Chapter 6: Applications of Derivatives

Ans: The area of a circle (A) with radius $(r)$ is $A=\pi {{r}^{2}}$.

Now, for a given radius, the change in the area of the circle is given by, 

$\frac{dA}{dr}=\frac{d}{dr}\left( \pi {{r}^{2}} \right)=2\pi r$ 

1. When $r=3cm$

$  \frac{dA}{dr}=2\pi (3) $  

$  =6\pi  $  

As a result, when the radius of the circle is 3cm, the area of the circle changes at a rate of $6\pi $ cm.

2. When $r=4cm$

$  \frac{dA}{dr}=2\pi (4) $  

$  =8\pi  $  

As a result, when the radius of the circle is 4cm, the area of the circle changes at a rate of $8\pi  $cm.

Ans:

Let $x$ be the length of a side, $v$ be the volume, and $s$ be the surface area of the cube. 

When $x$ is a function of time $t$, we have

$V={{x}^{3}}$ and $S=6{{x}^{2}}$

It is given that $\frac{dV}{dt}=8c{{m}^{3}}/s$ 

Then, by chain rule we have

$  \therefore 8=\frac{dv}{dt}=\left( {{x}^{3}} \right) $  

$  =\frac{d}{x}\left( x^3 \right).\frac{dx}{dt}=3{{x}^{2}}.\frac{dx}{dt} $  

$\Rightarrow \frac{dx}{dt}=\frac{8}{3{{x}^{2}}}$                                                                                    …… (1)

Now, $\Rightarrow \frac{ds}{dt}=\frac{d}{dt}\left( 6{{x}^{2}} \right)=\frac{d}{dx}\left( 6{{x}^{2}} \right).\frac{dx}{dt}$                                 (By chain rule)

$=12x.\frac{dx}{dt}=\frac{32}{x}$ 

Thus, when $x=12cm$,

$  \frac{dS}{dt}=\frac{32}{12}c{{m}^{2}}/s $

$  =\frac{8}{3}c{{m}^{2}}/s. $  

As a result, if the cube's edge length is $12cm$, the surface area is rising at a rate of $\frac{8}{3}c{{m}^{2}}/s.$

Ans: For the radius $(r)$, the area of a circle (A) is $A=\pi {{r}^{2}}$.

Now, for time ($t$), the rate of change of area (A) is given by,

$  \frac{dA}{dt}=\frac{d}{d}\left( \pi {{r}^{2}} \right) $  

$  =2\pi r\frac{dr}{dt} $                                                          (By chain rule)

It is given that the increase in radius of circle is,

$\frac{dr}{dt}=3cm/s$

$\frac{dA}{dt}=2\pi r\left( 3 \right)=6\pi r$ 

Thus, when $r=10cm$ 

$\frac{dA}{dt}=6\pi \left( 10 \right)=60\pi c{{m}^{2}}/s$ 

As a result, when the radius of the circle is $10cm$, the rate at which the area of the circle increases is $60\pi c{{m}^{2}}/s$.

Ans: Let $x$ be the length of a side and $v$ be the volume of the cube. Then,        $V={{x}^{3}}$ 

$\therefore \frac{dV}{dt}=3{{x}^{2}}.\frac{dx}{dt}$  (By chain rule)

It is given that,

$\frac{dx}{dt}=3cm/s$ 

$\therefore \frac{dV}{dt}=3{{x}^{2}}(3)=9{{x}^{2}}$ 

Then, when $x=10cm$,

$\frac{dV}{dt}=9{{\left( 10 \right)}^{2}}=900c{{m}^{3}}/s$  

As a result, when the edge is $10cm$ long, the volume of the cube increases at the rate of $900c{{m}^{3}}/s$.

Ans: The area of a circle (A) with radius $(r)$ is given by,

$A=\pi {{r}^{2}}$

Now, for time ($t$), the rate of change of area (A) is given by,

$  \therefore \frac{dA}{dt}=\frac{d}{dt}\left( \pi {{r}^{2}} \right) $

$  =\frac{d}{dr}\left( \pi {{r}^{2}} \right)\frac{dr}{dt}=2\pi r\frac{dr}{dt} $ (By chain rule)

It is given that $\frac{dr}{dt}=5cm/s$.

Thus, when $r=8cm$

$\frac{dA}{dt}=80\pi $

As a result, when the circular wave's radius is $8$cm, the enclosed area grows at a pace of $80\pi c{{m}^{2}}/s$.

