac{1}{1}=1
\end{aligned}
\end{aligned}\)
Using Technology
WolframAlpha evaluates the sines, cosines, and tangents of angles for us.
Go to www.wolframalpha.com.
Find \(\sin 45^{\circ}\) , \(\cos 45^{\circ}\) , and \(\tan 45^{\circ}\) .
Solution
To compute these ratios, enter Evaluate sin(45), cos(45), tan(45) into the entry field.
Separate the entries with commas. WA does not see spaces.
WolframAlpha tells you what it thinks you entered, then tells you its answers.
We conclude that \(\sin 45^{\circ}=\frac{1}{\sqrt{2}}\) , \(\cos 45^{\circ}=\frac{1}{\sqrt{2}}\) , and \(\tan 45^{\circ}=1\) .
WA also provides us with decimal approximations to these ratios.
\(\sin 45^{\circ}=0.7070107, \cos 45^{\circ}=0.7070107\), and \(\tan 45^{\circ}=1\)
Notice that these are the same values we got in Example 2.
Find \(\sin 30^{\circ}\), \(\sin 60^{\circ}\), \(\sin 90^{\circ}\).
Solution
To compute these ratios, enter Evaluate sin(30), sin(60), sin(90) into the entry field.
Separate the entries with commas. WA does not see spaces.
WolframAlpha tells you what it thinks you entered, then tells you its answers.
We conclude that \(\sin 30^{\circ}=\frac{1}{2}\), \(\sin 60^{\circ}=\frac{\sqrt{3}}{2}\) and \(\sin 90^{\circ}=1\) .
WA also provides us with decimal approximations to these ratios.
\(\sin 30^{\circ}=0.5\), \(\sin 60^{\circ}=0.866025\) and \(\sin 90^{\circ}=1\) .
Try these
Find \(\sin \theta \) , \(\cos \theta \), and \(\tan \theta\) for each triangle. Write your answers as decimal numbers rounded to 4 places.
- Answer
-
- \({\mathrm{sin} 45{}^\circ \ }=\frac{1}{\sqrt{2}}=0.7071,\ \) cos \(\ 45{}^\circ =\frac{1}{\sqrt{2}}=0.7071\) , tan \(\ 45{}^\circ =1\)
- \({\mathrm{sin} \theta \ }=\frac{4}{5}=0.8,\ \) cos \(\ \theta =\frac{3}{5}=0.6\) , tan \(\ \theta =\frac{4}{3}=1.3333\)
- \({\mathrm{sin} \theta \ }=\frac{\sqrt{5}}{3}=0.7454,\ \) cos \(\ \theta =\frac{2}{3}=0.6666\) , tan \(\ \theta =\frac{\sqrt{5}}{2}=1.1180\)
- \({\mathrm{sin} \theta \ }=\frac{2}{\sqrt{5}}=0.8944,\ \) cos \(\ \theta =\frac{1}{\sqrt{5}}=0.4472\) , tan \(\ \theta =\frac{2}{1}=2\)
- \({\mathrm{sin} \theta \ }=\frac{15}{17}=0.8834,\ \) cos \(\ \theta =\frac{8}{17}=0.4705\) , tan \(\ \theta =\frac{15}{8}=1.875\)
Find each value. Write your answers as decimal numbers rounded to 4 places.
- \(\sin 30^{\circ}\) , \(\cos 30^{\circ}\) , \(\tan 30^{\circ}\)
- \(\sin 90^{\circ}\) , \(\cos 90^{\circ}\)
- \(\sin 0^{\circ}\) , \(\cos 0^{\circ}\) , \(\tan 0^{\circ}\)
- \(\sin 180^{\circ}\) , \(\cos 180^{\circ}\)
- \(\sin 120^{\circ}\) , \(\cos 120^{\circ}\)
- Answer
-
- \({\mathrm{sin} 30{}^\circ =0.5\ }\) , cos \(\ 30{}^\circ =0.8661\) , tan \(\ 30{}^\circ =0.5774\)
- \({\mathrm{sin} 90{}^\circ =1\ }\) , cos \(\ 90{}^\circ =0\)
- \({\mathrm{sin} 0{}^\circ =0\ }\) , cos \(\ 0{}^\circ =1\) , tan \(\ 0{}^\circ =0\)
- \({\mathrm{sin} 180{}^\circ =0\ }\) , cos \(\ 180{}^\circ =-1\)
- \({\mathrm{sin} 120{}^\circ =0.8660\ }\) , cos \(\ 120{}^\circ =-0.5\) | libretexts | 2025-03-17T22:27:22.264910 | 2023-04-03T20:58:30 | {
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"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/05%3A_Some_Basic_Trigonometry/5.01%3A_The_Basic_Trigonometric_Functions",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "5.1: The Basic Trigonometric Functions",
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https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/05%3A_Some_Basic_Trigonometry/5.02%3A_Circular_Trigonometry | 5.2: Circular Trigonometry
The Sine Function on the Unit Circle
In computer games, objects typically move up-and-down and left-to-right. These movements can be produced using the sine and cosine functions.
Draw a circle with radius 1 unit and on its circumference, place a point, let’s call it \(P\) .
The circle centered at the origin with radius 1 is called the unit-circle.
From our presentation of the sine and cosine function using right triangles, we can see that
\[\sin \theta=\frac{\text { opposite }}{\text { hypotenuse }}=\frac{y}{1}=y \nonumber \]
\[y=\sin \theta \nonumber \]
This tells us that the sine of the angle \(\theta\) determines the vertical distance of the point \(P\) from the horizontal axis.
The Cosine Function on the Unit Circle
To define cosine function, place a point \(P(x,y)\) on the circumference of unit-circle.
Once again, from our presentation of the cosine functions using right triangles, we can see that
\[\cos \theta=\frac{\text { adjacent }}{\text { hypotenuse }}=\frac{x}{1}=x \nonumber \]
\[x=\cos \theta \nonumber \]
This tells us that the cosine of the angle \(\theta\) determines the horizontal distance of the point \(P\) from the vertical axis.
The Sine and Cosine Functions on any Circle
We can extend this idea by making the radius of the circle \(r\) units rather than just 1 unit.
Using the same reasoning we just used with the unit circle, we see that
\[\sin \theta=\frac{\text { opposite }}{\text { hypotenuse }}=\frac{y}{r} \rightarrow r \cdot \sin \theta=y \rightarrow y=r \cdot \sin \theta \nonumber \]
\[\cos \theta=\frac{\text { adjacent }}{\text { hypotenuse }}=\frac{x}{r} \rightarrow r \cdot \cos \theta \rightarrow x=r \cdot \cos \theta \nonumber \]
which, again, tells us that the sine of the angle \(\theta\) determines the vertical distance of the point \(P\) from the horizontal axis and that the cosine of the angle \(\theta\) determines the horizontal distance of the point \(P\) from the vertical axis.
If \(P\) represents an object, that object’s height \(y\) off the ground (the horizontal axis) is given by \(r\cdot \sin \theta \) , and that object’s horizontal distance \(x\) from some reference point is given by \(r\cdot \cos \theta \) . The height of the object is controlled by some number \(r\) times \(\sin \theta\) , and its horizontal distance is controlled by some number \(r\) times \(\cos \theta\) .
An object lies on the circumference of a unit circle. Find its coordinates if the line segment from the origin to the object makes angle of 30° with the horizontal.
Solution
Because the object is on the circumference of unit circle, we can use
\(\begin{gathered}
x=r \cos \theta \text { and } y=r \sin \theta, \text { with } r=1, \theta=30 \\
x=1 \cos 30^{\circ} \text { and } y=1 \sin 30^{\circ} \\
x=\cos 30^{\circ} \text { and } y=\sin 30^{\circ} \\
x=0.8660 \text { and } y=0.5
\end{gathered}\)
The coordinates of the object are (0.8660, 0.5).
An object lies on the circumference of a circle of radius 5 cm. Find its coordinates if the line segment from the origin to the object makes angle of 40° with the horizontal.
Solution
Because the object is on the circumference of circle of radius 5 cm, we can use
\(\begin{gathered}
x=r \cos \theta \text { and } y=r \sin \theta, \text { with } r=5, \theta=40^{\circ} \\
x=5 \cos 40^{\circ} \quad \text { and } y=5 \sin 40^{\circ} \\
x=5(0.7660) \text { and } y=5(0.6428) \\
x=3.8302 \quad \text { and } y=3.2139
\end{gathered}\)
The coordinates of the object are (3.8302, 3.2139).
The coordinates of an object are (2.1, 3.6373). Find its distance from the origin.
Solution
We can use the Pythagorean Theorem, \(a^2+b^2=c^2\) , where \(c\) is the hypotenuse, the radius of the circle in our case.
\(\begin{aligned}
& 2.1^2+3.6373^2=r^2 \\
& 4.41+13.2300=r^2 \\
& 17.64=r^2 \\
& \sqrt{17.64}=\sqrt{r^2} \\
& 4.2=r
\end{aligned}\)
We conclude that the object is about 4.2 cm from the origin.
Try these
An object lies on the circumference of a unit circle. Find its coordinates if the line segment from the origin to the object makes angle of 45° with the horizontal.
- Answer
-
(0.7071, 0.7071)
An object lies on the circumference of a unit circle. Find its coordinates if the line segment from the origin to the object makes angle of 5° with the horizontal.
- Answer
-
(0.9962, 0.0872)
An object lies on the circumference of a circle of radius 25 cm. Find its coordinates if the line segment from the origin to the object makes angle of 75° with the horizontal.
- Answer
-
(6.4705, 4.8396)
An object lies on the circumference of a circle of radius 10 feet. Find its coordinates if the line segment from the origin to the object makes angle of 135° with the horizontal.
