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1 | 2518-2521 | , if f and g are continuous functions, then
(f ± g) (x) = f (x) ± g(x) is continuous (f g) (x) = f (x) g(x) is continuous |
1 | 2519-2522 | (f g) (x) = f (x) g(x) is continuous ( )
( )
( )
f
f x
x
g
g x
=
(wherever g(x) ≠ 0) is continuous |
1 | 2520-2523 | g) (x) = f (x) g(x) is continuous ( )
( )
( )
f
f x
x
g
g x
=
(wherever g(x) ≠ 0) is continuous ® Every differentiable function is continuous, but the converse is not true |
1 | 2521-2524 | g(x) is continuous ( )
( )
( )
f
f x
x
g
g x
=
(wherever g(x) ≠ 0) is continuous ® Every differentiable function is continuous, but the converse is not true ® Chain rule is rule to differentiate composites of functions |
1 | 2522-2525 | ( )
( )
( )
f
f x
x
g
g x
=
(wherever g(x) ≠ 0) is continuous ® Every differentiable function is continuous, but the converse is not true ® Chain rule is rule to differentiate composites of functions If f = v o u, t = u (x)
and if both dt
dx and dv
dt exist then
df
dv dt
dx
=dt dx
⋅
® Following are some of the standard derivatives (in appropriate domains):
(
)
1
2
1
sin
1
d
x
dx
x
−
=
−
(
)
1
2
1
cos
1
d
x
dx
x
−
= −
−
(
)
1
2
1
tan
1
d
x
dx
x
−
=
+
(
x)
x
d
e
e
dx
=
(
)
1
log
d
x
dx
x
=
® Logarithmic differentiation is a powerful technique to differentiate functions
of the form f (x) = [u (x)]v (x) |
1 | 2523-2526 | ® Every differentiable function is continuous, but the converse is not true ® Chain rule is rule to differentiate composites of functions If f = v o u, t = u (x)
and if both dt
dx and dv
dt exist then
df
dv dt
dx
=dt dx
⋅
® Following are some of the standard derivatives (in appropriate domains):
(
)
1
2
1
sin
1
d
x
dx
x
−
=
−
(
)
1
2
1
cos
1
d
x
dx
x
−
= −
−
(
)
1
2
1
tan
1
d
x
dx
x
−
=
+
(
x)
x
d
e
e
dx
=
(
)
1
log
d
x
dx
x
=
® Logarithmic differentiation is a powerful technique to differentiate functions
of the form f (x) = [u (x)]v (x) Here both f (x) and u(x) need to be positive for
this technique to make sense |
1 | 2524-2527 | ® Chain rule is rule to differentiate composites of functions If f = v o u, t = u (x)
and if both dt
dx and dv
dt exist then
df
dv dt
dx
=dt dx
⋅
® Following are some of the standard derivatives (in appropriate domains):
(
)
1
2
1
sin
1
d
x
dx
x
−
=
−
(
)
1
2
1
cos
1
d
x
dx
x
−
= −
−
(
)
1
2
1
tan
1
d
x
dx
x
−
=
+
(
x)
x
d
e
e
dx
=
(
)
1
log
d
x
dx
x
=
® Logarithmic differentiation is a powerful technique to differentiate functions
of the form f (x) = [u (x)]v (x) Here both f (x) and u(x) need to be positive for
this technique to make sense —v
v
v
v
v—
Rationalised 2023-24
v With the Calculus as a key, Mathematics can be successfully applied
to the explanation of the course of Nature |
1 | 2525-2528 | If f = v o u, t = u (x)
and if both dt
dx and dv
dt exist then
df
dv dt
dx
=dt dx
⋅
® Following are some of the standard derivatives (in appropriate domains):
(
)
1
2
1
sin
1
d
x
dx
x
−
=
−
(
)
1
2
1
cos
1
d
x
dx
x
−
= −
−
(
)
1
2
1
tan
1
d
x
dx
x
−
=
+
(
x)
x
d
e
e
dx
=
(
)
1
log
d
x
dx
x
=
® Logarithmic differentiation is a powerful technique to differentiate functions
of the form f (x) = [u (x)]v (x) Here both f (x) and u(x) need to be positive for
this technique to make sense —v
v
v
v
v—
Rationalised 2023-24
v With the Calculus as a key, Mathematics can be successfully applied
to the explanation of the course of Nature ” — WHITEHEAD v
6 |
1 | 2526-2529 | Here both f (x) and u(x) need to be positive for
this technique to make sense —v
v
v
v
v—
Rationalised 2023-24
v With the Calculus as a key, Mathematics can be successfully applied
to the explanation of the course of Nature ” — WHITEHEAD v
6 1 Introduction
In Chapter 5, we have learnt how to find derivative of composite functions, inverse
trigonometric functions, implicit functions, exponential functions and logarithmic functions |
1 | 2527-2530 | —v
v
v
v
v—
Rationalised 2023-24
v With the Calculus as a key, Mathematics can be successfully applied
to the explanation of the course of Nature ” — WHITEHEAD v
6 1 Introduction
In Chapter 5, we have learnt how to find derivative of composite functions, inverse
trigonometric functions, implicit functions, exponential functions and logarithmic functions In this chapter, we will study applications of the derivative in various disciplines, e |
1 | 2528-2531 | ” — WHITEHEAD v
6 1 Introduction
In Chapter 5, we have learnt how to find derivative of composite functions, inverse