Ans. The circumference of a circle (C) with the radius $\left( r \right)$ is given by

$C=2\pi r$.

Now, the rate of change of circumference (C) for time ($t$) is given by,

$\frac{dC}{dt}=\frac{dC}{dr}.\frac{dr}{dt}$                                                          (By chain rule)

$  =\frac{d}{dr}\left( 2\pi r \right)\frac{dr}{dt} $  

$  =2\pi .\frac{dr}{dt} $

It is given that $\frac{dr}{dt}=0.7cm/s$.

Hence, the rate of increase of the circumference is $1.4\pi cm/s$.

Ans: It is given that the length $x$ is decreasing at the rate of $5cm/\min $ and the width $y$ is increasing at the rate of $4cm/\min $i.e,

$  \frac{dx}{dt}=-5cm/\min , $  

$  \frac{dy}{dt}=4cm/\min  $  

(a) The perimeter (P) of a rectangle is given by

$  P=2\left( x+y \right) $  

$  \therefore \frac{dP}{dt}=2\left( \frac{dx}{dt}+\frac{dy}{dt} \right) $  

$  =-2cm/\min  $  

Hence, the rate of decrease in the perimeter of rectangle is $2cm/\min $.

(b) The area (A) of a rectangle is given by,

$A=x\times y$

$  \frac{dA}{dt}=\frac{dx}{dt}.y+x.\frac{dy}{dt} $

$  =-5y+4x $  

When $x=8cm$ and $y=6cm$,

$  \frac{dA}{dt}=\left( -5\times 6+4\times 8 \right)c{{m}^{2}}/\min  $  

$  =2c{{m}^{2}}/\min  $  

Hence, the rate of increase in the area of the rectangle is $2c{{m}^{2}}/\min $.

Ans: The volume of a sphere (V) with radius ($r$) is given by,

$V=\frac{4}{3}\pi {{r}^{3}}$ 

For time ($t$), the rate of change of volume (V) is given by,

$\frac{dV}{dt}=\frac{dV}{dr}.\frac{dr}{dt}$                                                   (By chain rule)

$  =\frac{d}{dr}\left( \frac{4}{3}\pi {{r}^{3}} \right).\frac{dr}{dt} $  

$  =4\pi {{r}^{2}}.\frac{dr}{dt} $  

It is given that 

$  \frac{dv}{dt}=900c{{m}^{3}}/s $  

$  \therefore 900=4\pi {{r}^{2}}.\frac{dr}{dt} $  

$  \Rightarrow \frac{dr}{dt}=\frac{900}{4\pi {{r}^{2}}} $  

$  =\frac{225}{\pi {{r}^{2}}} $  

Therefore, when the radius is $15$cm,

$\frac{dr}{dt}=\frac{225}{\pi {{\left( 15 \right)}^{2}}}=\frac{1}{\pi }$ 

As a result, the rate at which the balloon's radius increases when the radius is $15cm$ is $\frac{1}{\pi }cm/\sec $.

Ans: The volume of a sphere (V) with radius ($r$) is given by,

$V=\frac{4}{3}\pi {{r}^{3}}$ 

Rate of change of volume (V) for time ($t$) is given by,

$\frac{dV}{dr}=\frac{d}{dr}\left( \frac{4}{3}\pi {{r}^{3}} \right) $  

$=4\pi {{r}^{2}} $  

Therefore, when the radius is $10cm$,

$  \frac{dV}{dr}=4\pi {{\left( 10 \right)}^{2}} $  

$  =400\pi  $  

Hence, the rate of increase in the volume of the balloon is $400\pi c{{m}^{3}}/\sec $.

Ans: Let the height of the wall be $y$m and the foot of the ladder be $x$m away from the wall.

Then, by Pythagoras Theorem, we have:

${{x}^{2}}+{{y}^{2}}=25$  (Length of the ladder is $5m$)

$\Rightarrow y=\sqrt{25-{{x}^{2}}}$ 

Then, the rate of change of height $y$for time $t$ is given by,

$\frac{dy}{dt}=\frac{-x}{\sqrt{25-{{x}^{2}}}}.\frac{dx}{dt}$ 

It is given that $\frac{dx}{dt}=2cm/s$ 

$\therefore \frac{dy}{dt}=\frac{-2x}{\sqrt{25-{{x}^{2}}}}$ 

Now, when $x=4m$ we have:

$\frac{dy}{dt}=\frac{8}{3}$ 

Hence, the rate of decrease in the height of the ladder on the wall is    $\frac{8}{2}cm/\sec $.