- Answer
-
(-7.0711, 7.0711)
How high above the ground is an object that makes an angle of 60° with a 4-foot-tall observer’s eyes and is 35 feet away from that observer’s eyes? Round to two decimals place.
- Answer
-
34.31 ft
The coordinates of an object are (5.682, 2.0521). Find its distance from the origin if it makes an angle of 60° with the horizontal.
- Answer
-
6 units | libretexts | 2025-03-17T22:27:22.347337 | 2023-04-03T20:58:32 | {
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"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/05%3A_Some_Basic_Trigonometry/5.02%3A_Circular_Trigonometry",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "5.2: Circular Trigonometry",
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https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/05%3A_Some_Basic_Trigonometry/5.03%3A_Graphs_of_the_Sine_Function | 5.3: Graphs of the Sine Function
Discrete Graph of the Sine Function from 0° to 90°
The graph of the sine function gives a visual illustration of how it determines the height of an object from a horizontal axis.
Imagine an object moving counterclockwise along the circumference of the unit circle. Start the object’s motion at the point (1,0), then measure its height from the horizontal axis as its angle from origin increases from 0° to 90°.
Graphs of the Heights
If the angle is between 0° and 90°, the graph of the heights looks like this
We can see that from 0° and 90°, as the angle from the observer to the object increases, the height of the object from the horizontal increases. That is, the object moves vertically upward.
If the angle goes past 90°, say all the way to 180°, the graph of the heights looks like this
The object moves vertically upward, then vertically downward.
The Continuous Sine Curve from 0° to 90°
If we plotted all the heights for all the infinitely many angles between 0° and 90°, we would get this continuous graph
The Continuous Sine Curve from 0° to 180°
If we plotted all the heights for all the angles between 0° and 180°, we would get the continuous graph below
The Continuous Sine Curve from 0° to 360°
If we were to let the object travel all the way around the circle, we get the graph of the sine curve from 0° to 360°. You can see that when the angle \(\theta\) is between 180° and 360°, the object is below the horizontal and may not be visible to an observer.
The Extended Sine Curve
If we were to let the object keep travelling around the circle, we would see that the height of the curve just oscillates between –1 and 1.
It may now be visually apparent that
The sine function controls the vertical distance of an object above or below the horizontal.
What to See
The graph shows how an object’s vertical distance from the horizontal changes as the angle of view increases. As the angle of view increases, the vertical distance from the horizontal increases and decreases.
What Not to See
The graph does not show how an object moves horizontally as the angle of view increases. The object is not moving up and down horizontally along the curve as time goes by. The horizontal axis is the angle of view, not time.
Try these
An object moves along the circumference of a unit circle. Find its height from the horizontal if the angle it makes from the origin is
- 225°
- 270°
- 315°
- 360°
- Answer
-
- -0.7071
- -1
- -0.7071
- 0
An object moves along the circumference of a unit circle. Find its height from the horizontal if the angle it makes from the origin is
- 390°
- 405°
- 420°
- 450°
- Answer
-
- 0.5
- 0.7071
- 0.8660
- 1
Determine if each statement is true or false.
- Height at 87° > height at 78°
- Height at 155° > height at 145°
- Height at 30° ≥ height at 150°
- Height at 90° ≥ height at 270°
- Answer
-
- True, since 0.9986 \(\mathrm{>}\) 0.9781
- False, since 0.4226 \(\mathrm{<}\) 0.5736
- True, since 0.5 = 0.5
- True, since \(1\ge -1\)
Keeping in mind that the sine function determines vertical distance, and the cosine function determines horizontal distance, determine if each statement is true or false. The observer is at the origin.
- Vertical height at 87° > horizontal distance at 87°
- Vertical height at 155° > horizontal distance at 55°
- Vertical height at 20° < horizontal distance at 20°
- Vertical height at 135° = horizontal distance at 315°
- Answer
-
- True, since \({\mathrm{sin} (87{}^\circ )=0.9986>\ {\mathrm{cos} (87{}^\circ )=0.0523\ }\ }\)
- False, since \({\mathrm{sin} (155{}^\circ )=0.4226<\ {\mathrm{cos} (55{}^\circ )=0.5736\ }\ }\)
- True, since \({\mathrm{sin} (20{}^\circ )=0.3420<\ {\mathrm{cos} (20{}^\circ )=0.9396\ }\ }\)
- True, since \({\mathrm{sin} (135{}^\circ )=0.7071=\ {\mathrm{cos} (315{}^\circ )=0.7071\ }\ }\) | libretexts | 2025-03-17T22:27:22.425029 | 2023-04-03T20:58:34 | {
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"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/05%3A_Some_Basic_Trigonometry/5.03%3A_Graphs_of_the_Sine_Function",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "5.3: Graphs of the Sine Function",
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https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/05%3A_Some_Basic_Trigonometry/5.04%3A_Graphs_of_the_Cosine_Function | 5.4: Graphs of the Cosine Function
Discrete Graph of the Cosine Function from 0° to 360°
Just as the sine function determines the vertical distance of an object from an observer, the cosine function determines the horizontal distance of an object from that observer.
Here is table of values for the cosine function for angles between 0° and 360° followed by a graph of the cosine function for all angles from 0° to 360°.
| Angle \(\theta\) | Cosine \(\theta\) (Horizontal Distance from Observer) |
|---|---|
| 0 \(\mathrm{{}^\circ}\) | 1 |
| 45 \(\mathrm{{}^\circ}\) | 0.7071 |
| 90 \(\mathrm{{}^\circ}\) | 0 |
| 135 \(\mathrm{{}^\circ}\) | –0.7071 |
| 180 \(\mathrm{{}^\circ}\) | –1 |
| 225 \(\mathrm{{}^\circ}\) | –0.7071 |
| 270 \(\mathrm{{}^\circ}\) | 0 |
| 315 \(\mathrm{{}^\circ}\) | 0.7071 |
| 360 \(\mathrm{{}^\circ}\) | 1 |
The positive cosine values indicate that the object is to the front of the observer whereas the negative values indicate that the object is to the back of the observer. For example, at an angle of 45° from the observer’s eye, the object is 0.7071 units in front of the observer. At an angle of 135° from the observer’s eye, the object is 0.7071 units behind the observer.
If you think that this graph looks like the graph of the sine function, but shifted to the left by 90°, you would be right.
The Extended Cosine Curve
Just as the sine curve does, the heights of the cosine curve oscillate between –1 and 1.
The graph of the cosine function gives a visual illustration of how it determines the horizontal distance of an object from a vertical axis.
It may now be visually apparent that
The cosine function determines the horizontal distance of an object to the left or right of an observer.
What to See
The graph shows how an object’s horizontal distance from the observer changes as the angle of view increases. As the angle of view increases, the horizontal distance from the vertical increases (moves away from the observer) and decreases (moves toward the observer).
What Not to See
The graph does not show how an object moves vertically as angle of view increases. The object is not moving along the curve.
Try these
An object moves along the circumference of a unit circle. Find its horizontal distance from an observer if the angle it makes from observer’s eye is
- 225°
- 270°
- 315°
- 360°
- Answer
-
- -0.7071
- 0
- 0.7071
- 1
An object moves along the circumference of a unit circle. Find its horizontal distance from an observer if the angle it makes from observer’s eye is
- 390°
- 405°
- 420°
- 450°
- Answer
-
- 0.8660
- 0.7071
- 0.5
- 0
Determine if each statement is true or false.
- Horizontal distance at 87° > Horizontal distance at 78°
- Horizontal distance at 45° > Horizontal distance at 145°
- Horizontal distance at 30° ≥ Horizontal distance at 150°
- Horizontal distance at 90° = Horizontal distance at 270°
- Answer
-
- False, since 0.0523 \(\mathrm{<}\) 0.2079
- False, since 0.7071 \(\mathrm{<}\) 0.9063 (Be careful here: 0.7071 \(\mathrm{>}\) –0.9063, but the negative sign tells us the object is the left of the observer. Think absolute value. At 45 \(\mathrm{{}^\circ}\) , the object is 0.7071 to the right of the observer. At 145 \(\mathrm{{}^\circ}\) , the object is 0.9063 units to the left of the observer, and, therefore, farther from the observer.)
- False, since is 0.8660 =-0.8660
- True, since 0 = 0 | libretexts | 2025-03-17T22:27:22.499317 | 2023-04-03T20:58:36 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/05%3A_Some_Basic_Trigonometry/5.04%3A_Graphs_of_the_Cosine_Function",
"book_url": "https://commons.libretexts.org/book/math-125024",
"title": "5.4: Graphs of the Cosine Function",
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https://math.libretexts.org/Bookshelves/Applied_Mathematics/Mathematics_for_Game_Developers_(Burzynski)/05%3A_Some_Basic_Trigonometry/5.05%3A_Amplitude_and_Period_of_the_Sine_and_Cosine_Functions | 5.5: Amplitude and Period of the Sine and Cosine Functions
Amplitude
We have seen how the graphs of both the sine function, \(y={\mathrm{sin} \theta \ }\) and the cosine function \(y={\mathrm{cos} \theta \ }\) , oscillate between \(-1\) and \(+1\) . That is, the heights oscillate between –1 and 1.
The height from the horizontal axis to the peak (or through) of a sine or cosine function is called the amplitude of the function. Each of the curves \(y=\sin \theta\) and \(y=\cos \theta\) has amplitude 1.
If we were to multiply the sine function \(y={\mathrm{sin} \theta \ }\ \) by \(3\) , getting \(y={3\mathrm{sin} \theta \ }\) , each of the sine values would be multiplied by 3, making each value 3 times what it was. Each height would be tripled. The amplitude of \(y={3\mathrm{sin} \theta \ }\) is 3.
