Solution

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

$A=\pi r^{2}$

Now, the rate of change of the area with respect to its radius is given by,

$\frac{d A}{d r}=\frac{d}{d r}(\pi r^{2})=2 \pi r$

1. When $r=3 cm$,

$\frac{d A}{d r}=2 \pi(3)=6 \pi$

Hence, the area of the circle is changing at the rate of $6 cm^{2} / s$ when its radius is $3 cm$.

2. When $r=4 cm$,

$\frac{d A}{d r}=2 \pi(4)=8 \pi$

Hence, the area of the circle is changing at the rate of $8 cm^{2} / s$ when its radius is $4 cm$.

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 $S=6 x^{2}$ where $x$ is a function of time $t$.

It is given that $\frac{d V}{d t}=8 cm^{3} / s$.

Then, by using the chain rule, we have:

$ \therefore=\frac{d V}{d t}=\frac{d}{d t}(x^{3})=\frac{d}{d x}(x^{3}) \cdot \frac{d x}{d t}=3 x^{2} \cdot \frac{d x}{d t} $

$ \begin{equation*} \Rightarrow \frac{d x}{d t}=\frac{8}{3 x^{2}} \tag{1} \end{equation*} $

$ \begin{aligned} & \text{ Now, } \frac{d S}{d t}=\frac{d}{d t}(6 x^{2})=\frac{d}{d x}(6 x^{2}) \cdot \frac{d x}{d t} \quad \text{ [By chain rule] } \\ & =12 x \cdot \frac{d x}{d t}=12 x \cdot(\frac{8}{3 x^{2}})=\frac{32}{x} \end{aligned} $

Hence, if the length of the edge of the cube is $12 cm$, then the surface area is increasing

at the rate of $\frac{8}{3} cm^{2} / s$.

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,

$\frac{d A}{d t}=\frac{d}{d t}(\pi r^{2}) \cdot \frac{d r}{d t}=2 \pi r \frac{d r}{d t} \quad$ [By chain rule]

It is given that,

$\frac{d r}{d t}=3 cm / s$ $\therefore \frac{d A}{d t}=2 \pi r(3)=6 \pi r$

Thus, when $r=10 cm$,

$\frac{d A}{d t}=6 \pi(10)=60 \pi cm^{2} / s$

Hence, the rate at which the area of the circle is increasing when the radius is $10 cm$ is $60 п cm^{2} / s$.

Solution

Let $x$ be the length of a side and $V$ be the volume of the cube. Then,

$V=x^{3}$.

$\therefore \frac{d V}{d t}=3 x^{2} \cdot \frac{d x}{d t}$ (By chain rule)

It is given that,

$\frac{d x}{d t}=3 cm / s$

$\therefore \frac{d V}{d t}=3 x^{2}(3)=9 x^{2}$

Thus, when $x=10 cm$,

$\frac{d V}{d t}=9(10)^{2}=900 cm^{3} / s$

Hence, the volume of the cube is increasing at the rate of $900 cm^{3} / s$ when the edge is 10 $cm$ long.

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,

$\frac{d A}{d t}=\frac{d}{d t}(\pi r^{2})=\frac{d}{d r}(\pi r^{2}) \frac{d r}{d t}=2 \pi r \frac{d r}{d t}$ [By chain rule]

It is given that $\frac{d r}{d t}=5 cm / s$.

Thus, when $r=8 cm$,

$\frac{d A}{d t}=2 \pi(8)(5)=80 \pi$

Hence, when the radius of the circular wave is $8 cm$, the enclosed area is increasing at the rate of $80 \pi cm^{2} / s$.

Solution

The circumference of a circle $(C)$ with radius $(r)$ is given by $C=2 \pi r$.

Therefore, the rate of change of circumference (C) with respect to time $(t)$ is given by, $\frac{d C}{d t}=\frac{d C}{d r} \cdot \frac{d r}{d t}$ (By chain rule) $=\frac{d}{d r}(2 \pi r) \frac{d r}{d t}$

$=2 \pi \cdot \frac{d r}{d t}$

It is given that $\frac{d r}{d t}=0.7 cm / s$.

Hence, the rate of increase of the circumference is $2 \pi(0.7)=1.4 \pi cm / s$.

