Question 1

Find the rate of change of the area of a circle with respect to its radius $r$ when

(a) $r=3 \mathrm{~cm}$

(b) $r=4 \mathrm{~cm}$

Accepted Answer

We know that the area of a circle, $A=\pi r^{2}$

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

$$

\begin{aligned}

\frac{d A}{d r} & =\frac{d}{d r}\left(\pi r^{2}ight) \

& =2 \pi r

\end{aligned}

$$

(a) When $r=3 \mathrm{~cm}$

Then,

$$

\begin{aligned}

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

& =6 \pi

\end{aligned}

$$

Thus, the area is changing at the rate of $6 \pi$.

(b) When $r=4 \mathrm{~cm}$

Then,

$$

\begin{aligned}

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

& =8 \pi

\end{aligned}

$$

Thus, the area is changing at the rate of $8 \pi$.

Question 2

The volume of a cube is increasing at the rate of $8 \mathrm{~cm}^{3} / \mathrm{s}$. How fast is the surface area increasing when the length of its edge is 12 cm ?

Accepted Answer

Let the side length, volume and surface area respectively be equal to $x, V$ and $S$. Hence, $V=x^{3}$ and $S=6 x^{2}$

We have,

$$

\frac{d V}{d t}=8 \mathrm{~cm}^{3} / \mathrm{s}

$$

Therefore,

$$

\begin{align*}

\frac{d V}{d t} & =\frac{d}{d t}\left(x^{3}ight) \

8 & =\frac{d}{d x}\left(x^{3}ight) \frac{d x}{d t} \

8 & =3 x^{2} \frac{d x}{d t} \

\frac{d x}{d t} & =\frac{8}{3 x^{2}} 	ag{1}

\end{align*}

$$

Now,

$$

\begin{aligned}

\frac{d S}{d t} & =\frac{d}{d t}\left(6 x^{2}ight) \

& =\frac{d}{d x}\left(6 x^{2}ight) \frac{d x}{d t} \

& =12 x \frac{d x}{d t} \

& =12 x\left(\frac{8}{3 x^{2}}ight) \

& =\frac{32}{x}

\end{aligned}

$$

So, when $x=12 \mathrm{~cm}$

Then,

$$

\begin{aligned}

\frac{d S}{d t} & =\frac{32}{12} \mathrm{~cm}^{2} / \mathrm{s} \

& =\frac{8}{3} \mathrm{~cm}^{2} / \mathrm{s}

\end{aligned}

$$

Question 3

The radius of a circle is increasing uniformly at the rate of $3 \mathrm{~cm} / \mathrm{s}$. Find the rate at which the area of the circle is increasing when the radius is $10 \mathrm{~cm} / \mathrm{s}$.

Accepted Answer

We know that $A=\pi r^{2}$

Now,

$$

\begin{aligned}

\frac{d A}{d t} & =\frac{d}{d r}\left(\pi r^{2}ight) \

& =2 \pi r \frac{d r}{d t}

\end{aligned}

$$

We have,

$$

\frac{d r}{d t}=3 \mathrm{~cm} / \mathrm{s}

$$

Hence,

$$

\begin{aligned}

\frac{d A}{d t} & =2 \pi r(3) \

& =6 \pi r

\end{aligned}

$$

So, when $r=10 \mathrm{~cm}$

Then,

$$

\begin{aligned}

\frac{d A}{d t} & =6 \pi(10) \

& =60 \pi \mathrm{~cm}^{2} / \mathrm{s}

\end{aligned}

$$

Question 4

An edge of a variable cube is increasing at the rate of $3 \mathrm{~cm} / \mathrm{s}$. How fast is the volume of the cube increasing when the edge is 10 cm long?

Accepted Answer

Let the length and the volume of the cube respectively be $x$ and $V$.

Hence, $V=x^{3}$

Now,

$$

\begin{aligned}

\frac{d V}{d t} & =\frac{d}{d t}\left(x^{3}ight) \

& =\frac{d}{d x}\left(x^{3}ight) \frac{d x}{d t} \

& =3 x^{2} \frac{d x}{d t}

\end{aligned}

$$

We have,

$$

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

$$

Hence,

$$

\begin{aligned}

\frac{d V}{d t} & =3 x^{2}(3) \

& =9 x^{2}

\end{aligned}

$$

So, when $x=10 \mathrm{~cm}$

Then,

$$

\begin{aligned}

\frac{d V}{d t} & =9(10)^{2} \

& =900 \mathrm{~cm}^{3} / \mathrm{s}

\end{aligned}

$$

Question 5

A stone is dropped into a quiet lake and waves move in circles at the speed of $5 \mathrm{~cm} / \mathrm{s}$. At the instant when the radius of the circular wave is 8 cm , how fast is the encoding area is increasing?