If we were to multiply the cosine function \(y=\mathrm{cos\ }\theta \ \ \) by \(1/3\) , getting \(y={1/3\mathrm{cos} \theta \ }\) ,each of the cosine values would be multiplied by 1/3 making each value 1/3 of what it was. Each height of \(y=\mathrm{cos}\theta \ \) would be 1/3 of what it was.The amplitude of \(y={1/3\mathrm{cos} \theta \ }\) is 1/3.
The Amplitude of \(y=A\mathrm{sin}\theta\) and \(y=A\mathrm{cos}\theta\)
Suppose \(A\) represents a positive number. Then the amplitude of both \(y=A\mathrm{sin}\theta\) and \(y=A\mathrm{cos}\theta\) is \(A\) and it represents height from the horizontal axis to the peak of the curve.
The amplitude of \(y=5/8\mathrm{sin}\theta\) is 5/8. This means that the peak of the curve is 5/8 of a unit above the horizontal axis.
The amplitude of \(y=3\mathrm{sin}\theta\) is 3. This means that the peak of the curve is 3 units above the horizontal axis.
Period
Both the sine function and cosine function, \(y=\mathrm{sin}\theta\) and \(y=\mathrm{cos}\theta ,\) go through exactly one cycle from 0 \(\mathrm{{}^\circ}\) to 360 \(\mathrm{{}^\circ}\) .
The period of the sine function and cosine functions, \(y=\mathrm{sin}\theta\) and \(y=\mathrm{cos}\theta ,\) is the “time” required for one complete cycle.
An interesting thing happens to the curves \(y=\mathrm{sin}\theta\) and \(y=\mathrm{cos}\theta\) when the angle \(\theta\) is multiplied by some positive number, \(B.\) If the number \(B\ \) is greater than 1, the number of cycles on 0 \(\mathrm{{}^\circ}\) to 360 \(\mathrm{{}^\circ}\) increases for both \(y=\mathrm{sin}\theta\) and \(y=\mathrm{cos}\theta\) . That is, the peaks of the curve are closer together, meaning their periods decrease. If the number \(B\ \) is strictly between 0 and 1, the peaks of the curve are farther apart, meaning their periods increase.
The Period of \(y=\mathrm{sin}\left(B\theta \right)\) and \(y=\mathrm{cos}\left(B\theta \right)\)
Suppose \(B\) represents a positive number. Then the period of both \(y=\mathrm{sin}\left(B\theta \right)\) and \(y=\mathrm{cos}\left(\mathrm{B}\theta \right)\) is \(\frac{360{}^\circ }{B}.\) As B gets bigger, \(\frac{360{}^\circ }{B}\) gets smaller and the period increases.
If we were to multiply the angle in the sine function \(y={\mathrm{sin} \theta \ }\ \) by \(3\) , getting \(y={\mathrm{sin} 3\theta \ }\) , each of the angle’s values would be multiplied by 3 making each value 3 times what it was. Each angle would be tripled and there would be 3 cycles in the interval 0 \(\mathrm{{}^\circ}\) to 360 \(\mathrm{{}^\circ}\) .
The period of \(y={\mathrm{sin} 3\theta \ }\) is \(\frac{360{}^\circ }{3}=120{}^\circ\) . The period of \(y={\mathrm{sin} 3\theta \ }\) is smaller than that of \(y={\mathrm{sin} \theta \ }\) .
If we were to multiply the angle in the sine function \(y={\mathrm{sin} \theta \ }\ \) by \(1/3\) , getting \(y=\mathrm{sin}\left(\frac{1}{3}\theta \right).\) Each of the angle’s values would be multiplied by 1/3 making each value 1/3 what it was and there would be only 1/3 of a cycle in the interval 0 \(\mathrm{{}^\circ}\) to 360 \(\mathrm{{}^\circ}\) .
The period of \(y=\mathrm{sin}\left(\frac{1}{3}\theta \right)\) is \(\frac{360{}^\circ }{1/3}=\ 360{}^\circ \times \ 3\) = \(1080{}^\circ\) . The period of \(y=\mathrm{sin}\left(\frac{1}{3}\theta \right)\) is greater than that of \(y={\mathrm{sin} \theta \ }.\)
Using Technology
We can use technology to help us construct the graph of a sine or cosine function.
Go to www.wolframalpha.com.
Plot two complete cycles of \(y={\mathrm{6sin} 2\theta \ }\) from 0 \(\mathrm{{}^\circ}\) to 360 \(\mathrm{{}^\circ}\) .
Solution
Type plot y = 6sin2x, x = 0..360 degrees in the entry field.
WolframAlpha tells you what it thinks you entered, then produces the graph.
You can see that WolframAlpha has plotted two complete cycles from 0 \(\mathrm{{}^\circ}\) to 360 \(\mathrm{{}^\circ}\) with amplitude 6.
Find the period of \(y={\mathrm{6sin} 8\theta \ }\) .
Solution
We just need to evaluate \(\frac{360{}^\circ }{B}\) with \(B=8\) .
\(\frac{360{}^\circ }{8}=45{}^\circ \)
The period of \(y={\mathrm{6sin} 8\theta \ }\) is \(45{}^\circ\)
The graph of \(y={\mathrm{6sin} 8\theta \ \ }\) helps us visualize this 45 \(\mathrm{{}^\circ}\) period. You can see that the peaks differ by 45 \(\mathrm{{}^\circ}\) .
Try these
Write the equation of each graph.
- Answer
-
- \(y=3\mathrm{sin}\mathrm{}(2x)\)
- \(y=2\mathrm{cos}(3x)\)
- \(y=7\mathrm{cos}\mathrm{}(x)\)
How many complete cycles are there in the graph of \(y=4 \cos (3 \theta)\) from 0° to 360°? What is the period and amplitude of this function?
- Answer
-
3 complete cycles. Period is \(\frac{360{}^\circ }{3}=120{}^\circ .\) Amplitude is 4.
How many complete cycles are there in the graph of \(y=5 \sin \left(\frac{4}{5} \theta\right)\) from 0° to 360°? What is the period and amplitude of this function?
- Answer
-
\(\frac{4}{5}\) of a complete cycle. Period is \(\frac{360{}^\circ }{4/5}=360{}^\circ \times \ \frac{5}{4}=450{}^\circ\) . Amplitude is 5.
Write the equation of a sine curve that has amplitude 15 and period 50°. You need to specify both \(A\) and \(B\) in \(y=A \sin (B \theta)\). Keep in mind that the period of this function is \(\frac{360^{\circ}}{B}\).
- Answer
-
\(y=15\mathrm{sin}\left(7.2\theta \right)\) , where \(\frac{360{}^\circ }{\ B}=50{}^\circ \to B=\ \frac{360{}^\circ }{\ 50{}^\circ }=7.2\)
Write the equation of a cosine curve that has amplitude 100 and period 12°. You need to specify both \(A\) and \(B\) in \(y=A \cos (B \theta)\). Keep in mind that the period of this function is \(\frac{360^{\circ}}{B}\).
- Answer
-
\(y=100\mathrm{cos}\left(30\theta \right)\) , where \(\frac{360{}^\circ }{\ B}=12{}^\circ \to B=\ \frac{360{}^\circ }{\ 12{}^\circ }=30\)
Write the equation of a cosine function that has amplitude 3 and makes two complete cycles from 0° to 180°.
- Answer
-
\(y=3\mathrm{cos}\mathrm{}(4\theta )\) We need to specify both \(A\ \mathrm{and}\ B\) in \(y=A\mathrm{cos}\left(B\theta \right)\) . Since the amplitude is 3 \(,\ \ A=3.\) Since the curve makes two complete cycles from 0 \(\mathrm{{}^\circ}\) to 180 \(\mathrm{{}^\circ}\) , it must make 4 complete cycles from 0 \(\mathrm{{}^\circ}\) to 360 \(\mathrm{{}^\circ}\) . So, \(\ B=4.\)
Write the equation of a sine function that has amplitude 4 and makes three complete cycles from 0° to 90°.
- Answer
-
\(y=4\mathrm{sin}\mathrm{}(12\theta )\) We need to specify both \(A\ \mathrm{and}\ B\) in \(y=A\mathrm{cos}\left(B\theta \right)\) . Since the amplitude is 4 \(,\ \ A=4.\) Since the curve makes three complete cycles from 0 \(\mathrm{{}^\circ}\) to 90 \(\mathrm{{}^\circ}\) , it must make 12 complete cycles from 0 \(\mathrm{{}^\circ}\) to 360 \(\mathrm{{}^\circ}\) . So, \(B=12\) . | libretexts | 2025-03-17T22:27:22.596820 | 2023-04-03T20:58:38 | {
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https://socialsci.libretexts.org/Courses/Sacramento_City_College/LIBR_324%3A_Critical_Thinking_and_Information_Literacy/3%3A_SLO_3/3.4%3A_Rhetorical_Modes | 3.4: Rhetorical Modes
Though there are as many ways to write a paper as there are papers, it is helpful to categorize different types of writing under specific names or genres. This section defines the most commonly used types (called modes) of writing that you may encounter in college and beyond. Each describes the way the piece is written, so that you can choose this form as a writer or identify it quickly as a reader. Knowing each mode helps organize thinking and writing to make concentrating on content a swifter, more instinctive process. | libretexts | 2025-03-17T22:27:23.393723 | 2019-09-23T22:10:32 | {
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"url": "https://socialsci.libretexts.org/Courses/Sacramento_City_College/LIBR_324%3A_Critical_Thinking_and_Information_Literacy/3%3A_SLO_3/3.4%3A_Rhetorical_Modes",
"book_url": "https://commons.libretexts.org/book/socialsci-23897",
"title": "3.4: Rhetorical Modes",
"author": null
} |
https://socialsci.libretexts.org/Courses/Sacramento_City_College/LIBR_324%3A_Critical_Thinking_and_Information_Literacy/3%3A_SLO_3/3.4%3A_Rhetorical_Modes/All_Writing_is_Argument | All Writing is Argument
Learning Objectives
- Define rhetoric and explain the term’s historical context related to persuasive writing
- Demonstrate the importance of research writing as a rhetorical, persuasive activity
This chapter is about rhetoric—the art of persuasion. Every time we write, we engage in argument. Through writing, we try to persuade and influence our readers, either directly or indirectly. We work to get them to change their minds, to do something, or to begin thinking in new ways. Therefore, every writer needs to know and be able to use principles of rhetoric. The first step towards such knowledge is learning to see the argumentative nature of all writing.