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:

$\frac{d x}{d t}=-5 cm / min$ and $\frac{d y}{d t}=4 cm / min$

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

$P=2(x+y)$

$\therefore \frac{d P}{d t}=2(\frac{d x}{d t}+\frac{d y}{d t})=2(-5+4)=-2 cm / min$

Hence, the perimeter is decreasing at the rate of $2 cm / min$.

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

$A=x \times y$

$\therefore \frac{d A}{d t}=\frac{d x}{d t} \cdot y+x \cdot \frac{d y}{d t}=-5 y+4 x$

When $x=8 cm$ and $y=6 cm, \frac{d A}{d t}=(-5 \times 6+4 \times 8) cm^{2} / min=2 cm^{2} / min$

Hence, the area of the rectangle is increasing at the rate of $2 cm^{2} / min$.

Solution

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

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

$\therefore$ Rate of change of volume $(V)$ with respect to time $(t)$ is given by,

$\frac{d V}{d t}=\frac{d V}{d r} \cdot \frac{d r}{d t} _{\text{[By chain rule] }}$

$=\frac{d}{d r}(\frac{4}{3} \pi r^{3}) \cdot \frac{d r}{d t}$

$=4 \pi r^{2} \cdot \frac{d r}{d t}$

It is given that $\frac{d V}{d t}=900 cm^{3} / s$.

$\therefore 900=4 \pi r^{2} \cdot \frac{d r}{d t}$

$\Rightarrow \frac{d r}{d t}=\frac{900}{4 \pi r^{2}}=\frac{225}{\pi r^{2}}$

Therefore, when radius $=15 cm$,

$\frac{d r}{d t}=\frac{225}{\pi(15)^{2}}=\frac{1}{\pi}$

Hence, the rate at which the radius of the balloon increases when the radius is $15 cm$

$1 cm / s$.

is $\pi$

Solution

The volume of a sphere $(V)$ with radius $(r)$ is given by $V=\frac{4}{3} \pi r^{3}$.

Rate of change of volume $(V)$ with respect to its radius $(r)$ is given by,

$\frac{d V}{d r}=\frac{d}{d r}(\frac{4}{3} \pi r^{3})=\frac{4}{3} \pi(3 r^{2})=4 \pi r^{2}$

Therefore, when radius $=10 cm$,

$\frac{d V}{d r}=4 \pi(10)^{2}=400 \pi$

Hence, the volume of the balloon is increasing at the rate of $400 cm^{3} / s$.

Solution

Let $y m$ be the height of the wall at which the ladder touches. Also, let the foot of the ladder be $x$ maway from the wall.

Then, by Pythagoras theorem, we have:

$x^{2}+y^{2}=25$ [Length of the ladder $=5 m$ ]

$\Rightarrow y=\sqrt{25-x^{2}}$

Then, the rate of change of height $(y)$ with respect to time $(t)$ is given by,

$\frac{d y}{d t}=\frac{-x}{\sqrt{25-x^{2}}} \cdot \frac{d x}{d t}$

It is given that $\frac{d x}{d t}=2 cm / s$. $\therefore \frac{d y}{d t}=\frac{-2 x}{\sqrt{25-x^{2}}}$

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

$\frac{d y}{d t}=\frac{-2 \times 4}{\sqrt{25-4^{2}}}=-\frac{8}{3}$

Hence, the height of the ladder on the wall is decreasing at the rate of $\frac{8}{3} cm / s$.

Solution

The equation of the curve is given as:

$6 y=x^{3}+2$

The rate of change of the position of the particle with respect to time $(t)$ is given by,

$6 \frac{d y}{d t}=3 x^{2} \frac{d x}{d t}+0$

$\Rightarrow 2 \frac{d y}{d t}=x^{2} \frac{d x}{d t}$

When the $y$-coordinate of the particle changes 8 times as fast as the

$x$-coordinate i.e., $(\frac{d y}{d t}=8 \frac{d x}{d t})$, we have:

$2(8 \frac{d x}{d t})=x^{2} \frac{d x}{d t}$

$\Rightarrow 16 \frac{d x}{d t}=x^{2} \frac{d x}{d t}$

$\Rightarrow(x^{2}-16) \frac{d x}{d t}=0$

$\Rightarrow x^{2}=16$

$\Rightarrow x= \pm 4$

When $x=4, y=\frac{4^{3}+2}{6}=\frac{66}{6}=11$.