trigonometric functions, implicit functions, exponential functions and logarithmic functions In this chapter, we will study applications of the derivative in various disciplines, e g |
1 | 2529-2532 | 1 Introduction
In Chapter 5, we have learnt how to find derivative of composite functions, inverse
trigonometric functions, implicit functions, exponential functions and logarithmic functions In this chapter, we will study applications of the derivative in various disciplines, e g , in
engineering, science, social science, and many other fields |
1 | 2530-2533 | In this chapter, we will study applications of the derivative in various disciplines, e g , in
engineering, science, social science, and many other fields For instance, we will learn
how the derivative can be used (i) to determine rate of change of quantities, (ii) to find
the equations of tangent and normal to a curve at a point, (iii) to find turning points on
the graph of a function which in turn will help us to locate points at which largest or
smallest value (locally) of a function occurs |
1 | 2531-2534 | g , in
engineering, science, social science, and many other fields For instance, we will learn
how the derivative can be used (i) to determine rate of change of quantities, (ii) to find
the equations of tangent and normal to a curve at a point, (iii) to find turning points on
the graph of a function which in turn will help us to locate points at which largest or
smallest value (locally) of a function occurs We will also use derivative to find intervals
on which a function is increasing or decreasing |
1 | 2532-2535 | , in
engineering, science, social science, and many other fields For instance, we will learn
how the derivative can be used (i) to determine rate of change of quantities, (ii) to find
the equations of tangent and normal to a curve at a point, (iii) to find turning points on
the graph of a function which in turn will help us to locate points at which largest or
smallest value (locally) of a function occurs We will also use derivative to find intervals
on which a function is increasing or decreasing Finally, we use the derivative to find
approximate value of certain quantities |
1 | 2533-2536 | For instance, we will learn
how the derivative can be used (i) to determine rate of change of quantities, (ii) to find
the equations of tangent and normal to a curve at a point, (iii) to find turning points on
the graph of a function which in turn will help us to locate points at which largest or
smallest value (locally) of a function occurs We will also use derivative to find intervals
on which a function is increasing or decreasing Finally, we use the derivative to find
approximate value of certain quantities 6 |
1 | 2534-2537 | We will also use derivative to find intervals
on which a function is increasing or decreasing Finally, we use the derivative to find
approximate value of certain quantities 6 2 Rate of Change of Quantities
Recall that by the derivative ds
dt , we mean the rate of change of distance s with
respect to the time t |
1 | 2535-2538 | Finally, we use the derivative to find
approximate value of certain quantities 6 2 Rate of Change of Quantities
Recall that by the derivative ds
dt , we mean the rate of change of distance s with
respect to the time t In a similar fashion, whenever one quantity y varies with another
quantity x, satisfying some rule
( )
y
=f x
, then dy
dx (or f′(x)) represents the rate of
change of y with respect to x and
dy
dx
x x
= 0
(or f′(x0)) represents the rate of change
of y with respect to x at
0
x
=x |
1 | 2536-2539 | 6 2 Rate of Change of Quantities
Recall that by the derivative ds
dt , we mean the rate of change of distance s with
respect to the time t In a similar fashion, whenever one quantity y varies with another
quantity x, satisfying some rule
( )
y
=f x
, then dy
dx (or f′(x)) represents the rate of
change of y with respect to x and
dy
dx
x x
= 0
(or f′(x0)) represents the rate of change
of y with respect to x at
0
x
=x if Further, if two variables x and y are varying with respect to another variable t, i |
1 | 2537-2540 | 2 Rate of Change of Quantities
Recall that by the derivative ds
dt , we mean the rate of change of distance s with
respect to the time t In a similar fashion, whenever one quantity y varies with another
quantity x, satisfying some rule
( )
y
=f x
, then dy
dx (or f′(x)) represents the rate of
change of y with respect to x and
dy
dx
x x
= 0