I have two goals in this chapter: to explain the term rhetoric and to give you some historical perspective on its origins and development; and to demonstrate the importance of seeing research writing as a rhetorical, persuasive activity.
As consumers of written texts, we are often tempted to divide writing into two categories: argumentative and non-argumentative. According to this view, in order to be argumentative, writing must have the following qualities. It has to defend a position in a debate between two or more opposing sides; it must be on a controversial topic; and the goal of such writing must be to prove the correctness of one point of view over another.
On the other hand, this view goes, non-argumentative texts include narratives, descriptions, technical reports, news stories, and so on. When deciding to which category a given piece of writing belongs, we sometimes look for familiar traits of argument, such as the presence of a thesis statement, of “factual” evidence, and so on.
Research writing is often categorized as “non-argumentative.” This happens because of the way in which we learn about research writing. Most of us do that through the traditional research report, the kind which focuses too much on information-gathering and note cards and not enough on constructing engaging and interesting points of view for real audiences. It is the gathering and compiling of information, and not doing something productive and interesting with this information, that become the primary goals of this writing exercise. Generic research papers are also often evaluated on the quantity and accuracy of external information that they gather, rather on the persuasive impact they make and the interest they generate among readers.
Having written countless research reports, we begin to suspect that all research-based writing is non-argumentative. Even when explicitly asked to construct a thesis statement and support it through researched evidence, beginning writers are likely to pay more attention to such mechanics of research as finding the assigned number and kind of sources and documenting them correctly, than to constructing an argument capable of making an impact on the reader.
ARGUMENTS AREN’T VERBAL FIGHTS
We often have narrow concept of the word “argument.” In everyday life, argument often implies a confrontation, a clash of opinions and personalities, or just a plain verbal fight. It implies a winner and a loser, a right side and a wrong one. Because of this understanding of the word “argument,” the only kind of writing seen as argumentative is the debate-like “position” paper, in which the author defends his or her point of view against other, usually opposing points of view.
Such an understanding of argument is narrow because arguments come in all shapes and sizes. I invite you to look at the term “argument” in a new way. What if we think of “argument” as an opportunity for conversation, for sharing with others our point of view on something, for showing others our perspective of the world? What if we see it as the opportunity to tell our stories, including our life stories? What if we think of “argument” as an opportunity to connect with the points of view of others rather than defeating those points of view?
Some years ago, I heard a conference speaker define argument as the opposite of “beating your audience into rhetorical submission.” I still like that definition because it implies gradual and even gentle explanation and persuasion instead of coercion. It implies effective use of details, and stories, including emotional ones. It implies the understanding of argument as an explanation of one’s world view.
Arguments then, can be explicit and implicit, or implied. Explicit arguments contain noticeable and definable thesis statements and lots of specific proofs. Implicit arguments, on the other hand, work by weaving together facts and narratives, logic and emotion, personal experiences and statistics. Unlike explicit arguments, implicit ones do not have a one-sentence thesis statement. Instead, authors of implicit arguments use evidence of many different kinds in effective and creative ways to build and convey their point of view to their audience. Research is essential for creative effective arguments of both kinds.
To consider the many types and facets of written argumentation, consider the following exploration activity.
WRITING ACTIVITY: ANALYZING WRITING SITUATIONS
Working individually or in small groups, consider the following writing situations. Are these situations opportunities for argumentative writing? If so, what elements of argument do you see? Use your experience as a reader and imagine the kinds of published texts that might result from these writing situations. Apply the ideas about argument mentioned so far in this chapter, including the “explicit” and “implicit” arguments
• A group of scientists develops a hypothesis and conducts a series of experiments to test it. After obtaining the results from those experiments, they decide to publish their findings in a scientific journal. However, the data can be interpreted in two ways. The authors can use a long-standing theory with which most of his colleagues agree. But they can also use a newer and more ambitious theory on which there is no consensus in the field, but which our authors believe to be more comprehensive and up-to-date. Using different theories will produce different interpretations of the data and different pieces of writing. Are both resulting texts arguments? Why or why not?
• An author wants to write a memoir. She is particularly interested in her relationship with her parents as a teenager. In order to focus on that period of her life, she decides to omit other events and time periods from the memoir. The finished text is a combination of stories, reflections, and facts. This text does not have a clear thesis statement or proofs. Could this “selective” memory” writing be called an argument? What are the reasons for your decision?
• A travel writer who is worried about global warming goes to Antarctica and observes the melting of the ice there. Using her observations, interviews with scientists, and secondary research, she then prepares an article about her trip for The National Geographic magazine or a similar publication. Her piece does not contain a one-sentence thesis statement or a direct call to fight global warming. At the same time, her evidence suggests that ice in the Arctic melts faster than it used to. Does this writer engage in argument? Why or why not? What factors influenced your decision?
• A novelist writes a book based on the events of the American Civil War. He recreates historical characters from archival research, but adds details, descriptions, and other characters to his book that are not necessarily historic. The resulting novel is in the genre known as “historical fiction.” Like all works of fiction, the book does not have a thesis statement or explicit proofs. It does, however, promote a certain view of history, some of which is based on the author’s research and some—on his imagination and creative license. Is this a representation of history, an argument, or a combination of both? Why or why not?
You can probably think of many more examples when argument in writing is expressed through means other than the traditional thesis statement and proofs. As you work through this book, continue to think about the nature of argument in writing and discuss it with your classmates and your instructor.
DEFINITIONS OF RHETORIC AND THE RHETORICAL SITUATION
The art of creating effective arguments is explained and systematized by a discipline called rhetoric. Writing is about making choices, and knowing the principles of rhetoric allows a writer to make informed choices about various aspects of the writing process. Every act of writing takes places in a specific rhetorical situation. The three most basic and important components of a rhetorical situations are:
- Purpose of writing
- Intended audience,
- Occasion, or context in which the text will be written and read
These factors help writers select their topics, arrange their material, and make other important decisions about their work.
Before looking closely at different definitions and components of rhetoric, let us try to understand what rhetoric is not. In recent years, the word “rhetoric” has developed a bad reputation in American popular culture. In the popular mind, the term “rhetoric” has come to mean something negative and deceptive. Open a newspaper or turn on the television, and you are likely to hear politicians accusing each other of “too much rhetoric and not enough substance.” According to this distorted view, rhetoric is verbal fluff, used to disguise empty or even deceitful arguments.
Examples of this misuse abound. Here are some examples.
A 2003 CNN news article “North Korea Talks On Despite Rhetoric” describes the decision by the international community to continue the talks with North Korea about its nuclear arms program despite what the author sees as North Koreans’ “rhetorical blast” at a US official taking part in the talks. The implication here is that that, by verbally attacking the US official, the North Koreans attempted to hide the lack of substance in their argument. The word “rhetoric” in this context implies a strategy to deceive or distract.
Another example is the title of the now-defunct political website “Spinsanity: Countering Rhetoric with Reason.” The website’s authors state that “engaged citizenry, active press and strong network of fact-checking websites and blogs can help turn the tide of deception that we now see.” ( http://www.spinsanity.org ). What this statement implies, of course, is that rhetoric is “spin” and that it is the opposite of truth.
Rhetoric is not a dirty trick used by politicians to conceal and obscure, but an art, which, for many centuries, has had many definitions. Perhaps the most popular and overreaching definition comes to us from the Ancient Greek thinker Aristotle. Aristotle defined rhetoric as “the faculty of observing in any given case
the available means of persuasion” (Ch.2). Aristotle saw primarily as a practical tool, indispensable for civic discourse.
ELEMENTS OF THE RHETORICAL SITUATION
When composing, every writer must take into account the conditions under which the writing is produced and will be read. It is customary to represent the three key elements of the rhetorical situation as a triangle of writer, reader, and text, or, as “communicator,” “audience,” and “message.”
The three elements of the rhetorical situation are in a constant and dynamic interrelation. All three are also necessary for communication through writing to take place. For example, if the writer is taken out of this equation, the text will not be created. Similarly, eliminating the text itself will leave us with the reader and writer, but without any means of conveying ideas between them, and so on.
Moreover, changing on or more characteristics of any of the elements depicted in the figure above will change the other elements as well. For example, with the change in the beliefs and values of the audience, the message will also likely change to accommodate those new beliefs, and so on.
In his discussion of rhetoric, Aristotle states that writing’s primary purpose is persuasion. Other ancient rhetoricians’ theories expand the scope of rhetoric by adding new definitions, purposes, and methods. For example, another Greek philosopher and rhetorician Plato saw rhetoric as a means of discovering the truth, including personal truth, through dialog and discussion. According to Plato, rhetoric can be directed outward (at readers or listeners), or inward (at the writer him or herself). In the latter case, the purpose of rhetoric is to help the author discover something important about his or her own experience and life.
The third major rhetorical school of Ancient Greece whose views have profoundly influenced our understanding of rhetoric were the Sophists. The Sophists were teachers of rhetoric for hire. The primary goal of their activities was to teach skills and strategies for effective speaking and writing. Many Sophists claimed that they could make anyone into an effective rhetorician. In their most extreme variety, Sophistic rhetoric claims that virtually anything could be proven if the rhetorician has the right skills. The legacy of Sophistic rhetoric is controversial. Some scholars, including Plato himself, have accused the Sophists of bending ethical standards in order to achieve their goals, while others have praised them for promoting democracy and civic participation through argumentative discourse.
