In this page we have *NCERT Solutions for Class 12 Maths Chapter 6: Application of Derivatives* for
EXERCISE 6.1 . Hope you like them and do not forget to like , social share
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Find the rate of change of the area of a circle with respect to its radius

i) a = 3cm,

ii) a = 4cm

Area of circle with radius ‘r’ is given by

A = $\pi r^{2}$

The rate of change of area w.r.t ‘r’ is,

$\frac{\mathrm{d} A}{\mathrm{d} r} = \frac{\mathrm{d} (\pi r^{2})}{\mathrm{d} r} = 2\pi r$

At r = 3cm

$\frac{\mathrm{d} A}{\mathrm{d} r} = 2\pi (3) = 6\pi$

Thus, $6\pi(cm^{2} / cm)$ is the rate of change of area of circle, when a = 3cm.

At r = 4cm

$\frac{\mathrm{d} A}{\mathrm{d} r} = 2\pi (4) = 8\pi$

Thus, $8\pi(cm^{2} /cm )$ is the rate of change of area of circle, when a = 4cm.

The volume of a cube is increasing at the rate of 8 cm

Let’s assume the length of edge is y

V = $y^{3}$

S = $y^{2}$

y is a function of time x.

Given that

$\frac{\mathrm{d} V}{\mathrm{d} x} = 8(cm^{3}/s)$.

Applying chain rule, we get:

$8 = \frac{\mathrm{d} V}{\mathrm{d} x} = \frac{\mathrm{d} (y^{3})}{\mathrm{d} x} = \frac{\mathrm{d}( y^{3})}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 3y^{2}*\frac{\mathrm{d} y}{\mathrm{d} x}$

$\frac{\mathrm{d} y}{\mathrm{d} x} = 8/3y^{2}$……………..(a)

Applying chain rule on surface area, we get

$\frac{\mathrm{d} S}{\mathrm{d} x} = \frac{\mathrm{d} (6y^{2})}{\mathrm{d} x} = \frac{\mathrm{d} (6y^{2})}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 12y*\frac{\mathrm{d} y}{\mathrm{d} x}$

$=12y( 8/3y^{2}) = 32/y$

Here its is given that y = 18cm.

$ \frac{\mathrm{d} S}{\mathrm{d} x} = 32/12 = 8/3 cm^{2}/s$

Thus, the surface area of a circle is increasing at the rate of $8/3cm^{2}/s$ when the edge length is 12cm.

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.

We know that the area of circle with radius ‘y’ is $\pi y^{2}$.

i.e. A = $\pi y^{2}$

The rate of change of area w.r.t time ‘x’ is,

Applying chain rule in above equation, we get

$\frac{\mathrm{d} A}{\mathrm{d} x} = \frac{\mathrm{d} (\pi y^{2})}{\mathrm{d} x} = \frac{\mathrm{d}(\pi y^{2}) }{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 2\pi y\frac{\mathrm{d} y}{\mathrm{d} x}$

Here, given that, $\frac{\mathrm{d} y}{\mathrm{d} x} = 3cm/s$

$ \frac{\mathrm{d} A}{\mathrm{d} x} = 2\pi y(3) = 6\pi y$

Also given that radius y = 10cm

$ \frac{\mathrm{d} A}{\mathrm{d} x} = 6\pi (10) = 60cm^{2}/s$

Thus, $60cm^{2} /s$ is the rate of change of area of circle, when y = 10cm.

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?

Let’s assume the length of edge is y

So, V = $y^{3}$ ; y is function of time ‘x’.

Then, By applying chain rule we get,

$ \frac{\mathrm{d} V}{\mathrm{d} x} = 3y^{2}*\frac{\mathrm{d} y}{\mathrm{d} x}$

It is given in question

$\frac{\mathrm{d} y}{\mathrm{d} x} = 3cm/s$

So we can say that

$ \frac{\mathrm{d} V}{\mathrm{d} x} = 3y^{2}(3) = 9y^{2}$

Also given that edge length is y = 10cm

$ \frac{\mathrm{d} V}{\mathrm{d} x} = 9(10^{2}) = 900cm^{3}/s$

Thus, the Volume of a variable cube is increasing at the rate of $900cm^{3}/s$ when the edge length is 10cm.