(or f′(x0)) represents the rate of change
of y with respect to x at
0
x
=x if Further, if two variables x and y are varying with respect to another variable t, i e |
1 | 2538-2541 | In a similar fashion, whenever one quantity y varies with another
quantity x, satisfying some rule
( )
y
=f x
, then dy
dx (or f′(x)) represents the rate of
change of y with respect to x and
dy
dx
x x
= 0
(or f′(x0)) represents the rate of change
of y with respect to x at
0
x
=x if Further, if two variables x and y are varying with respect to another variable t, i e ,
( )
x
=f t
and
( )
y
=g t
, then by Chain Rule
dy
dx = dy
dx
dt
dt , if
0
dx
dt ≠
Chapter 6
APPLICATION OF
DERIVATIVES
Rationalised 2023-24
MATHEMATICS
148
Thus, the rate of change of y with respect to x can be calculated using the rate of
change of y and that of x both with respect to t |
1 | 2539-2542 | if Further, if two variables x and y are varying with respect to another variable t, i e ,
( )
x
=f t
and
( )
y
=g t
, then by Chain Rule
dy
dx = dy
dx
dt
dt , if
0
dx
dt ≠
Chapter 6
APPLICATION OF
DERIVATIVES
Rationalised 2023-24
MATHEMATICS
148
Thus, the rate of change of y with respect to x can be calculated using the rate of
change of y and that of x both with respect to t Let us consider some examples |
1 | 2540-2543 | e ,
( )
x
=f t
and
( )
y
=g t
, then by Chain Rule
dy
dx = dy
dx
dt
dt , if
0
dx
dt ≠
Chapter 6
APPLICATION OF
DERIVATIVES
Rationalised 2023-24
MATHEMATICS
148
Thus, the rate of change of y with respect to x can be calculated using the rate of
change of y and that of x both with respect to t Let us consider some examples Example 1 Find the rate of change of the area of a circle per second with respect to
its radius r when r = 5 cm |
1 | 2541-2544 | ,
( )
x
=f t
and
( )
y
=g t
, then by Chain Rule
dy
dx = dy
dx
dt
dt , if
0
dx
dt ≠
Chapter 6
APPLICATION OF
DERIVATIVES
Rationalised 2023-24
MATHEMATICS
148
Thus, the rate of change of y with respect to x can be calculated using the rate of
change of y and that of x both with respect to t Let us consider some examples Example 1 Find the rate of change of the area of a circle per second with respect to
its radius r when r = 5 cm Solution The area A of a circle with radius r is given by A = πr2 |
1 | 2542-2545 | Let us consider some examples Example 1 Find the rate of change of the area of a circle per second with respect to
its radius r when r = 5 cm Solution The area A of a circle with radius r is given by A = πr2 Therefore, the rate
of change of the area A with respect to its radius r is given by
2
A
(
)
2
d
d
r
r
dr
=dr
π
= π |
1 | 2543-2546 | Example 1 Find the rate of change of the area of a circle per second with respect to
its radius r when r = 5 cm Solution The area A of a circle with radius r is given by A = πr2 Therefore, the rate
of change of the area A with respect to its radius r is given by
2
A
(
)
2
d
d
r
r
dr
=dr
π
= π When r = 5 cm,
A
10
d
dr =
π |
1 | 2544-2547 | Solution The area A of a circle with radius r is given by A = πr2 Therefore, the rate
of change of the area A with respect to its radius r is given by
2
A
(
)
2
d
d
r
r
dr
=dr
π
= π When r = 5 cm,
A
10
d
dr =
π Thus, the area of the circle is changing at the rate of
10π cm2/s |
1 | 2545-2548 | Therefore, the rate
of change of the area A with respect to its radius r is given by
2
A
(
)
2
d
d
r
r
dr
=dr
π
= π When r = 5 cm,
A
10
d
dr =
π Thus, the area of the circle is changing at the rate of
10π cm2/s Example 2 The volume of a cube is increasing at a rate of 9 cubic centimetres per
second |
1 | 2546-2549 | When r = 5 cm,
A
10
d
dr =
π Thus, the area of the circle is changing at the rate of
10π cm2/s Example 2 The volume of a cube is increasing at a rate of 9 cubic centimetres per
second How fast is the surface area increasing when the length of an edge is 10
centimetres |
1 | 2547-2550 | Thus, the area of the circle is changing at the rate of
10π cm2/s Example 2 The volume of a cube is increasing at a rate of 9 cubic centimetres per
second How fast is the surface area increasing when the length of an edge is 10
centimetres Solution Let x be the length of a side, V be the volume and S be the surface area of
the cube |
1 | 2548-2551 | Example 2 The volume of a cube is increasing at a rate of 9 cubic centimetres per
second How fast is the surface area increasing when the length of an edge is 10
centimetres Solution Let x be the length of a side, V be the volume and S be the surface area of