What do these various definitions of rhetoric have to do with research writing? Everything! If you have ever had trouble with a writing assignment, chances are it was because you could not figure out the assignment’s purpose. Or, perhaps you did not understand very well whom your writing was supposed to appeal to. It is hard to commit to purposeless writing done for no one in particular.
Research is not a very useful activity if it is done for its own sake. If you think of a situation in your own life where you had to do any kind of research, you probably
had a purpose that the research helped you to accomplish. You could, for example, have been considering buying a car and wanted to know which make and model would suite you best. Or, you could have been looking for an apartment to rent and wanted to get the best deal for your money. Or, perhaps your family was planning a vacation and researched the best deals on hotels, airfares, and rental cars. Even in these simple examples of research that are far simpler than research most writers conduct, you as a researcher were guided by some overriding purpose. You researched because you had a purpose to accomplish.
HOW TO APPROACH WRITING TASKS RHETORICALLY
The three main elements of rhetorical theory are purpose, audience, and occasion. We will look at these elements primarily through the lens of Classical Rhetoric, the rhetoric of Ancient Greece and Rome. Principles of classical rhetoric (albeit some of them modified) are widely accepted across the modern Western civilization. Classical rhetoric provides a solid framework for analysis and production of effective texts in a variety of situations.
PURPOSE
Good writing always serves a purpose. Texts are created to persuade, entertain, inform, instruct, and so on. In a real writing situation, these discrete purposes are often combined.
Exercise \(\PageIndex{1}\): Analyzing Purpose
Recall any text you wrote, in or outside of school. Think not only of school papers, but also of letters to relatives and friends, e-mails, shopping lists, online postings, and so on. Consider the following questions.
- Was the purpose of the writing well defined for you in the assignment, or did you have to define it yourself?
- What did you have to do in order to understand or create your purpose?
- Did you have trouble articulating and fulfilling your writing purpose?
Be sure to record your answers and share them with your classmates and/or instructor.
AUDIENCE
The second key element of the rhetorical approach to writing is audience-awareness. As you saw from the rhetorical triangle earlier in this chapter, readers are an indispensable part of the rhetorical equation, and it is essential for every writer to understand their audience and tailor his or her message to the audience’s needs.
The key principles that every writer needs to follow in order to reach and affect his or her audience are as follows:
- Have a clear idea about who your readers will be.
- Understand your readers’ previous experiences, knowledge, biases, and expectations and how these factors can influence their reception of your argument.
- When writing, keep in mind not only those readers who are physically present or whom you know (your classmates and instructor), but all readers who would benefit from or be influenced by your argument.
- Choose a style, tone, and medium of presentation appropriate for your intended audience.
Exercise \(\PageIndex{2}\): Analyzing Audience
Every writer needs to consider his or her audience carefully when writing. Otherwise, you writing will be directed at no one in particular. As a result, your purpose will become unclear and your work will lose its effectiveness.
Consider any recent writing task that you faced. As with all the exploration activities included in this chapter, do not limit yourself to school writing assignments. Include letters, e-mails, notes, and any other kinds of writing you may do.
- Did you have a clearly defined audience?
- If not, what measures did you take to define and understand your audience?
- How did you know who your readers were?
- Did your writing purpose fit what your intended audience needed or wanted to hear?
- What were the best ways to appeal to your audience (both logical and emotional)?
- How did your decision to use or not to use external research influence the reception of your argument by your audience?
OCCASION
Occasion is an important part of the rhetorical situation. It is a part of the writing context that was mentioned earlier in the chapter. Writers do not work in a vacuum. Instead, the content, form and reception of their work by readers are heavily influenced by the conditions in society as well as by personal situations of their readers. These conditions in which texts are created and read affect every aspect of writing and every stage of the writing process, from topic selection, to decisions about what kinds of arguments used and their arrangement, to the writing style, voice, and persona which the writer wishes to project in his or her writing. All elements of the rhetorical situation work together in a dynamic relationship. Therefore, awareness of rhetorical occasion and other elements of the context of your writing will also help you refine your purpose and understand your audience better. Similarly having a clear purpose in mind when writing and knowing your audience will help you understand the context in which you are writing and in which your work will be read better.
One aspect of writing where you can immediately benefit from understanding occasion and using it to your rhetorical advantage is the selection of topics for your compositions. Any topic can be good or bad, and a key factor in deciding on whether it fits the occasion. In order to understand whether a particular topic is suitable for a composition, it is useful to analyze whether the composition would address an issue, or a rhetorical exigency when created. The writing activity below can help you select topics and issues for written arguments.
Exercise \(\PageIndex{3}\): Analyzing Rhetorical Exigency
- If you are considering a topic for a paper, think whether the paper would address a specific problem or issue. In other words, will it address a real exigency, something that needs to be solved or discussed?
- Who are the people with interests and stakes in the problem?
- What are your limitations? Can you hope to solve the problem once and for all, or should your goals be more modest? Why or why not?
Share your results with your classmates and instructor.
To understand how writers can study and use occasion in order to make effective arguments, let us examine another ancient rhetorical concept. Kairos is one of the most fascinating terms from Classical rhetoric. It signifies the right, or opportune moment for an argument to be made. It is such a moment or time when the subject of the argument is particularly urgent or important and when audiences are more likely to be persuaded by it. Ancient rhetoricians believed that if the moment for the argument is right, for instance if there are conditions in society which would make the audience more receptive to the argument, the rhetorician would have more success persuading such an audience.
For example, as I write this text, a heated debate about the war on terrorism and about the goals and methods of this war is going on in the US. It is also the year of the Presidential Election, and political candidates try to use the war on terrorism to their advantage when they debate each other. These are topics of high public interested, with print media, television, radio, and the Internet constantly discussing them. Because there is an enormous public interest in the topic of terrorism, well-written articles and reports on the subject will not fall on deaf ears. Simply put, the moment, or occasion, for the debate is right, and it will continue until public interest in the subject weakens or disappears.
RHETORICAL APPEALS
In order to persuade their readers, writers must use three types of proofs or rhetorical appeals. They are logos, or logical appeal; pathos, or emotional appeal; and ethos, or ethical appeal, or appeal based on the character and credibility of the author. It is easy to notice that modern words “logical,” “pathetic,” and “ethical” are derived from those Greek words. In his work Rhetoric, Aristotle writes that the three appeals must be used together in every piece of persuasive discourse. An argument based on the appeal to logic, or emotions alone will not be an effective one.
Understanding how logos, pathos, and ethos should work together is very important for writers wh use research. Often, research writing assignment are written in a way that seems to emphasize logical proofs over emotional or ethical ones. Such logical proofs in research papers typically consist of factual information, statistics, examples, and other similar evidence. According to this view, writers of academic papers need to be unbiased and objective, and using logical proofs will help them to be that way.
Because of this emphasis on logical proofs, you may be less familiar with the kinds of pathetic and ethical proofs available to you. Pathetic appeals, or appeals to emotions of the audience were considered by ancient rhetoricians as important as logical proofs. Yet, writers are sometimes not easily convinced to use pathetic appeals in their writing. As modern rhetoricians and authors of the influential book Classical Rhetoric for the Modern Student (1998), Edward P.J. Corbett and Robert Connors said, “People are rather sheepish about acknowledging that their opinions can be affected by their emotions” (86). According to Corbett, many of us think that there may be something wrong about using emotions in argument. But, I agree with Corbett and Connors, pathetic proofs are not only admissible in argument, but necessary (86-89). The most basic way of evoking appropriate emotional responses in your audience, according to Corbett, is the use of vivid descriptions (94).
Using ethical appeals, or appeals based on the character of the writer, involves establishing and maintaining your credibility in the eyes of your readers. In other words, when writing, think about how you are presenting yourself to your audience. Do you give your readers enough reasons to trust you and your argument, or do you give them reasons to doubt your authority and your credibility? Consider all the times when your decision about the merits of a given argument was affected by the person or people making the argument. For example, when watching television news, are you predisposed against certain cable networks and more inclined towards others because you trust them more?
So, how can a writer establish a credible persona for his or her audience? One way to do that is through external research. Conducting research and using it well in your writing help with you with the factual proofs (logos), but it also shows your readers that you, as the author, have done your homework and know what you are talking about. This knowledge, the sense of your authority that this creates among your readers, will help you be a more effective writer.
The logical, pathetic, and ethical appeals work in a dynamic combination with one another. It is sometimes hard to separate one kind of proof from another and the methods by which the writer achieved the desired rhetorical effect. If your research contains data which is likely to cause your readers to be emotional, it data can enhance the pathetic aspect of your argument. The key to using the three appeals, is to use them in combination with each other, and in moderation. It is impossible to construct a successful argument by relying too much on one or two appeals while neglecting the others.
RESEARCH WRITING AS CONVERSATION
Writing is a social process. Texts are created to be read by others, and in creating those texts, writers should be aware of not only their personal assumptions, biases, and tastes, but also those of their readers. Writing, therefore, is an interactive process. It is a conversation, a meeting of minds, during which ideas are exchanged, debates and discussions take place and, sometimes, but not always, consensus is reached. You may be familiar with the famous quote by the 20th century rhetorician Kenneth Burke who compared writing to a conversation at a social event. In his 1974 book The Philosophy of Literary Form Burke writes,
Imagine that you enter a parlor. You come late. When you arrive, others have long preceded you, and they are engaged in a heated discussion, a discussion too heated for them to pause and tell you exactly what it is about. In fact, the discussion had already begun long before any of them got there, so that no one present is qualified to retrace for you all the steps that had gone before. You listen for a while, until you decide that you have caught the tenor of the argument; then you put in your oar. Someone answers; you answer him, another comes to your defense; another aligns himself against you, to either the embarrassment of gratification of your opponent, depending upon the quality of your ally’s assistance. However, the discussion is interminable. The hour grows late, you must depart. And you do depart, with the discussion still vigorously in progress (110-111).
