Chapter 6
APPLICATION OF
DERIVATIVES
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.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. Finally, we use the derivative to find
approximate value of certain quantities.
6.2 Rate of Change of Quantities
ds
Recall that by the derivative , we mean the rate of change of distance s with
dt
respect to the time t. In a similar fashion, whenever one quantity y varies with another
dy
quantity x, satisfying some rule y = f ( x ) , then (or f ′(x)) represents the rate of
dx
dy
change of y with respect to x and dx (or f ′(x0)) represents the rate of change
x = x0
of y with respect to x at x = x0 .
Further, if two variables x and y are varying with respect to another variable t, i.e.,
if x = f (t ) and y = g (t ) , then by Chain Rule
dy dy dx dx
= , if ≠0
dx dt dt dt
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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. Therefore, the rate
dA d
of change of the area A with respect to its radius r is given by = ( π r 2 ) = 2π r .
dr dr
dA
When r = 5 cm, = 10π . Thus, the area of the circle is changing at the rate of
dr
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. Then, V = x3 and S = 6x2, where x is a function of time t.
dV
Now = 9cm3/s (Given)
dt
dV d 3 d dx
Therefore 9= = ( x ) = ( x3 ) ⋅ (By Chain Rule)
dt dt dx dt
dx
= 3x ⋅
2
dt
dx 3
or = 2 ... (1)
dt x
dS d d dx
Now = (6 x 2 ) = (6 x 2 ) ⋅ (By Chain Rule)
dt dt dx dt
3 36
= 12x ⋅ 2 = (Using (1))
x x
dS
Hence, when x = 10 cm, = 3.6 cm2 /s
dt
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, 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 d d dr dr
= (π r 2 ) = (π r 2 ) ⋅ = 2π r (By Chain Rule)
dt dt dr dt dt
dr
It is given that = 4cm/s
dt
dA
Therefore, when r = 10 cm, = 2π (10) (4) = 80π
dt
Thus, the enclosed area is increasing at the rate of 80π cm2/s, when r = 10 cm.
dy
A Note
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
dx dy
= −3 cm/min and = 2 cm/min
dt dt
(a) The perimeter P of a rectangle is given by
P = 2 (x + y)
dP dx dy
Therefore = 2 + = 2 ( −3 + 2) = −2 cm/min
dt dt dt
(b) The area A of the rectangle is given by
A=x . y
dA dx dy
Therefore = ⋅ y + x⋅
dt dt dt
= – 3(6) + 10(2) (as x = 10 cm and y = 6 cm)
= 2 cm2/min
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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
dC
Marginal cost (MC) = = 0.005(3x 2 ) − 0.02(2 x ) + 30
dx
x = 3, MC = 0.015(3 ) − 0.04(3) + 30
2
When
= 0.135 – 0.12 + 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. 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
dR
Marginal Revenue (MR) = = 6 x + 36
dx
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. 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. 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?
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?
Reprint 2024-25
APPLICATION OF
DERIVATIVES
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.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. Finally, we use the derivative to find
approximate value of certain quantities.
6.2 Rate of Change of Quantities
ds
Recall that by the derivative , we mean the rate of change of distance s with
dt
respect to the time t. In a similar fashion, whenever one quantity y varies with another
dy
quantity x, satisfying some rule y = f ( x ) , then (or f ′(x)) represents the rate of
dx
dy
change of y with respect to x and dx (or f ′(x0)) represents the rate of change
x = x0
of y with respect to x at x = x0 .
Further, if two variables x and y are varying with respect to another variable t, i.e.,
if x = f (t ) and y = g (t ) , then by Chain Rule
dy dy dx dx
= , if ≠0
dx dt dt dt
Reprint 2024-25
,148 MATHEMATICS
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. Therefore, the rate
dA d
of change of the area A with respect to its radius r is given by = ( π r 2 ) = 2π r .
dr dr
dA
When r = 5 cm, = 10π . Thus, the area of the circle is changing at the rate of
dr
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. Then, V = x3 and S = 6x2, where x is a function of time t.
dV
Now = 9cm3/s (Given)
dt
dV d 3 d dx
Therefore 9= = ( x ) = ( x3 ) ⋅ (By Chain Rule)
dt dt dx dt
dx
= 3x ⋅
2
dt
dx 3
or = 2 ... (1)
dt x
dS d d dx
Now = (6 x 2 ) = (6 x 2 ) ⋅ (By Chain Rule)
dt dt dx dt
3 36
= 12x ⋅ 2 = (Using (1))
x x
dS
Hence, when x = 10 cm, = 3.6 cm2 /s
dt
Reprint 2024-25
, 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 d d dr dr
= (π r 2 ) = (π r 2 ) ⋅ = 2π r (By Chain Rule)
dt dt dr dt dt
dr
It is given that = 4cm/s
dt
dA
Therefore, when r = 10 cm, = 2π (10) (4) = 80π
dt
Thus, the enclosed area is increasing at the rate of 80π cm2/s, when r = 10 cm.
dy
A Note
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
dx dy
= −3 cm/min and = 2 cm/min
dt dt
(a) The perimeter P of a rectangle is given by
P = 2 (x + y)
dP dx dy
Therefore = 2 + = 2 ( −3 + 2) = −2 cm/min
dt dt dt
(b) The area A of the rectangle is given by
A=x . y
dA dx dy
Therefore = ⋅ y + x⋅
dt dt dt
= – 3(6) + 10(2) (as x = 10 cm and y = 6 cm)
= 2 cm2/min
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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
dC
Marginal cost (MC) = = 0.005(3x 2 ) − 0.02(2 x ) + 30
dx
x = 3, MC = 0.015(3 ) − 0.04(3) + 30
2
When
= 0.135 – 0.12 + 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. 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
dR
Marginal Revenue (MR) = = 6 x + 36
dx
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. 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. 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?
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?
Reprint 2024-25