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?

We know that the area of circle with radius ‘y’ is $\pi y^{2}$.

i.e. A = $\pi y^{2}$

The rate of change of area w.r.t time ‘x’ is,

Applying chain rule in above equation, we get

$\frac{\mathrm{d} A}{\mathrm{d} x} = \frac{\mathrm{d} (\pi y^{2})}{\mathrm{d} x} = \frac{\mathrm{d}(\pi y^{2}) }{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 2\pi y\frac{\mathrm{d} y}{\mathrm{d} x}$

Here, given that, $\frac{\mathrm{d} y}{\mathrm{d} x} = 5cm/s$

$ \frac{\mathrm{d} A}{\mathrm{d} x} = 2\pi y(5) =10\pi y$

Now we need to find the increase at radius y = 8cm

$ \frac{\mathrm{d} A}{\mathrm{d} x} = 10\pi (8) = 80cm^{2}/s$

The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference

We know that the circumference of circle with radius ‘y’ is

C = $2\pi y$

The rate of change of circumference w.r.t time ‘x’ is obtained by applying chain rule,

$\frac{\mathrm{d} C}{\mathrm{d} x} = \frac{\mathrm{d} (2\pi y)}{\mathrm{d} x} = \frac{\mathrm{d} (2\pi y)}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 2\pi \frac{\mathrm{d} y}{\mathrm{d} x}$

Given $\frac{\mathrm{d} y}{\mathrm{d} x} = .7 cm/s$

Therefore

$ \frac{\mathrm{d} C}{\mathrm{d} x} = 2\pi (0.7) = 1.4cm/s$

The length

the rates of change of (a) the perimeter, and (b) the area of the rectangle

Let a and b are the length and width of the rectangle

Then given in question

$\frac{\mathrm{d} a}{\mathrm{d} x} = -5cm/min\; and \; \frac{\mathrm{d} b}{\mathrm{d} x} = 4cm/min$

Perimeter of rectangle, P = 2(a + b)

The rate of change of perimeter w.r.t time ‘x’ is,

$ \frac{\mathrm{d} P}{\mathrm{d} x} = 2(\frac{\mathrm{d} a}{\mathrm{d} x} + \frac{\mathrm{d} b}{\mathrm{d} x}) = 2(-5 +4) = -2cm/min$

Thus, the perimeter of a rectangle is decreasing at 4cm/min.

Area of rectangle, A = a*b

The rate of change of area w.r.t time ‘x’ is,

$ \frac{\mathrm{d} A}{\mathrm{d} x} = \frac{\mathrm{d} a}{\mathrm{d} x}*b + \frac{\mathrm{d} a}{\mathrm{d} x}*a = -5b + 4a$

Here, when a = 8cm and b = 6cm,

$ \frac{\mathrm{d} A}{\mathrm{d} x} = -5(6) +4(8) = -2cm^{2}/min$

Thus, the area of rectangle is decreasing at $12cm^{2}/min$.

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.

As we know the shape of a balloon is spherical.

The volume(V) of Balloon is given by,

$V = \frac{4}{3}\pi y^{3}$

By applying chain rule in above equation by differentiating w.r.t time ‘x’, we get

$\frac{\mathrm{d} V}{\mathrm{d} x} = \frac{\mathrm{d} (\frac{4}{3}\pi y^{3})}{\mathrm{d} x} = \frac{\mathrm{d} (\frac{4}{3}\pi y^{3})}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 4\pi y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}$

Given in question , $\frac{\mathrm{d} V}{\mathrm{d} x} = 900cc/s$ (given)

$ 900 = 4\pi y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}$

$ \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{900}{4\pi y^{2}} = \frac{225}{\pi y^{2}}$

Here radius y = 15cm (given)

$ \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{225}{\pi 15^{2}} = \frac{1}{\pi }cm/s$

Thus, radius of football is increasing at$\frac{1}{\pi }cm/s$ when radius y = 15cm .

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.