This passage by Burke is extremely popular among writers because it captures the interactive nature of writing so precisely. Reading Burke’s words carefully, we will notice that the interaction between readers and writers is continuous. A writer always enters a conversation in progress. In order to participate in the discussion, just like in real life, you need to know what your interlocutors have been talking about. So you listen (read). Once you feel you have got the drift of the conversation, you say (write) something. Your text is read by others who respond to your ideas, stories, and arguments with their own. This interaction never ends!
To write well, it is important to listen carefully and understand the conversations that are going on around you. Writers who are able to listen to these conversations and pick up important topics, themes, and arguments are generally more effective at reaching and impressing their audiences. It is also important to treat research, writing, and every occasion for these activities as opportunities to participate in the on-going conversation of people interested in the same topics and questions which interest you.
Our knowledge about our world is shaped by the best and most up-to-date theories available to them. Sometimes these theories can be experimentally tested and proven, and sometimes, when obtaining such proof is impossible, they are based on consensus reached as a result of conversation and debate. Even the theories and knowledge that can be experimentally tested (for example in sciences) do not become accepted knowledge until most members of the scientific community accept them. Other members of this community will help them test their theories and hypotheses, give them feedback on their writing, and keep them searching for the best answers to their questions. As Burke says in his famous passage, the interaction between the members of intellectual communities never ends. No piece of writing, no argument, no theory or discover is ever final. Instead, they all are subject to discussion, questioning, and improvement.
A simple but useful example of this process is the evolution of humankind’s understanding of their planet Earth and its place in the Universe. As you know, in Medieval Europe, the prevailing theory was that the Earth was the center of the Universe and that all other planets and the Sun rotated around it. This theory was the result of the church’s teachings, and thinkers who disagreed with it were pronounced heretics and often burned. In 1543, astronomer Nikolaus Kopernikus argued that the Sun was at the center of the solar system and that all planets of the system rotate around the Sun. Later, Galieo experimentally proved Kopernikus’ theory with the help of a telescope. Of course, the Earch did not begin to rotate around the Sun with this discovery. Yet, Kopernikus’ and Galileo’s theories of the Universe went against the Catholic Church’s teachings which dominated the social discourse of Medieval Europe. The Inquisition did not engage in debate with the two scientists. Instead, Kopernikus was executed for his views and Galileo was sentenced to house arrest for his views.
Although in the modern world, dissenting thinkers are unlikely to suffer such harsh punishment, the examples of Kopernikus and Galileo teach us two valuable lessons about the social nature of knowledge. Firstly, Both Kopernikus and Galileo tried to improve on an existing theory of the Universe that placed our planet at the center. They did not work from nothing but used beliefs that already existed in their society and tried to modify and disprove those beliefs. Time and later scientific research proved that they were right. Secondly, even after Galileo was able to prove the structure of the Solar system experimentally, his theory did not become widely accepted until the majority of people in society assimilated it. Therefore, new findings do not become accepted knowledge until they penetrate the fabric of social discourse and until enough people accept them as true.
Exercise \(\PageIndex{4}\): Finding the Origins of Knowledge
- Seeing writing as an exchange of ideas means seeing all new theories, ideas, and beliefs as grounded in pre-existing knowledge. Therefore, when beginning a new writing project, writers never work “from scratch.” Instead, they tap into the resources of their community for ideas, inspiration, and research leads. Keeping these statements in mind, answer the following questions. Apply your answers to one of the research projects described in this book. Be sure to record your answers.
- Consider a possible research project topic. What do you know about your topic before you begin to write?
- Where did this knowledge come from? Be sure to include both secondary sources (books, websites, etc.) and primary ones (people, events, personal memories). Is this knowledge socially created? What communities or groups or people created it, how, and why?
- What parts of your current knowledge about your subject could be called “fact” and what parts could be called “opinion?”
- How can your current knowledge about the topic help you in planning and conducting the research for the project?
Share your thoughts with your classmates and instructor.
CONCLUSIONS
In this chapter, we have learned the definition of rhetoric and the basic differences between several important rhetorical schools. We have also discussed how to key elements of the rhetorical situation: purpose, audience, and context. As you work on the research writing projects presented throughout this book, be sure to revisit this chapter often. Everything that you have read about here and every activity you have completed as you worked through this chapter is applicable to all research writing projects in this book and beyond. Most school writing assignments give you direct instructions about your purpose, intended audience, and rhetorical occasion. Truly proficient and independent writers, however, learn to define their purpose, audiences, and contexts of their writing, on their own. The material in this chapter is designed to enable to become better at those tasks.
When you receive a writing assignment, it is very tempting to see it as just another hoop to jump through and not as a genuine rhetorical situation, an opportunity to influence others with your writing. It is certainly tempting to see yourself writing only for the teacher, without a real purpose and oblivious of the context of your writing.
The material of this chapter as well as the writing projects presented throughout this book are designed to help you think of writing as a persuasive, rhetorical activity. Conducting research and incorporating its results into your paper is a part of this rhetorical process.
References
- Pavel Zemilansky
- Aristotle. “Rhetoric.” Aristotle’s Rhetoric. June 21, 2004. www.public.iastate.edu/~honeyl/Rhetoric/. April 21, 2008.
- Burke, Kenneth. The Philosophy of Literary Form. Berkeley: University of California Press, 1941.
- CNN. “N. Korea Talks On Despite Rhetoric.” CNN.com. August 3, 2003. www.cnn.com/2003/WORLD/asiapc...lks/index.html…. April 21, 2008.
- Corbett, Edward, P.J and Connors, Robert. Classical Rhetoric for the Modern Student. Oxford University Press, USA; 4 edition, 1998.
- Fritz, Ben et al. “About Spinsanity.” Spinsanity. 2001-2005. http://www.spinsanity.org . April 21, 2008.
- Papakyriakou/Anagnostou, Ellen. Kairos. Ancient Greek Cities. 1998. http://www.sikyon.com/sicyon/Lysippo...sip_kairos.jpg . April 21, 2008.
- Rouzie, Albert. “The Rhetorical Triangle.” Rhetoric Resources. 1998. www-as.phy.ohiou.edu/~rouzie/.../rhetriang.gif. April 21, 2008. | libretexts | 2025-03-17T22:27:23.479593 | 2019-09-23T22:10:36 | {
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"title": "All Writing is Argument",
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https://socialsci.libretexts.org/Courses/Sacramento_City_College/LIBR_324%3A_Critical_Thinking_and_Information_Literacy/3%3A_SLO_3/3.4%3A_Rhetorical_Modes/Cause_and_Effect | Cause and Effect
Learning Objectives
- Determine the purpose and structure of cause and effect in writing.
- Understand how to write a cause-and-effect essay.
THE PURPOSE OF CAUSE AND EFFECT IN WRITING
It is often considered human nature to ask, “why?” and “how?” We want to know how our child got sick so we can better prevent it from happening in the future, or why our colleague a pay raise because we want one as well. We want to know how much money we will save over the long term if we buy a hybrid car. These examples identify only a few of the relationships we think about in our lives, but each shows the importance of understanding cause and effect.
A cause is something that produces an event or condition; an effect is what results from an event or condition. The purpose of the cause-and-effect essay is to determine how various phenomena relate in terms of origins and results. Sometimes the connection between cause and effect is clear, but often determining the exact relationship between the two is very difficult. For example, the following effects of a cold may be easily identifiable: a sore throat, runny nose, and a cough. But determining the cause of the sickness can be far more difficult. A number of causes are possible, and to complicate matters, these possible causes could have combined to cause the sickness. That is, more than one cause may be responsible for any given effect. Therefore, cause-and-effect discussions are often complicated and frequently lead to debates and arguments.
tip
Use the complex nature of cause and effect to your advantage. Often it is not necessary, or even possible, to find the exact cause of an event or to name the exact effect. So, when formulating a thesis, you can claim one of a number of causes or effects to be the primary, or main, cause or effect. As soon as you claim that one cause or one effect is more crucial than the others, you have developed a thesis.
Exercise \(\PageIndex{1}\)
Consider the causes and effects in the following thesis statements. List a cause and effect for each one on your own sheet of paper.
- The growing childhood obesity epidemic is a result of technology.
- Much of the wildlife is dying because of the oil spill.
- The town continued programs that it could no longer afford, so it went bankrupt.
- More young people became politically active as use of the Internet spread throughout society.
- While many experts believed the rise in violence was due to the poor economy, it was really due to the summer-long heat wave.
Exercise \(\PageIndex{2}\)
Write three cause-and-effect thesis statements of your own for each of the following five broad topics.
- Health and nutrition
- Sports
- Media
- Politics
- History
THE STRUCTURE OF A CAUSE-AND-EFFECT ESSAY
The cause-and-effect essay opens with a general introduction to the topic, which then leads to a thesis that states the main cause, main effect, or various causes and effects of a condition or event.
The cause-and-effect essay can be organized in one of the following two primary ways:
- Start with the cause and then talk about the effects.
- Start with the effect and then talk about the causes.
For example, if your essay were on childhood obesity, you could start by talking about the effect of childhood obesity and then discuss the cause or you could start the same essay by talking about the cause of childhood obesity and then move to the effect.
Regardless of which structure you choose, be sure to explain each element of the essay fully and completely. Explaining complex relationships requires the full use of evidence, such as scientific studies, expert testimony, statistics, and anecdotes.
