Chapter 6: Applications of Derivatives

Ans: Given the equation of the curve as $6y={{x}^{3}}+2$.

The rate of change of the position of the particle for time

(t)  is given by,

$  6\frac{dy}{dt}=3{{x}^{2}}\frac{dx}{dt}+0 $

$  \Rightarrow 2\frac{dy}{dt}={{x}^{2}}\frac{dx}{dt} $

When the particle's y-coordinate changes $8$ times as fast as its x-coordinate i.e., $\left( \frac{dy}{dt}=8\frac{dx}{dt} \right)$ , we have:

$  2\left( 8\frac{dx}{dt} \right)={{x}^{2}}\frac{dx}{dt} $  

$  \Rightarrow 16\frac{dx}{dt}={{x}^{2}}\frac{dx}{dt} $

$  \Rightarrow \left( {{x}^{2}}-16 \right)\frac{dx}{dt}=0 $  

$  \Rightarrow {{x}^{2}}=16 $  

$  \Rightarrow x=\pm 4 $  

When $x=-4,$ 

$y=\frac{-31}{3}$

and when $x=4,$ 

$y=11$ 

Hence, the points required on the curve are $\left( 4,11 \right)$ and $\left( -4,\frac{-31}{3} \right)$.

Ans: The air bubble is in the shape of a sphere.

The volume of an air bubble (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,

$\frac{dV}{dt}=\frac{dV}{dr}.\frac{dr}{dt}$                                                   (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 $\frac{dr}{dt}=\frac{1}{2}cm/s$ 

Therefore, when r is $1$cm,

$  \frac{dV}{dt}=4\pi {{\left( 1 \right)}^{2}}\left( \frac{1}{2} \right) $  

$  =2\pi c{{m}^{3}}/s $  

Hence, the volume of the bubble increases at the rate of $2\pi $cm3/s.

Ans: The volume of a sphere $\left( V \right)$ with radius \[\left( r \right)\] is given by,

$V=\frac{4}{3}\pi {{r}^{2}}$.

Diameter is $\frac{3}{2}\left( 2x+1 \right)$

$  r=\frac{3}{4}\left( 2x+1 \right) $  

$  V=\frac{4}{3}\pi {{r}^{3}} $  

$  =\frac{9}{16}\pi {{\left( 2x+1 \right)}^{3}} $  

Hence, the rate of change of volume with respect to $x$ is as

$\frac{dV}{dt}=\frac{9}{16}\pi \frac{d}{dt}{{\left( 2x+1 \right)}^{3}}$  

$=\frac{27}{8}\pi {{\left( 2x+1 \right)}^{3}}$.

Ans: The volume of a cone ($V$) with radius and height ($h$) is given by.

$V=\frac{1}{3}\pi {{r}^{2}}h$ 

It is provided that,

$  h=\frac{1}{6}r $  

$  \Rightarrow r=6h $  

$  \therefore V=\frac{1}{3}\pi {{\left( 6h \right)}^{2}}h $  

$  =12\pi {{h}^{3}} $  

The rate of change of volume for time ($t$) is given by

$\frac{dV}{dt}=12\pi \frac{d}{dh}\left( {{h}^{3}} \right).\frac{dh}{dt}$      (by chain rule)

$  =12\pi \left( 3{{h}^{2}} \right)\frac{dh}{dt} $

$  =36\pi {{h}^{2}}\frac{dh}{dt} $  

It is provided that $\frac{dV}{dt}=12c{{m}^{2}}/s$.

Therefore, when $h=4cm$ we have:

$  12=36\pi {{\left( 4 \right)}^{2}}\frac{dh}{dt} $  

$  \Rightarrow \frac{dh}{dt}=\frac{12}{36\pi \left( 16 \right)} $  

$  \frac{dh}{dt}=\frac{1}{48\pi } $  

Hence, when the height of the sand cone is $4$ cm, its height is increasing at the rate of $\frac{1}{48}cm/\sec $.

Ans: The rate of change in total cost for output is known as marginal cost.

$M C=\frac{d C}{d x}=0.007\left(3 x^{2}\right)-0.003(2 x)+15$

$=0.021 x^{2}-0.006 x+15$

When 

$  x=17,MC=0.021{{\left( 17 \right)}^{2}}-0.006\left( 17 \right)+15 $  

$  =0.21\left( 289 \right)-0.006\left( 17 \right)+15 $  

$  =6.069-0.102+15 $  

$  =20.967 $  

Hence, the marginal cost when $17$ units are produced is $Rs.20.967$.

Ans: The rate of change in total cost for the number of units sold is known as marginal cost. Let the number of units sold be x.

$  MR=\frac{dR}{dx}=13\left( 2x \right)+26 $  

$  =26x+26 $

When $x=7$,

$MR=208$ 

Hence, the required marginal revenue is $Rs.208$.

Ans: The area of a circle $\left( A \right)$ with radius $\left( r \right)$ is given by, $A=\pi {{r}^{2}}$

As a result, the area's rate of change in relation to its radius r is  $\frac{dA}{dt}=\frac{d}{dr}\left( \pi {{r}^{2}} \right)=2\pi r$ 

Therefore, when $r=6cm$,

$  \frac{dA}{dr}=2\pi \times 6 $  

$  =12\pi c{{m}^{2}}/s $  

Hence, the required rate of change of the area of a circle is $12\pi c{{m}^{2}}/s$.

Therefore, option B is correct.

Ans: The rate of change in total cost for the number of units sold is known as marginal cost.

$\therefore MR=\frac{dR}{dx}=6x+36$ 

Therefore, when $x=15$,

$MR=90+36=126$ 

Hence, the required marginal revenue is $Rs.126$.

The correct option is D.