Accepted Answer

We know that $A=\pi r^{2}$

Now,

$$

\begin{aligned}

\frac{d A}{d t} & =\frac{d}{d t}\left(\pi r^{2}ight) \

& =2 \pi r \frac{d r}{d t}

\end{aligned}

$$

We have,

$$

\frac{d r}{d t}=5 \mathrm{~cm} / \mathrm{s}

$$

Hence,

$$

\begin{aligned}

\frac{d A}{d t} & =2 \pi r(5) \

& =10 \pi r

\end{aligned}

$$

So, when $r=8 \mathrm{~cm}$

Then,

$$

\begin{aligned}

\frac{d A}{d t} & =10 \pi(8) \

& =80 \pi \mathrm{~cm}^{2} / \mathrm{s}

\end{aligned}

$$

Question 6

The radius of a circle is increasing at the rate of $0.7 \mathrm{~cm} / \mathrm{s}$. What is the rate of increase of its circumference?

Accepted Answer

We know that $C=2 \pi r$

Now,

$$

\begin{aligned}

\frac{d C}{d t} & =\frac{d}{d t}(2 \pi r) \

& =\frac{d}{d r}(2 \pi r) \frac{d r}{d t} \

& =2 \pi \frac{d r}{d t}

\end{aligned}

$$

We have,

$$

\frac{d r}{d t}=0.7 \pi \mathrm{~cm} / \mathrm{s}

$$

Hence,

$$

\begin{aligned}

\frac{d C}{d t} & =2 \pi(0.7) \

& =1.4 \pi \mathrm{~cm} / \mathrm{s}

\end{aligned}

$$

Question 7

The length $x$ of a rectangle is decreasing at the rate of $5 \mathrm{~cm} /$ minute and the width $y$ is increasing at the rate of $4 \mathrm{~cm} /$ minute . When 8 cm and $y=6 \mathrm{~cm}$, find the rate of change of (a) the perimeter and (b) the area of the rectangle.

Accepted Answer

It is given that $\frac{d x}{d t}=-5 \mathrm{~cm} /$ minute $\frac{d x}{d t}=4 \mathrm{~cm} /$ minute,$x=8 \mathrm{~cm}$ and $y=6 \mathrm{~cm}$

(a) The perimeter of a rectangle is given by $P=2(x+y)$

Therefore,

$$

\begin{aligned}

\frac{d P}{d t} & =2\left(\frac{d x}{d t}+\frac{d y}{d t}ight) \

& =2(-5+4) \

& =-2 \mathrm{~cm} / 	ext { minute }

\end{aligned}

$$

\begin{aligned}

\frac{d P}{d t} & =2\left(\frac{d x}{d t}+\frac{d y}{d t}ight) \

& =2(-5+4) \

& =-2 \mathrm{~cm} / 	ext { minute }

\end{aligned}

$$

(b) The area of a rectangle is given by $A=x y$

Therefore,

$$

\begin{aligned}

\frac{d A}{d t} & =\frac{d x}{d t} y+x \frac{d y}{d t} \

& =-5 y+4 x \

\frac{d A}{d t} & =\frac{d x}{d t} y+x \frac{d y}{d t} \

& =-5 y+4 x

\end{aligned}

$$

When $x=8 \mathrm{~cm}$ and $y=6 \mathrm{~cm}$

Then,

$$

\begin{aligned}

\frac{d A}{d t} & =(-5 	imes 6+4 	imes 8) \mathrm{cm}^{2} / 	ext { minute } \

& =2 \mathrm{~cm}^{2} / 	ext { minute }

\end{aligned}

$$

Question 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 .

Accepted Answer

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

Hence,

$$

\begin{aligned}

\frac{d V}{d t} & =\frac{d V}{d t}\left(\frac{4}{3} \pi r^{3}ight) \

& =\frac{d}{d r}\left(\frac{4}{3} \pi r^{3}ight) \frac{d r}{d t} \

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

\end{aligned}

$$

We have,

$$

\frac{d V}{d t}=900 \mathrm{~cm}^{2} / \mathrm{s}

$$

Therefore,

$$

\begin{aligned}

4 \pi r^{2} \frac{d r}{d t} & =900 \

\frac{d r}{d t} & =\frac{900}{4 \pi r^{2}} \

& =\frac{225}{\pi r^{2}}

\end{aligned}

$$

When radius, $r=15 \mathrm{~cm}$

Then,

$$

\begin{aligned}

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

& =\frac{1}{\pi} \mathrm{~cm} / \mathrm{s}

\end{aligned}

$$

6. Applications of Deivatives

6.1

6.2

6.3

6.4

6.5

Miscellaneous