the cube Then, V = x3 and S = 6x2, where x is a function of time t |
1 | 2549-2552 | How fast is the surface area increasing when the length of an edge is 10
centimetres Solution Let x be the length of a side, V be the volume and S be the surface area of
the cube Then, V = x3 and S = 6x2, where x is a function of time t Now
dV
dt = 9cm3/s (Given)
Therefore
9 =
3
3
V
(
)
(
)
d
d
d
dx
x
x
dt
dt
dx
dt
=
=
⋅
(By Chain Rule)
=
32
dx
x
dt
⋅
or
dx
dt =
x32 |
1 | 2550-2553 | Solution Let x be the length of a side, V be the volume and S be the surface area of
the cube Then, V = x3 and S = 6x2, where x is a function of time t Now
dV
dt = 9cm3/s (Given)
Therefore
9 =
3
3
V
(
)
(
)
d
d
d
dx
x
x
dt
dt
dx
dt
=
=
⋅
(By Chain Rule)
=
32
dx
x
dt
⋅
or
dx
dt =
x32 (1)
Now
dS
dt =
2
2
(6
)
(6
)
d
d
dx
x
x
dt
dx
dt
=
⋅
(By Chain Rule)
=
32
36
12x
x
x
⋅
=
(Using (1))
Hence, when
x = 10 cm,
2
3 |
1 | 2551-2554 | Then, V = x3 and S = 6x2, where x is a function of time t Now
dV
dt = 9cm3/s (Given)
Therefore
9 =
3
3
V
(
)
(
)
d
d
d
dx
x
x
dt
dt
dx
dt
=
=
⋅
(By Chain Rule)
=
32
dx
x
dt
⋅
or
dx
dt =
x32 (1)
Now
dS
dt =
2
2
(6
)
(6
)
d
d
dx
x
x
dt
dx
dt
=
⋅
(By Chain Rule)
=
32
36
12x
x
x
⋅
=
(Using (1))
Hence, when
x = 10 cm,
2
3 6 cm /s
dS
dt =
Rationalised 2023-24
APPLICATION OF DERIVATIVES
149
Example 3 A stone is dropped into a quiet lake and waves move in circles at a speed
of 4cm per second |
1 | 2552-2555 | Now
dV
dt = 9cm3/s (Given)
Therefore
9 =
3
3
V
(
)
(
)
d
d
d
dx
x
x
dt
dt
dx
dt
=
=
⋅
(By Chain Rule)
=
32
dx
x
dt
⋅
or
dx
dt =
x32 (1)
Now
dS
dt =
2
2
(6
)
(6
)
d
d
dx
x
x
dt
dx
dt
=
⋅
(By Chain Rule)
=
32
36
12x
x
x
⋅
=
(Using (1))
Hence, when
x = 10 cm,
2
3 6 cm /s
dS
dt =
Rationalised 2023-24
APPLICATION OF DERIVATIVES
149
Example 3 A stone is dropped into a quiet lake and waves move in circles at a speed
of 4cm per second At the instant, when the radius of the circular wave is 10 cm, how
fast is the enclosed area increasing |
1 | 2553-2556 | (1)
Now
dS
dt =
2
2
(6
)
(6
)
d
d
dx
x
x
dt
dx
dt
=
⋅
(By Chain Rule)
=
32
36
12x
x
x
⋅
=
(Using (1))
Hence, when
x = 10 cm,
2
3 6 cm /s
dS
dt =
Rationalised 2023-24
APPLICATION OF DERIVATIVES
149
Example 3 A stone is dropped into a quiet lake and waves move in circles at a speed
of 4cm per second At the instant, when the radius of the circular wave is 10 cm, how
fast is the enclosed area increasing Solution The area A of a circle with radius r is given by A = πr2 |
1 | 2554-2557 | 6 cm /s
dS
dt =
Rationalised 2023-24
APPLICATION OF DERIVATIVES
149
Example 3 A stone is dropped into a quiet lake and waves move in circles at a speed
of 4cm per second At the instant, when the radius of the circular wave is 10 cm, how
fast is the enclosed area increasing Solution The area A of a circle with radius r is given by A = πr2 Therefore, the rate
of change of area A with respect to time t is
dA
dt =
2
2
(
)
(
)
d
d
dr
r
r
dt
dr
dt
π
=
π
⋅
= 2π r dr
dt
(By Chain Rule)
It is given that
dr
dt = 4cm/s
Therefore, when r = 10 cm,
dA
dt = 2π(10) (4) = 80π
Thus, the enclosed area is increasing at the rate of 80π cm2/s, when r = 10 cm |
1 | 2555-2558 | At the instant, when the radius of the circular wave is 10 cm, how
fast is the enclosed area increasing Solution The area A of a circle with radius r is given by A = πr2 Therefore, the rate
of change of area A with respect to time t is
dA
dt =
2
2
(
)
(
)
d
d
dr
r
r
dt
dr
dt
π
=
π
⋅
= 2π r dr
dt
(By Chain Rule)
It is given that
dr
dt = 4cm/s
Therefore, when r = 10 cm,
dA
dt = 2π(10) (4) = 80π
Thus, the enclosed area is increasing at the rate of 80π cm2/s, when r = 10 cm ANote dy
dx is positive if y increases as x increases and is negative if y decreases
as x increases |
1 | 2556-2559 | Solution The area A of a circle with radius r is given by A = πr2 Therefore, the rate
of change of area A with respect to time t is
dA
dt =
2
2
(
)
(
)
d
d
dr
r
r
dt
dr
dt
π
=
π
⋅
= 2π r dr
dt
(By Chain Rule)
It is given that
dr
dt = 4cm/s
Therefore, when r = 10 cm,
dA
dt = 2π(10) (4) = 80π
Thus, the enclosed area is increasing at the rate of 80π cm2/s, when r = 10 cm ANote dy
dx is positive if y increases as x increases and is negative if y decreases
as x increases Example 4 The length x of a rectangle is decreasing at the rate of 3 cm/minute and
the width y is increasing at the rate of 2cm/minute |
1 | 2557-2560 | Therefore, the rate
of change of area A with respect to time t is
dA
dt =
2
2
(
)
(
)
d
d
dr
r
r
dt
dr
dt
π
=
π
⋅
= 2π r dr
dt
(By Chain Rule)