The volume(V) of sphere is given by,

$V = \frac{4}{3}\pi x^{3}$

By applying chain rule in above equation by differentiating w.r.t radius ‘x’, we get

$\frac{\mathrm{d} V}{\mathrm{d} x} = \frac{\mathrm{d} (\frac{4}{3}\pi x^{3})}{\mathrm{d} x} = 4\pi x^{2}$

We need to find the increase at r=10 cm

$ \frac{\mathrm{d} V}{\mathrm{d} x} = 4\pi (10)^{2} = 400\pi \;cm^{3}/s$

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 ?

Let a be the distance from bottom of ladder to wall and b be the height of the wall where the ladder is in contact with the wall.

Thus, we can apply Pythagoras theorem in this case, then we get

$a^{2}+ b^{2} = 25$

( As ladder is of 5m length)

$ b = \sqrt{25 – a^{2}}$

By differentiating above equation w.r.t time ‘x’ we will get rate of change of height,

$\frac{\mathrm{d} b}{\mathrm{d} x} = \frac{-a}{\sqrt{25 – a^{2}}}*\frac{\mathrm{d} a}{\mathrm{d} x}$

Here we have $\frac{\mathrm{d} a}{\mathrm{d} x} = 2cm/s$

$ \frac{\mathrm{d} b}{\mathrm{d} x} = \frac{-2a}{\sqrt{25 – a^{2}}}$

Here, the foot of the ladder is 4m far from the wall, a = 4m

$ \frac{\mathrm{d} b}{\mathrm{d} x} = \frac{-2(4)}{\sqrt{25 – (4)^{2}}} = \frac{8}{3} = \frac{-8}{3}$

Thus, the height of the Ladder is decreasing at the rate of (6/3)cm/s.

A particle moves along the curve 6

$6y = x^{3}+ 2$

By differentiating above equation w.r.t. time ‘t, we get

$6\frac{\mathrm{d} y}{\mathrm{d} t} = 3x^{2}\frac{\mathrm{d} x}{\mathrm{d} t} + 0$

$2\frac{\mathrm{d} y}{\mathrm{d} x} = x^{2}\frac{\mathrm{d} x}{\mathrm{d} t}$………………………….(1)

Given

i.e. $\frac{1}{8}\frac{\mathrm{d} y}{\mathrm{d} t} = \frac{\mathrm{d} x}{\mathrm{d} t}$

$ \frac{\mathrm{d} y}{\mathrm{d} t} = 8\frac{\mathrm{d} x}{\mathrm{d} t}$

Putting this value in equation (1)

$ 2(8\frac{\mathrm{d} x}{\mathrm{d} t}) =x^{2} \frac{\mathrm{d} x}{\mathrm{d} t}$ $ 16\frac{\mathrm{d} x}{\mathrm{d} t} = a^{2} \frac{\mathrm{d} x}{\mathrm{d} t}$

$ (x^{2} – 16 ) \frac{\mathrm{d} x}{\mathrm{d} t} = 0$

$ x^{2} = 16$

$ x = \pm 4$

Now, for x = 4

$ y=11$

For, x = -4

$ y=-31/3$

Thus, the required co-ordinate or points are $\left ( -4,\frac{-31}{3} \right )\; and \left ( 4,11 \right )$.

The radius of an air bubble is increasing at the rate of ½ cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

As we know the shape of an air bubble is spherical.

The volume(V) of sphere with radius ‘y’ is given by,

$V = \frac{4}{3}\pi y^{3}$

By applying chain rule in above equation by differentiating w.r.t time ‘x’, we get

$\frac{\mathrm{d} V}{\mathrm{d} x} = \frac{\mathrm{d} (\frac{4}{3}\pi y^{3})}{\mathrm{d} x} = \frac{\mathrm{d} (\frac{4}{3}\pi y^{3})}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 4\pi y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}$

Here, given that,

$\frac{\mathrm{d} y}{\mathrm{d} x} = 1/2cm/s$

$ \frac{\mathrm{d} V}{\mathrm{d} x} =2\pi y^{2}cm^{3}/s$

Also given that the radius y = 1cm,

$ \frac{\mathrm{d} V}{\mathrm{d} x} = 2\pi (1)^{2} = 2\pi cm^{3}/s$

A balloon, which always remains spherical, has a variable diameter (3/2)(2x+1) Find the rate of change of its volume with respect to

We know the shape of a balloon is spherical.

