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$\\$ $(a) r = 3 cm (b) r= 4 cm$

Solution :

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 :

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 $\dfrac{8}{3} cm/s$