Because cause-and-effect essays determine how phenomena are linked, they make frequent use of certain words and phrases that denote such linkage. See Table 10.4 “Phrases of Causation” for examples of such terms.
| as a result | consequently | because | due to |
| hence | since | thus | therefore |
The conclusion should wrap up the discussion and reinforce the thesis, leaving the reader with a clear understanding of the relationship that was analyzed.
tip
Be careful of resorting to empty speculation. In writing, speculation amounts to unsubstantiated guessing. Writers are particularly prone to such trappings in cause-and-effect arguments due to the complex nature of finding links between phenomena. Be sure to have clear evidence to support the claims that you make.
Exercise \(\PageIndex{3}\)
Look at some of the cause-and-effect relationships from Exercise 2. Outline the links you listed. Outline one using a cause-then-effect structure. Outline the other using the effect-then-cause structure.
WRITING A CAUSE-AND-EFFECT ESSAY
Choose an event or condition that you think has an interesting cause-and-effect relationship. Introduce your topic in an engaging way. End your introduction with a thesis that states the main cause, the main effect, or both.
Organize your essay by starting with either the cause-then-effect structure or the effect-then-cause structure. Within each section, you should clearly explain and support the causes and effects using a full range of evidence. If you are writing about multiple causes or multiple effects, you may choose to sequence either in terms of order of importance. In other words, order the causes from least to most important (or vice versa), or order the effects from least important to most important (or vice versa).
Use the phrases of causation when trying to forge connections between various events or conditions. This will help organize your ideas and orient the reader. End your essay with a conclusion that summarizes your main points and reinforces your thesis.
Exercise \(\PageIndex{4}\)
Choose one of the ideas you outlined in “Exercise 3” and write a full cause-and-effect essay. Be sure to include an engaging introduction, a clear thesis, strong evidence and examples, and a thoughtful conclusion.
key takeaways
- The purpose of the cause-and-effect essay is to determine how various phenomena are related.
- The thesis states what the writer sees as the main cause, main effect, or various causes and effects of a condition or event.
- The cause-and-effect essay can be organized in one of these two primary ways:
- Start with the cause and then talk about the effect.
- Start with the effect and then talk about the cause.
- Strong evidence is particularly important in the cause-and-effect essay due to the complexity of determining connections between phenomena.
- Phrases of causation are helpful in signaling links between various elements in the essay.
References
- This section was originally from Writing for Success, found at the University of Minnesota open textbook project. Full license information: This is a derivative of Writing for Success by a publisher who has requested that they and the original author not receive attribution, originally released and is used under CC BY-NC-SA. This work, unless otherwise expressly stated, is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. | libretexts | 2025-03-17T22:27:23.548943 | 2019-09-23T22:10:40 | {
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"book_url": "https://commons.libretexts.org/book/socialsci-23897",
"title": "Cause and Effect",
"author": null
} |
https://socialsci.libretexts.org/Courses/Sacramento_City_College/LIBR_324%3A_Critical_Thinking_and_Information_Literacy/3%3A_SLO_3/3.4%3A_Rhetorical_Modes/Classification | Classification
Learning Objectives
- Determine the purpose and structure of the classification essay.
- Understand how to write a classification essay.
THE PURPOSE OF CLASSIFICATION IN WRITING
The purpose of classification is to break down broad subjects into smaller, more manageable, more specific parts. We classify things in our daily lives all the time, often without even thinking about it. Cell phones, for example, have now become part of a broad category. They can be classified as feature phones, media phones, and smartphones.
Smaller categories, and the way in which these categories are created, help us make sense of the world. Keep both of these elements in mind when writing a classification essay.
tip
Choose topics that you know well when writing classification essays. The more you know about a topic, the more you can break it into smaller, more interesting parts. Adding interest and insight will enhance your classification essays.
Exercise \(\PageIndex{1}\)
On a separate sheet of paper, break the following categories into smaller classifications.
- The United States
- Colleges and universities
- Beverages
- Fashion
THE STRUCTURE OF A CLASSIFICATION ESSAY
The classification essay opens with an introductory paragraph that introduces the broader topic. The thesis should then explain how that topic is divided into subgroups and why. Take the following introductory paragraph, for example:
When people think of New York, they often think of only New York City. But New York is actually a diverse state with a full range of activities to do, sights to see, and cultures to explore. In order to better understand the diversity of New York state, it is helpful to break it into these five separate regions: Long Island, New York City, Western New York, Central New York, and Northern New York.
The underlined thesis explains not only the category and subcategory but also the rationale for breaking it into those categories. Through this classification essay, the writer hopes to show his or her readers a different way of considering the state.
Each body paragraph of a classification essay is dedicated to fully illustrating each of the subcategories. In the previous example, then, each region of New York would have its own paragraph.
The conclusion should bring all the categories and subcategories back together again to show the reader the big picture. In the previous example, the conclusion might explain how the various sights and activities of each region of New York add to its diversity and complexity.
tip
To avoid settling for an overly simplistic classification, make sure you break down any given topic at least three different ways. This will help you think outside the box and perhaps even learn something entirely new about a subject.
Exercise \(\PageIndex{2}\)
Using your classifications from Exercise 1, write a brief paragraph explaining why you chose to organize each main category in the way that you did.
WRITING A CLASSIFICATION ESSAY
Start with an engaging opening that will adequately introduce the general topic that you will be dividing into smaller subcategories. Your thesis should come at the end of your introduction. It should include the topic, your subtopics, and the reason you are choosing to break down the topic in the way that you are. Use the following classification thesis equation:
\[\text{topic} + \text{subtopics} + \text{rationale for the subtopics} = \text{thesis} \ldotp\]
The organizing strategy of a classification essay is dictated by the initial topic and the subsequent subtopics. Each body paragraph is dedicated to fully illustrating each of the subtopics. In a way, coming up with a strong topic pays double rewards in a classification essay. Not only do you have a good topic, but you also have a solid organizational structure within which to write.
Be sure you use strong details and explanations for each subcategory paragraph that help explain and support your thesis. Also, be sure to give examples to illustrate your points. Finally, write a conclusion that links all the subgroups together again. The conclusion should successfully wrap up your essay by connecting it to your topic initially discussed in the introduction.
Exercise \(\PageIndex{3}\)
Building on Exercise 1 and Exercise 2, write a five-paragraph classification essay about one of the four original topics. In your thesis, make sure to include the topic, subtopics, and rationale for your breakdown. And make sure that your essay is organized into paragraphs that each describes a subtopic.
key takeaways
- The purpose of classification is to break a subject into smaller, more manageable, more specific parts.
- Smaller subcategories help us make sense of the world, and the way in which these subcategories are created also helps us make sense of the world.
- A classification essay is organized by its subcategories.
References
- This section was originally from Writing for Success, found at the University of Minnesota open textbook project. Full license information: This is a derivative of Writing for Success by a publisher who has requested that they and the original author not receive attribution, originally released and is used under CC BY-NC-SA. This work, unless otherwise expressly stated, is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. | libretexts | 2025-03-17T22:27:23.610972 | 2019-09-23T22:10:37 | {
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"book_url": "https://commons.libretexts.org/book/socialsci-23897",
"title": "Classification",
"author": null
} |
https://socialsci.libretexts.org/Courses/Sacramento_City_College/LIBR_324%3A_Critical_Thinking_and_Information_Literacy/3%3A_SLO_3/3.4%3A_Rhetorical_Modes/Comparison_and_Contrast | Comparison and Contrast
Learning Objectives
- Determine the purpose and structure of comparison and contrast in writing.
- Explain organizational methods used when comparing and contrasting.
- Understand how to write a compare-and-contrast essay.
THE PURPOSE OF COMPARISON AND CONTRAST IN WRITING
Comparison in writing discusses elements that are similar, while contrast in writing discusses elements that are different. A compare-and-contrast essay, then, analyzes two subjects by comparing them, contrasting them, or both.
The key to a good compare-and-contrast essay is to choose two or more subjects that connect in a meaningful way. The purpose of conducting the comparison or contrast is not to state the obvious but rather to illuminate subtle differences or unexpected similarities. For example, if you wanted to focus on contrasting two subjects you would not pick apples and oranges; rather, you might choose to compare and contrast two types of oranges or two types of apples to highlight subtle differences. For example, Red Delicious apples are sweet, while Granny Smiths are tart and acidic. Drawing distinctions between elements in a similar category will increase the audience’s understanding of that category, which is the purpose of the compare-and-contrast essay.
Similarly, to focus on comparison, choose two subjects that seem at first to be unrelated. For a comparison essay, you likely would not choose two apples or two oranges because they share so many of the same properties already. Rather, you might try to compare how apples and oranges are quite similar. The more divergent the two subjects initially seem, the more interesting a comparison essay will be.
WRITING AT WORK
Comparing and contrasting is also an evaluative tool. In order to make accurate evaluations about a given topic, you must first know the critical points of similarity and difference. Comparing and contrasting is a primary tool for many workplace assessments. You have likely compared and contrasted yourself to other colleagues. Employee advancements, pay raises, hiring, and firing are typically conducted using comparison and contrast. Comparison and contrast could be used to evaluate companies, departments, or individuals.
Exercise \(\PageIndex{1}\)
Brainstorm an essay that leans toward contrast. Choose one of the following three categories. Pick two examples from each. Then come up with one similarity and three differences between the examples.
- Romantic comedies
- Internet search engines
- Cell phones
Exercise \(\PageIndex{2}\)
Brainstorm an essay that leans toward comparison. Choose one of the following three items. Then come up with one difference and three similarities.
- Department stores and discount retail stores
- Fast food chains and fine dining restaurants
- Dogs and cats
THE STRUCTURE OF A COMPARISON AND CONTRAST ESSAY
The compare-and-contrast essay starts with a thesis that clearly states the two subjects that are to be compared, contrasted, or both and the reason for doing so. The thesis could lean more toward comparing, contrasting, or both. Remember, the point of comparing and contrasting is to provide useful knowledge to the reader. Take the following thesis as an example that leans more toward contrasting.