When $x=-4, y=\frac{(-4)^{3}+2}{6}=-\frac{62}{6}=-\frac{31}{3}$.

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

Solution

The air bubble is in the shape of a sphere.

Now, the volume of an air bubble $(V)$ with radius $(r)$ is given by,

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

The rate of change of volume $(V)$ with respect to time $(t)$ is given by,

$ \begin{matrix} \frac{d V}{d t} & =\frac{4}{3} \pi \frac{d}{d r}(r^{3}) \cdot \frac{d r}{d t} & \text{ [By chain rule] } \\ & =\frac{4}{3} \pi(3 r^{2}) \frac{d r}{d t} & \\ & =4 \pi r^{2} \frac{d r}{d t} & \end{matrix} $

It is given that $\frac{d r}{d t}=\frac{1}{2} cm / s$.

Therefore, when $r=1 cm$,

$\frac{d V}{d t}=4 \pi(1)^{2}(\frac{1}{2})=2 \pi cm^{3} / s$

Hence, the rate at which the volume of the bubble increases is $2 \pi cm^{3} / s$.

Solution

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

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

It is given that:

Diameter $=\frac{3}{2}(2 x+1)$

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

$\therefore V=\frac{4}{3} \pi(\frac{3}{4})^{3}(2 x+1)^{3}=\frac{9}{16} \pi(2 x+1)^{3}$

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

$ \frac{d V}{d x}=\frac{9}{16} \pi \frac{d}{d x}(2 x+1)^{3}=\frac{9}{16} \pi \times 3(2 x+1)^{2} \times 2=\frac{27}{8} \pi(2 x+1)^{2} . $

Solution

The volume of a cone $(V)$ with radius $(r)$ and height $(h)$ is given by,

$V=\frac{1}{3} \pi r^{2} h$

It is given that,

$h=\frac{1}{6} r \Rightarrow r=6 h$

$\therefore V=\frac{1}{3} \pi(6 h)^{2} h=12 \pi h^{3}$

The rate of change of volume with respect to time $(t)$ is given by,

$\frac{d V}{d t}=12 \pi \frac{d}{d h}(h^{3}) \cdot \frac{d h}{d t}$ [By chain rule]

$=12 \pi(3 h^{2}) \frac{d h}{d t}$

$=36 \pi h^{2} \frac{d h}{d t}$

It is also given that $\frac{d V}{d t}=12 cm^{3} / s$.

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

$ \begin{aligned} & 12=36 \pi(4)^{2} \frac{d h}{d t} \\ & \Rightarrow \frac{d h}{d t}=\frac{12}{36 \pi(16)}=\frac{1}{48 \pi} \end{aligned} $

Hence, when the height of the sand cone is $4 cm$, its height is increasing at the rate

of $\frac{1}{48 \pi} cm / s$.

Solution

Marginal cost is the rate of change of total cost with respect to output.

$\therefore$ Marginal cost (MC) $=\frac{d C}{d x}=0.007(3 x^{2})-0.003(2 x)+15$

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

When $x=17, M C=0.021(17^{2})-0.006(17)+15$ $=0.021(289)-0.006(17)+15$

$=6.069-0.102+15$

$=20.967$

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

Solution

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

$\therefore$ Marginal Revenue (MR) $=\frac{d R}{d x}=13(2 x)+26=26 x+26$

When $x=7$,

$M R=26(7)+26=182+26=208$

Hence, the required marginal revenue is Rs 208.

Solution

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

$A=\pi r^{2}$

Therefore, the rate of change of the area with respect to its radius $r$ is

$\frac{d A}{d r}=\frac{d}{d r}(\pi r^{2})=2 \pi r$. $\therefore$ When $r=6 cm$,

$\frac{d A}{d r}=2 \pi \times 6=12 \pi cm^{2} / s$

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

The correct answer is $B$.

Solution

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

$\therefore$ Marginal Revenue (MR) $=\frac{d R}{d x}=3(2 x)+36=6 x+36$

$\therefore$ When $x=15$,

$M R=6(15)+36=90+36=126$

Hence, the required marginal revenue is Rs 126 .

The correct answer is D.