# Application of Derivatives

## Class 12 NCERT

### NCERT

1   Find the rate of change to the area of a circle with respect to its radius $r$ when$\\$ $(a) r = 3 cm (b) r= 4 cm$

The area of a circle $( A )$ with radius $( r )$ is given by, $A =\pi r ^2$$\\ Now, the area of the circle is changing of the area with respect to its radius is given by,\\ \dfrac{dA}{dr}=\dfrac{d}{dr}(\pi r^2) =2 \pi r$$\\$ 1.When $r=3cm$,$\dfrac{dA}{dr}=2\pi (3)=6\pi$$\\ Hence, the area of the circle is changing at the rate of 6\pi cm when its radius is 3 cm .\\ 2. When r = 4 cm ,$$\dfrac{dA}{dr} =2\pi (4)=8 \pi$$\\ Hence, the area of the circle is changing at the rate of 8\pi cm when its radius is 4 cm . 2 The volume of a cube is increasing at the rate of 8 cm^ 3 /s. How fast is the surface area increasing when the length of an edge is 12 cm? ##### Solution : Let x be the length of a side, v be the volume, and s be the surface area of the cube. Then, V = x^ 3 and S =6 x^ 2 when x is a function of time t .\\ It is given that \dfrac{dv}{dt}=8 cm^3/s$$\\$ Then, by using the chain rule, we have:$\\$ $\therefore 8=\dfrac{dv}{dt}=(x^3)=\dfrac{d}{x}(x).\dfrac{dx}{dt}=3x^2.\dfrac{dx}{dt}\\ \implies \dfrac{dx}{dt}=\dfrac{8}{3x^2} ............(1)$$\\ Now, \implies \dfrac{ds}{dt}=\dfrac{d}{dt}(6x^2)=\dfrac{d}{dt}(6x^2).\dfrac{d}{dt} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [By chain rule]\\ =12x.\dfrac{dx}{dt}=12x.(\dfrac{8}{3x^2})=\dfrac{32}{x}$$\\$ Thus,when $x=12 cm,\dfrac{dS}{dt}=\dfrac{32}{12}cm^2/s=\dfrac{8}{3}cm^2/s$$\\ Hence, if the length of the edge of the cube is 12 cm , then the surface area is increasing at the rate of \dfrac{8}{3} cm^2/s 3 The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm. ##### Solution : The area of a circle ( A ) with radius ( r ) is given by,\\ A=\pi r^2$$\\$ Now, the rate of change of area (A ) with respect to time ( t ) is given by,$\\$ $\dfrac{dA}{dt}=\dfrac{d}{d}(\pi r^2).\dfrac{dr}{dx}=2\pi r\dfrac{dr}{dt}$ [By chain rule]$\\$ It is given that,$\\$ $\dfrac{dr}{dt}=3 cm/s\\ \therefore \dfrac{dA}{dt}=2\pi r(3)=6 \pi r$$\\ Thus , when r=10 cm,\\ \dfrac{dA}{dt}=6\pi (10)=60\pi cm^2/s$$\\$ Hence, the rate at which the area of the circle is increasing when the radius is $10 cm$ is $60 \pi cm ^2 / s .$

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?

##### Solution :

Let $x$ be the length of a side and $v$ be the volume of the cube. Then, $V = x^ 3\\ \therefore \dfrac{dV}{dt}=3x^2.\dfrac{dx}{dt}$ $\ \ \ \ \ \ \ \ \$(by chain rule) $\\$ It is given that,$\\$ $\dfrac{dV}{dt}=3x^2(3)=9x^2$$\\ Then, when x =10 cm ,\\ \dfrac{dV}{dt}=9(10)^2=900 cm^3/s$$\\$ Hence, the volume of the cube is increasing at the rate of $900 cm ^3 /s$ 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?

##### Solution :

The area of a circle (A) with radius (r) is given by $A =\pi r^ 2$$\\ Therefore, the rate of change of area ( A ) with respect to time (t ) is given by, \\ \therefore \dfrac{dA}{dt}=\dfrac{d}{dt}(\pi r^2)=\dfrac{d}{dr}(\pi r^2)\dfrac{dr}{dt}=2 \pi r\dfrac{dr}{dt} \ \ \ \ \ \ \ \ \ \ (by chain rule)\\ Thus, when r = 8 cm ,\\ \dfrac{dA}{dt}=2\pi(8)(5)=80\pi$$\\$ Hence, when the radius of the circular wave is $8 cm$, the enclosed area is increasing at the rate of $80 \pi cm ^2 / s .$

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?

##### Solution :

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

##### Solution :

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