It is given that
dr
dt = 4cm/s
Therefore, when r = 10 cm,
dA
dt = 2π(10) (4) = 80π
Thus, the enclosed area is increasing at the rate of 80π cm2/s, when r = 10 cm ANote dy
dx is positive if y increases as x increases and is negative if y decreases
as x increases Example 4 The length x of a rectangle is decreasing at the rate of 3 cm/minute and
the width y is increasing at the rate of 2cm/minute When x =10cm and y = 6cm, find
the rates of change of (a) the perimeter and (b) the area of the rectangle |
1 | 2558-2561 | ANote dy
dx is positive if y increases as x increases and is negative if y decreases
as x increases Example 4 The length x of a rectangle is decreasing at the rate of 3 cm/minute and
the width y is increasing at the rate of 2cm/minute When x =10cm and y = 6cm, find
the rates of change of (a) the perimeter and (b) the area of the rectangle Solution Since the length x is decreasing and the width y is increasing with respect to
time, we have
3 cm/min
dx
dt = −
and
2 cm/min
dy
dt =
(a)
The perimeter P of a rectangle is given by
P = 2 (x + y)
Therefore
dP
dt = 2
2
3
2
2
dx
dt
dy
+dt
=
− +
= −
(
)
cm/min
(b)
The area A of the rectangle is given by
A = x |
1 | 2559-2562 | Example 4 The length x of a rectangle is decreasing at the rate of 3 cm/minute and
the width y is increasing at the rate of 2cm/minute When x =10cm and y = 6cm, find
the rates of change of (a) the perimeter and (b) the area of the rectangle Solution Since the length x is decreasing and the width y is increasing with respect to
time, we have
3 cm/min
dx
dt = −
and
2 cm/min
dy
dt =
(a)
The perimeter P of a rectangle is given by
P = 2 (x + y)
Therefore
dP
dt = 2
2
3
2
2
dx
dt
dy
+dt
=
− +
= −
(
)
cm/min
(b)
The area A of the rectangle is given by
A = x y
Therefore
dA
dt = dx
dy
y
x
dt
dt
⋅
+
⋅
= – 3(6) + 10(2)
(as x = 10 cm and y = 6 cm)
= 2 cm2/min
Rationalised 2023-24
MATHEMATICS
150
Example 5 The total cost C(x) in Rupees, associated with the production of x units of
an item is given by
C(x) = 0 |
1 | 2560-2563 | When x =10cm and y = 6cm, find
the rates of change of (a) the perimeter and (b) the area of the rectangle Solution Since the length x is decreasing and the width y is increasing with respect to
time, we have
3 cm/min
dx
dt = −
and
2 cm/min
dy
dt =
(a)
The perimeter P of a rectangle is given by
P = 2 (x + y)
Therefore
dP
dt = 2
2
3
2
2
dx
dt
dy
+dt
=
− +
= −
(
)
cm/min
(b)
The area A of the rectangle is given by
A = x y
Therefore
dA
dt = dx
dy
y
x
dt
dt
⋅
+
⋅
= – 3(6) + 10(2)
(as x = 10 cm and y = 6 cm)
= 2 cm2/min
Rationalised 2023-24
MATHEMATICS
150
Example 5 The total cost C(x) in Rupees, associated with the production of x units of
an item is given by
C(x) = 0 005 x3 – 0 |
1 | 2561-2564 | Solution Since the length x is decreasing and the width y is increasing with respect to
time, we have
3 cm/min
dx
dt = −
and
2 cm/min
dy
dt =
(a)
The perimeter P of a rectangle is given by
P = 2 (x + y)
Therefore
dP
dt = 2
2
3
2
2
dx
dt
dy
+dt
=
− +
= −
(
)
cm/min
(b)
The area A of the rectangle is given by
A = x y
Therefore
dA
dt = dx
dy
y
x
dt
dt
⋅
+
⋅
= – 3(6) + 10(2)
(as x = 10 cm and y = 6 cm)
= 2 cm2/min
Rationalised 2023-24
MATHEMATICS
150
Example 5 The total cost C(x) in Rupees, associated with the production of x units of
an item is given by
C(x) = 0 005 x3 – 0 02 x2 + 30x + 5000
Find the marginal cost when 3 units are produced, where by marginal cost we
mean the instantaneous rate of change of total cost at any level of output |
1 | 2562-2565 | y
Therefore
dA
dt = dx
dy
y
x
dt
dt
⋅
+
⋅
= – 3(6) + 10(2)
(as x = 10 cm and y = 6 cm)
= 2 cm2/min
Rationalised 2023-24
MATHEMATICS
150
Example 5 The total cost C(x) in Rupees, associated with the production of x units of
an item is given by
C(x) = 0 005 x3 – 0 02 x2 + 30x + 5000
Find the marginal cost when 3 units are produced, where by marginal cost we
mean the instantaneous rate of change of total cost at any level of output Solution Since marginal cost is the rate of change of total cost with respect to the
output, we have
Marginal
cost (MC) =
0 |
1 | 2563-2566 | 005 x3 – 0 02 x2 + 30x + 5000
Find the marginal cost when 3 units are produced, where by marginal cost we
mean the instantaneous rate of change of total cost at any level of output Solution Since marginal cost is the rate of change of total cost with respect to the
output, we have
Marginal
cost (MC) =
0 005(32
)
0 |
1 | 2564-2567 | 02 x2 + 30x + 5000
