# 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?

The circumference of a circle $(C )$ with radius $(r )$ is given by $C =2 nr .$$\\ Therefore, the rate of change of circumference (C) with respect to time ( t ) is given by,\\ \dfrac{\mathrm{d}C}{\mathrm{d}t} =\dfrac{\mathrm{d}C}{\mathrm{d}r}.\dfrac{\mathrm{d}r}{\mathrm{d}t} \ \ \ \ \ \ (\text{by chain rule})\\ =\dfrac{\mathrm{d}}{\mathrm{d}r}(2\pi r)\dfrac{\mathrm{d}r}{\mathrm{d}t}\\ =2 \pi .\dfrac{\mathrm{d}r}{\mathrm{d}t}$$\\$ It is given that $\dfrac{\mathrm{d}r}{\mathrm{d}t}=0.7 cm /s$$\\ Hence, the rate of increase of the circumference is \\ 2\pi(0.7) =1.4 \pi cm/s 7 The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x= 8 cm and y = 6 cm , find the rates of change of (a ) the perimeter, and ( b ) the area of the rectangle. ##### Solution : Since the length ( x ) is decreasing at the rate of 5 cm / minute and the width y is increasing at the rate of 4 cm / minute , we have:\\ \dfrac{\mathrm{d}t}{\mathrm{d}y}=-5 cm / min and \dfrac{\mathrm{d}y}{\mathrm{d}t} =4 cm /min$$\\$ (a) The perimeter (P ) of a rectangle is given by, $P = 2 ( x + y )\\ \therefore \dfrac{\mathrm{d}P}{\mathrm{d}t}=2(\dfrac{\mathrm{d}x}{\mathrm{d}t}+\dfrac{\mathrm{d}y}{\mathrm{d}t}) \\ =2(-5+4 )=-2 cm/ min$ Hence, the perimeter is decreasing at the rate of $2 cm/min,$$\\ \\ (b) The area ( A ) of a rectangle is given by, A = x * y\\ \dfrac{\mathrm{d}A}{\mathrm{d}t} =\dfrac{\mathrm{d}x}{\mathrm{d}t}.y+x.\dfrac{\mathrm{d}y}{\mathrm{d}t}=-5y+4x$$\\$ When $x=8 cm$ and $y=6 cm,\dfrac{\mathrm{d}A}{\mathrm{d}t} =(-5*6+4*8) cm^2 / min =2 cm^2 /min$$\\ Hence, the area of the rectangle is increasing at the rate of 2 cm^ 2 /min . 8 A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm. ##### Solution : The volume of a sphere ( V ) with radius (r ) is given by,\\ V=\dfrac{4}{3} \pi r^3$$\\$ $\therefore$ Rate of change of volume ($V$) with respect to time ($t$) is given by , $\\$ $\dfrac{\mathrm dV}{\mathrm dt}=\dfrac{\mathrm dV}{\mathrm dr}.\dfrac{ \mathrm dr}{\mathrm dt} \quad \text{(by chain rule)}\\ =\dfrac{\mathrm d}{\mathrm dr}(\dfrac{4}{3}\pi r^3).\dfrac{\mathrm dr}{\mathrm dt}\\ =4\pi r^2.\dfrac{\mathrm dr}{\mathrm dt}$$\\ It is given that\\ \dfrac{\mathrm dv}{\mathrm dt}=900 cm^3/s\\ \therefore 900 = 4 \pi r^2.\dfrac{\mathrm dr}{\mathrm dt}\\ \Rightarrow \dfrac{\mathrm dr}{\mathrm dt}=\dfrac{900}{4 \pi r^2}=\dfrac{225}{\pi r^2}$$\\$ Therefore,where radius =$15 cm$$\\ \dfrac{\mathrm dr}{\mathrm dt}=\dfrac{225}{\pi(15)^2}=\dfrac{1}{\pi}$$\\$ Hence, the rate at which the radius of the balloon increases when the radius is $15 cm$ is $\dfrac{1}{\pi} cm/s$

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 $10 cm.$

##### Solution :

The volume of a sphere ( v ) with radius (r ) is given by$\\$ $V=\dfrac{4}{3}\pi r^2$$\\ Rate of change of volume (v) with respect to its radius (r) is given by ,\\ \dfrac{\mathrm dV}{\mathrm dr} =\dfrac{\mathrm d}{\mathrm dr}(\dfrac{4}{3}\pi r^3)=\dfrac{4}{3}\pi (3r^2)=4\pi r^2$$\\$ Therefore, when radius $= 10 cm,$$\\ \dfrac{\mathrm d V}{\mathrm dr} =4 \pi (10)^2=400 \pi$$\\$ Hence, the volume of the balloon is increasing at the rate of $400 \pi cm^3/s$

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 $2 cm/s$. How fast is its height on the wall decreasing when the foot of the ladder is $4 m$ away from the wall?

##### Solution :

Let $y \ m$ be the height of the wall at which the ladder touches. Also, let the foot of the ladder be $x \ m$ away from the wall.$\\$ Then, by Pythagoras Theorem, we have:$\\$ $x^2+y^2=25 \qquad \quad \text{(Length of the ladder= 5m)}\\ \Rightarrow y=\sqrt{25-x^2}$$\\ Then, the rate of change of height ( y) with respect to time (t ) is given by,\\ \dfrac{\mathrm dy}{\mathrm dt}=\dfrac{-x}{\sqrt{25-x^2}}.\dfrac{\mathrm dx}{\mathrm dt}$$\\$ It is given that $\dfrac{\mathrm dx}{\mathrm dt}=2 cm/s$$\\ \therefore \dfrac{\mathrm dy}{\mathrm dt}=\dfrac{-2x}{\sqrt{25-x^2}}$$\\$ Now,when $x=4m,$ we have:$\\$ $\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{-2*4}{\sqrt{25-4^2}}=\dfrac{8}{3}$$\\$ Hence, the height of the ladder on the wall is decreasing at the rate of $\dfrac{8}{3} cm/s$