Thesis statement: Organic vegetables may cost more than those that are conventionally grown, but when put to the test, they are definitely worth every extra penny.
Here the thesis sets up the two subjects to be compared and contrasted (organic versus conventional vegetables), and it makes a claim about the results that might prove useful to the reader.
You may organize compare-and-contrast essays in one of the following two ways:
- According to the subjects themselves, discussing one then the other
- According to individual points, discussing each subject in relation to each point
See Figure 10.1 “Comparison and Contrast Diagram”, which diagrams the ways to organize our organic versus conventional vegetables thesis.
Figure 10.1
The organizational structure you choose depends on the nature of the topic, your purpose, and your audience.
Given that compare-and-contrast essays analyze the relationship between two subjects, it is helpful to have some phrases on hand that will cue the reader to such analysis. See Table 10.3 “Phrases of Comparison and Contrast” for examples.
| Comparison | Contrast |
|---|---|
| one similarity |
one difference |
| another similarity | another difference |
| both | conversely |
|
like |
in contrast |
| likewise | unlike |
| similarly | while |
|
in a similar fashion |
whereas |
Exercise \(\PageIndex{3}\)
Create an outline for each of the items you chose in Exercise 1 & 2. Use the point-by-point organizing strategy for one of them, and use the subject organizing strategy for the other.
WRITING A COMPARISON AND CONTRAST ESSAY
First choose whether you want to compare seemingly disparate subjects, contrast seemingly similar subjects, or compare and contrast subjects. Once you have decided on a topic, introduce it with an engaging opening paragraph. Your thesis should come at the end of the introduction, and it should establish the subjects you will compare, contrast, or both as well as state what can be learned from doing so.
The body of the essay can be organized in one of two ways: by subject or by individual points. The organizing strategy that you choose will depend on, as always, your audience and your purpose. You may also consider your particular approach to the subjects as well as the nature of the subjects themselves; some subjects might better lend themselves to one structure or the other. Make sure to use comparison and contrast phrases to cue the reader to the ways in which you are analyzing the relationship between the subjects.
After you finish analyzing the subjects, write a conclusion that summarizes the main points of the essay and reinforces your thesis.
WRITING AT WORK
Many business presentations are conducted using comparison and contrast. The organizing strategies—by subject or individual points—could also be used for organizing a presentation. Keep this in mind as a way of organizing your content the next time you or a colleague have to present something at work.
Exercise \(\PageIndex{4}\)
Choose one of the outlines you created in Exercise 3, and write a full compare-and-contrast essay. Be sure to include an engaging introduction, a clear thesis, well-defined and detailed paragraphs, and a fitting conclusion that ties everything together.
key takeaways
- A compare-and-contrast essay analyzes two subjects by either comparing them, contrasting them, or both.
- The purpose of writing a comparison or contrast essay is not to state the obvious but rather to illuminate subtle differences or unexpected similarities between two subjects.
- The thesis should clearly state the subjects that are to be compared, contrasted, or both, and it should state what is to be learned from doing so.
- There are two main organizing strategies for compare-and-contrast essays.
- Organize by the subjects themselves, one then the other.
- Organize by individual points, in which you discuss each subject in relation to each point.
- Use phrases of comparison or phrases of contrast to signal to readers how exactly the two subjects are being analyzed.
References
- This section was originally from Writing for Success, found at the University of Minnesota open textbook project. Full license information: This is a derivative of Writing for Success by a publisher who has requested that they and the original author not receive attribution, originally released and is used under CC BY-NC-SA. This work, unless otherwise expressly stated, is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. | libretexts | 2025-03-17T22:27:23.679791 | 2019-09-23T22:10:39 | {
"license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/",
"url": "https://socialsci.libretexts.org/Courses/Sacramento_City_College/LIBR_324%3A_Critical_Thinking_and_Information_Literacy/3%3A_SLO_3/3.4%3A_Rhetorical_Modes/Comparison_and_Contrast",
"book_url": "https://commons.libretexts.org/book/socialsci-23897",
"title": "Comparison and Contrast",
"author": null
} |
https://socialsci.libretexts.org/Courses/Sacramento_City_College/LIBR_324%3A_Critical_Thinking_and_Information_Literacy/3%3A_SLO_3/3.4%3A_Rhetorical_Modes/Definition | Definition
Learning Objectives
- Determine the purpose and structure of the definition essay.
- Understand how to write a definition essay.
THE PURPOSE OF DEFINITION IN WRITING
The purpose of a definition essay may seem self-explanatory: the purpose of the definition essay is to simply define something. But defining terms in writing is often more complicated than just consulting a dictionary. In fact, the way we define terms can have far-reaching consequences for individuals as well as collective groups.
Take, for example, a word like alcoholism. The way in which one defines alcoholism depends on its legal, moral, and medical contexts. Lawyers may define alcoholism in terms of its legality; parents may define alcoholism in terms of its morality; and doctors will define alcoholism in terms of symptoms and diagnostic criteria. Think also of terms that people tend to debate in our broader culture. How we define words, such asmarriage and climate change, has enormous impact on policy decisions and even on daily decisions. Think about conversations couples may have in which words like commitment, respect, or love need clarification.
Defining terms within a relationship, or any other context, can at first be difficult, but once a definition is established between two people or a group of people, it is easier to have productive dialogues. Definitions, then, establish the way in which people communicate ideas. They set parameters for a given discourse, which is why they are so important.
tip
When writing definition essays, avoid terms that are too simple, that lack complexity. Think in terms of concepts, such as hero, immigration, or loyalty, rather than physical objects. Definitions of concepts, rather than objects, are often fluid and contentious, making for a more effective definition essay.
WRITING AT WORK
Definitions play a critical role in all workplace environments. Take the term sexual harassment, for example. Sexual harassment is broadly defined on the federal level, but each company may have additional criteria that define it further. Knowing how your workplace defines and treats all sexual harassment allegations is important. Think, too, about how your company defines lateness, productivity, or contributions.
Exercise \(\PageIndex{1}\)
On a separate sheet of paper, write about a time in your own life in which the definition of a word, or the lack of a definition, caused an argument. Your term could be something as simple as the category of an all-star in sports or how to define a good movie. Or it could be something with higher stakes and wider impact, such as a political argument. Explain how the conversation began, how the argument hinged on the definition of the word, and how the incident was finally resolved.
Collaboration
Please share with a classmate and compare your responses.
THE STRUCTURE OF A DEFINITION ESSAY
The definition essay opens with a general discussion of the term to be defined. You then state as your thesis your definition of the term.
The rest of the essay should explain the rationale for your definition. Remember that a dictionary’s definition is limiting, and you should not rely strictly on the dictionary entry. Instead, consider the context in which you are using the word. Context identifies the circumstances, conditions, or setting in which something exists or occurs. Often words take on different meanings depending on the context in which they are used. For example, the ideal leader in a battlefield setting could likely be very different than a leader in an elementary school setting. If a context is missing from the essay, the essay may be too short or the main points could be confusing or misunderstood.
The remainder of the essay should explain different aspects of the term’s definition. For example, if you were defining a good leader in an elementary classroom setting, you might define such a leader according to personality traits: patience, consistency, and flexibility. Each attribute would be explained in its own paragraph.
WRITING AT WORK
It is a good idea to occasionally assess your role in the workplace. You can do this through the process of definition. Identify your role at work by defining not only the routine tasks but also those gray areas where your responsibilities might overlap with those of others. Coming up with a clear definition of roles and responsibilities can add value to your résumé and even increase productivity in the workplace.
Exercise \(\PageIndex{2}\)
On a separate sheet of paper, define each of the following items in your own terms. If you can, establish a context for your definition.
- Bravery
- Adulthood
- Consumer culture
- Violence
- Art
WRITING A DEFINITION ESSAY
Choose a topic that will be complex enough to be discussed at length. Choosing a word or phrase of personal relevance often leads to a more interesting and engaging essay.
After you have chosen your word or phrase, start your essay with an introduction that establishes the relevancy of the term in the chosen specific context. Your thesis comes at the end of the introduction, and it should clearly state your definition of the term in the specific context. Establishing a functional context from the beginning will orient readers and minimize misunderstandings.
The body paragraphs should each be dedicated to explaining a different facet of your definition. Make sure to use clear examples and strong details to illustrate your points. Your concluding paragraph should pull together all the different elements of your definition to ultimately reinforce your thesis.
Exercise \(\PageIndex{3}\)
Create a full definition essay from one of the items you already defined in Exercise 2” Be sure to include an interesting introduction, a clear thesis, a well-explained context, distinct body paragraphs, and a conclusion that pulls everything together.
key takeaways
- Definitions establish the way in which people communicate ideas. They set parameters for a given discourse.
- Context affects the meaning and usage of words.
- The thesis of a definition essay should clearly state the writer’s definition of the term in the specific context.
- Body paragraphs should explain the various facets of the definition stated in the thesis.
- The conclusion should pull all the elements of the definition together at the end and reinforce the thesis.
References
- This section was originally from Writing for Success, found at the University of Minnesota open textbook project. Full license information: This is a derivative of Writing for Success by a publisher who has requested that they and the original author not receive attribution, originally released and is used under CC BY-NC-SA. This work, unless otherwise expressly stated, is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. | libretexts | 2025-03-17T22:27:23.744679 | 2019-09-23T22:10:37 | {
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"url": "https://socialsci.libretexts.org/Courses/Sacramento_City_College/LIBR_324%3A_Critical_Thinking_and_Information_Literacy/3%3A_SLO_3/3.4%3A_Rhetorical_Modes/Definition",
"book_url": "https://commons.libretexts.org/book/socialsci-23897",
"title": "Definition",
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Subsets and Splits