Find the marginal cost when 3 units are produced, where by marginal cost we
mean the instantaneous rate of change of total cost at any level of output Solution Since marginal cost is the rate of change of total cost with respect to the
output, we have
Marginal
cost (MC) =
0 005(32
)
0 02(2 )
30
dC
x
x
dx =
−
+
When
x = 3, MC =
0 |
1 | 2565-2568 | Solution Since marginal cost is the rate of change of total cost with respect to the
output, we have
Marginal
cost (MC) =
0 005(32
)
0 02(2 )
30
dC
x
x
dx =
−
+
When
x = 3, MC =
0 015(3 )2
0 |
1 | 2566-2569 | 005(32
)
0 02(2 )
30
dC
x
x
dx =
−
+
When
x = 3, MC =
0 015(3 )2
0 04(3)
30
−
+
= 0 |
1 | 2567-2570 | 02(2 )
30
dC
x
x
dx =
−
+
When
x = 3, MC =
0 015(3 )2
0 04(3)
30
−
+
= 0 135 – 0 |
1 | 2568-2571 | 015(3 )2
0 04(3)
30
−
+
= 0 135 – 0 12 + 30 = 30 |
1 | 2569-2572 | 04(3)
30
−
+
= 0 135 – 0 12 + 30 = 30 015
Hence, the required marginal cost is ` 30 |
1 | 2570-2573 | 135 – 0 12 + 30 = 30 015
Hence, the required marginal cost is ` 30 02 (nearly) |
1 | 2571-2574 | 12 + 30 = 30 015
Hence, the required marginal cost is ` 30 02 (nearly) Example 6 The total revenue in Rupees received from the sale of x units of a product
is given by R(x) = 3x2 + 36x + 5 |
1 | 2572-2575 | 015
Hence, the required marginal cost is ` 30 02 (nearly) Example 6 The total revenue in Rupees received from the sale of x units of a product
is given by R(x) = 3x2 + 36x + 5 Find the marginal revenue, when x = 5, where by
marginal revenue we mean the rate of change of total revenue with respect to the
number of items sold at an instant |
1 | 2573-2576 | 02 (nearly) Example 6 The total revenue in Rupees received from the sale of x units of a product
is given by R(x) = 3x2 + 36x + 5 Find the marginal revenue, when x = 5, where by
marginal revenue we mean the rate of change of total revenue with respect to the
number of items sold at an instant Solution Since marginal revenue is the rate of change of total revenue with respect to
the number of units sold, we have
Marginal Revenue
(MR) =
R
6
36
d
dx =x
+
When
x = 5, MR = 6(5) + 36 = 66
Hence, the required marginal revenue is ` 66 |
1 | 2574-2577 | Example 6 The total revenue in Rupees received from the sale of x units of a product
is given by R(x) = 3x2 + 36x + 5 Find the marginal revenue, when x = 5, where by
marginal revenue we mean the rate of change of total revenue with respect to the
number of items sold at an instant Solution Since marginal revenue is the rate of change of total revenue with respect to
the number of units sold, we have
Marginal Revenue
(MR) =
R
6
36
d
dx =x
+
When
x = 5, MR = 6(5) + 36 = 66
Hence, the required marginal revenue is ` 66 EXERCISE 6 |
1 | 2575-2578 | Find the marginal revenue, when x = 5, where by
marginal revenue we mean the rate of change of total revenue with respect to the
number of items sold at an instant Solution Since marginal revenue is the rate of change of total revenue with respect to
the number of units sold, we have
Marginal Revenue
(MR) =
R
6
36
d
dx =x
+
When
x = 5, MR = 6(5) + 36 = 66
Hence, the required marginal revenue is ` 66 EXERCISE 6 1
1 |
1 | 2576-2579 | Solution Since marginal revenue is the rate of change of total revenue with respect to
the number of units sold, we have
Marginal Revenue
(MR) =
R
6
36
d
dx =x
+
When
x = 5, MR = 6(5) + 36 = 66
Hence, the required marginal revenue is ` 66 EXERCISE 6 1
1 Find the rate of change of the area of a circle with respect to its radius r when
(a)
r = 3 cm
(b)
r = 4 cm
2 |
1 | 2577-2580 | EXERCISE 6 1
1 Find the rate of change of the area of a circle with respect to its radius r when
(a)
r = 3 cm
(b)
r = 4 cm
2 The volume of a cube is increasing at the rate of 8 cm3/s |
1 | 2578-2581 | 1
1 Find the rate of change of the area of a circle with respect to its radius r when
(a)
r = 3 cm
(b)
r = 4 cm
2 The volume of a cube is increasing at the rate of 8 cm3/s How fast is the
surface area increasing when the length of an edge is 12 cm |
1 | 2579-2582 | Find the rate of change of the area of a circle with respect to its radius r when
(a)
r = 3 cm
(b)
r = 4 cm
2 The volume of a cube is increasing at the rate of 8 cm3/s How fast is the
surface area increasing when the length of an edge is 12 cm 3 |
1 | 2580-2583 | The volume of a cube is increasing at the rate of 8 cm3/s How fast is the
surface area increasing when the length of an edge is 12 cm 3 The radius of a circle is increasing uniformly at the rate of 3 cm/s |
1 | 2581-2584 | How fast is the