The volume(V) of sphere with radius ‘y’ is given by,

$V = \frac{4}{3}\pi y^{3}$

Here it is given that y = $\frac{3}{2}(2x + 1)$

$ V = \frac{4}{3}\pi (\frac{3}{2}(2x + 1))^{3} = \frac{9}{16}\pi (4a + 1)^{3}$

Differentiating above equation w.r.t ‘x’, we get,

$ \frac{\mathrm{d} V}{\mathrm{d} x} = \frac{\mathrm{d} \frac{9}{16}\pi(2x + 1)^{3}}{\mathrm{d} x} = \frac{27}{8}\pi (2x + 1)^{2}$

Sand is pouring from a pipe at the rate of 12 cm

Let V be the volume of cone, x be the radius of base of cone and y be the height of the cone which is increasing.

So, volume of cone is given by,

$V = \frac{1}{3}\pi x^{2}y$

Given , y = (1/6)x or 6y=x

$ V = \frac{1}{3}\pi (6y)^{2}y = 12\pi y^{3}$

Now, the rate of increase in the height of cone is obtained by differentiating above equation w.r.t. time ‘t’

$\frac{\mathrm{d} V}{\mathrm{d} t} = 12\pi \frac{\mathrm{d} y^{3}}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} t} = 36\pi y^{2}\frac{\mathrm{d} y}{\mathrm{d} t}$

Here, $\frac{\mathrm{d} V}{\mathrm{d} t} = 24cm^{3}/s$

$ 12 = 36\pi y^{2}\frac{\mathrm{d} y}{\mathrm{d} t}$

Also the y = 4cm is given,

$ 12 = 36\pi (4)^{2}\frac{\mathrm{d} y}{\mathrm{d} t}$

$ \frac{\mathrm{d} y}{\mathrm{d} t} = \frac{1}{48\pi }$

Thus, the height of wheat flour cone is increasing at the rate of $\frac{1}{48\pi } cm/s$.

The total cost C (

C(

Find the marginal cost when 17 units are produced.

MC = $\frac{\mathrm{d}C }{\mathrm{d} x} = \frac{\mathrm{d} (0.007x^{3} – 0.003x^{2} + 15x + 4000)}{\mathrm{d} x} = 0.021x^{2} – 0.006x +15$

Here, a = 17 is given.

$ MC = \frac{\mathrm{d}C }{\mathrm{d} a} = 0.021x^{2} – 0.006x +15 = 0.021(17)^{2} – 0.003(17) + 15 = 20.967$

Thus, marginal cost for 17 units manufactured is Rs. 20.967.

The total revenue in Rupees received from the sale of

R(

Find the marginal revenue when

Marginal Revenue(MR) is defined as the rate of change of total revenue generated w.r.t. numbers of units sold.

$MR = \frac{\mathrm{d} R}{\mathrm{d} x} = \frac{\mathrm{d} (13x^{2} + 26x + 15)}{\mathrm{d} x} = 26x + 26$

Here, x = 7 units (given)

MR = 26(7) + 26 = 208

The rate of change of the area of a circle with respect to its radius

(A) 10p

(B) 12p

(C) 8p

(D) 11p

We know that the area of circle with radius ‘r’ is $\pi r^{2}$.

A = $\pi r^{2}$

The rate of change of area w.r.t ‘r’ is,

$\frac{\mathrm{d} A}{\mathrm{d} r} = \frac{\mathrm{d} (\pi r^{2})}{\mathrm{d} r} = 2\pi r$

Here r = 6cm is given,

$ \frac{\mathrm{d} A}{\mathrm{d} a} = 2\pi (6) = 12\pi cm^{2}/s$

Thus, Option (B) is correct.

The total revenue in Rupees received from the sale of

R(

The marginal revenue, when

(A) 116

(B) 96

(C) 90

(D) 126

Marginal Revenue(MR) is defined as the rate of change of total revenue generated w.r.t. numbers of units sold.

$MR = \frac{\mathrm{d} R}{\mathrm{d}xy} = \frac{\mathrm{d} (3x^{2} + 36x + 5)}{\mathrm{d} x} = 6x + 36$

Here, x = 15 units (given)

MR = 6(15) + 36 = 126

Thus, marginal revenue for 15 units sold is Rs.126

Thus, option (D) is correct.

Class 12 Maths Class 12 Physics