surface area increasing when the length of an edge is 12 cm 3 The radius of a circle is increasing uniformly at the rate of 3 cm/s Find the rate
at which the area of the circle is increasing when the radius is 10 cm |
1 | 2582-2585 | 3 The radius of a circle is increasing uniformly at the rate of 3 cm/s Find the rate
at which the area of the circle is increasing when the radius is 10 cm 4 |
1 | 2583-2586 | The radius of a circle is increasing uniformly at the rate of 3 cm/s Find the rate
at which the area of the circle is increasing when the radius is 10 cm 4 An edge of a variable cube is increasing at the rate of 3 cm/s |
1 | 2584-2587 | Find the rate
at which the area of the circle is increasing when the radius is 10 cm 4 An edge of a variable cube is increasing at the rate of 3 cm/s How fast is the
volume of the cube increasing when the edge is 10 cm long |
1 | 2585-2588 | 4 An edge of a variable cube is increasing at the rate of 3 cm/s How fast is the
volume of the cube increasing when the edge is 10 cm long 5 |
1 | 2586-2589 | An edge of a variable cube is increasing at the rate of 3 cm/s How fast is the
volume of the cube increasing when the edge is 10 cm long 5 A stone is dropped into a quiet lake and waves move in circles at the speed of
5 cm/s |
1 | 2587-2590 | How fast is the
volume of the cube increasing when the edge is 10 cm long 5 A stone is dropped into a quiet lake and waves move in circles at the speed of
5 cm/s At the instant when the radius of the circular wave is 8 cm, how fast is
the enclosed area increasing |
1 | 2588-2591 | 5 A stone is dropped into a quiet lake and waves move in circles at the speed of
5 cm/s At the instant when the radius of the circular wave is 8 cm, how fast is
the enclosed area increasing Rationalised 2023-24
APPLICATION OF DERIVATIVES
151
6 |
1 | 2589-2592 | A stone is dropped into a quiet lake and waves move in circles at the speed of
5 cm/s At the instant when the radius of the circular wave is 8 cm, how fast is
the enclosed area increasing Rationalised 2023-24
APPLICATION OF DERIVATIVES
151
6 The radius of a circle is increasing at the rate of 0 |
1 | 2590-2593 | At the instant when the radius of the circular wave is 8 cm, how fast is
the enclosed area increasing Rationalised 2023-24
APPLICATION OF DERIVATIVES
151
6 The radius of a circle is increasing at the rate of 0 7 cm/s |
1 | 2591-2594 | Rationalised 2023-24
APPLICATION OF DERIVATIVES
151
6 The radius of a circle is increasing at the rate of 0 7 cm/s What is the rate of
increase of its circumference |
1 | 2592-2595 | The radius of a circle is increasing at the rate of 0 7 cm/s What is the rate of
increase of its circumference 7 |
1 | 2593-2596 | 7 cm/s What is the rate of
increase of its circumference 7 The length x of a rectangle is decreasing at the rate of 5 cm/minute and the
width y is increasing at the rate of 4 cm/minute |
1 | 2594-2597 | What is the rate of
increase of its circumference 7 The length x of a rectangle is decreasing at the rate of 5 cm/minute and the
width y is increasing at the rate of 4 cm/minute When x = 8cm and y = 6cm, find
the rates of change of (a) the perimeter, and (b) the area of the rectangle |
1 | 2595-2598 | 7 The length x of a rectangle is decreasing at the rate of 5 cm/minute and the
width y is increasing at the rate of 4 cm/minute When x = 8cm and y = 6cm, find
the rates of change of (a) the perimeter, and (b) the area of the rectangle 8 |
1 | 2596-2599 | The length x of a rectangle is decreasing at the rate of 5 cm/minute and the
width y is increasing at the rate of 4 cm/minute When x = 8cm and y = 6cm, find
the rates of change of (a) the perimeter, and (b) the area of the rectangle 8 A balloon, which always remains spherical on inflation, is being inflated by pumping
in 900 cubic centimetres of gas per second |
1 | 2597-2600 | When x = 8cm and y = 6cm, find
the rates of change of (a) the perimeter, and (b) the area of the rectangle 8 A balloon, which always remains spherical on inflation, is being inflated by pumping
in 900 cubic centimetres of gas per second Find the rate at which the radius of
the balloon increases when the radius is 15 cm |
1 | 2598-2601 | 8 A balloon, which always remains spherical on inflation, is being inflated by pumping
in 900 cubic centimetres of gas per second Find the rate at which the radius of
the balloon increases when the radius is 15 cm 9 |
1 | 2599-2602 | A balloon, which always remains spherical on inflation, is being inflated by pumping
in 900 cubic centimetres of gas per second Find the rate at which the radius of
the balloon increases when the radius is 15 cm 9 A balloon, which always remains spherical has a variable radius |
1 | 2600-2603 | Find the rate at which the radius of
the balloon increases when the radius is 15 cm 9 A balloon, which always remains spherical has a variable radius Find the rate at
which its volume is increasing with the radius when the later is 10 cm |
1 | 2601-2604 | 9 A balloon, which always remains spherical has a variable radius Find the rate at
which its volume is increasing with the radius when the later is 10 cm 10 |
1 | 2602-2605 | A balloon, which always remains spherical has a variable radius Find the rate at
which its volume is increasing with the radius when the later is 10 cm 10 A ladder 5 m long is leaning against a wall |
1 | 2603-2606 | Find the rate at
which its volume is increasing with the radius when the later is 10 cm 10 A ladder 5 m long is leaning against a wall The bottom of the ladder is pulled
along the ground, away from the wall, at the rate of 2cm/s |
1 | 2604-2607 | 10 A ladder 5 m long is leaning against a wall The bottom of the ladder is pulled
along the ground, away from the wall, at the rate of 2cm/s How fast is its height
on the wall decreasing when the foot of the ladder is 4 m away from the wall |
1 | 2605-2608 | A ladder 5 m long is leaning against a wall The bottom of the ladder is pulled
along the ground, away from the wall, at the rate of 2cm/s How fast is its height
on the wall decreasing when the foot of the ladder is 4 m away from the wall 11 |
1 | 2606-2609 | The bottom of the ladder is pulled
along the ground, away from the wall, at the rate of 2cm/s How fast is its height
on the wall decreasing when the foot of the ladder is 4 m away from the wall 11 A particle moves along the curve 6y = x3 +2 |
1 | 2607-2610 | How fast is its height
on the wall decreasing when the foot of the ladder is 4 m away from the wall 11 A particle moves along the curve 6y = x3 +2 Find the points on the curve at
which the y-coordinate is changing 8 times as fast as the x-coordinate |
1 | 2608-2611 | 11 A particle moves along the curve 6y = x3 +2 Find the points on the curve at
which the y-coordinate is changing 8 times as fast as the x-coordinate 12 |
1 | 2609-2612 | A particle moves along the curve 6y = x3 +2 Find the points on the curve at
which the y-coordinate is changing 8 times as fast as the x-coordinate 12 The radius of an air bubble is increasing at the rate of 1
2 cm/s |
1 | 2610-2613 | Find the points on the curve at
which the y-coordinate is changing 8 times as fast as the x-coordinate 12 The radius of an air bubble is increasing at the rate of 1
2 cm/s At what rate is the
volume of the bubble increasing when the radius is 1 cm |
1 | 2611-2614 | 12 The radius of an air bubble is increasing at the rate of 1
2 cm/s At what rate is the
volume of the bubble increasing when the radius is 1 cm 13 |
1 | 2612-2615 | The radius of an air bubble is increasing at the rate of 1
2 cm/s At what rate is the
volume of the bubble increasing when the radius is 1 cm 13 A balloon, which always remains spherical, has a variable diameter 3 (2
1)
2
x + |
1 | 2613-2616 | At what rate is the
volume of the bubble increasing when the radius is 1 cm 13 A balloon, which always remains spherical, has a variable diameter 3 (2
1)
2
x + Find the rate of change of its volume with respect to x |
1 | 2614-2617 | 13 A balloon, which always remains spherical, has a variable diameter 3 (2
1)
2
x + Find the rate of change of its volume with respect to x 14 |
1 | 2615-2618 | A balloon, which always remains spherical, has a variable diameter 3 (2
1)
2
x + Find the rate of change of its volume with respect to x 14 Sand is pouring from a pipe at the rate of 12 cm3/s |
1 | 2616-2619 | Find the rate of change of its volume with respect to x 14 Sand is pouring from a pipe at the rate of 12 cm3/s The falling sand forms a cone
on the ground in such a way that the height of the cone is always one-sixth of the
radius of the base |
1 | 2617-2620 | 14 Sand is pouring from a pipe at the rate of 12 cm3/s The falling sand forms a cone
on the ground in such a way that the height of the cone is always one-sixth of the
radius of the base How fast is the height of the sand cone increasing when the
height is 4 cm |
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