Find an angle $\theta$, where $0<\theta<\frac{\pi}{2}$, which increases twice as fast as its sine.
Let $\theta$ increases twice as fast as its sine.
$$\Rightarrow \quad \theta=2 \sin \theta$$
Now, on differentiating both sides w.r.t. $t$, we get
$$ \begin{aligned} \frac{d \theta}{d t} & =2 \cdot \cos \theta \cdot \frac{d \theta}{d t} \Rightarrow 1=2 \cos \theta \\ \Rightarrow\quad\frac{1}{2} & =\cos \theta \Rightarrow \cos \theta=\cos \frac{\pi}{3} \\ \therefore\quad\theta & =\frac{\pi}{3} \end{aligned}$$
So, the required angle is $\frac{\pi}{3}$.
Find the approximate value of $(1.999)^5$.
Let $$x=2$$
and $\Delta x=-0.001\quad$ $$ [\because 2-0.001=1.999]$$
Let $$y=x^5$$
On differentiating both sides w.r.t. $x$, we get
$$\frac{d y}{d x}=5 x^4$$
$$\begin{aligned} \text { Now, } \quad \Delta y & =\frac{d y}{d x} \cdot \Delta x=5 x^4 \times \Delta x \\ & =5 \times 2^4 \times[-0.001] \\ & =-80 \times 0.001=-0.080 \\ \therefore \quad(1.999)^5 & =y+\Delta y \\ & =2^5+(-0.080) \\ & =32-0.080=31.920 \end{aligned}$$
Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm , respectively.
$$\begin{aligned} & \text { Let internal radius }=r \text { and external radius }=R \\ & \therefore \text { Volume of hollow spherical shell, } V=\frac{4}{3} \pi\left(R^3-r^3\right) \\ & \Rightarrow \quad V=\frac{4}{3} \pi\left[(3.0005)^3-(3)^3\right]\quad\text{.... (i)} \end{aligned}$$
Now, we shall use differentiation to get approximate value of $(3.0005)^3$.
Let $$\quad(3.0005)^3=y+\Delta y$$
$$\begin{aligned} \text{and }\quad & x=3, \Delta x=0.0005 \\ \text{Also, let}\quad & y=x^3 \end{aligned}$$
$$\begin{aligned} &\text { On differentiating both sides w.r.t. } x \text {, we get }\\ &\begin{aligned} \frac{d y}{d x} & =3 x^2 \\ \therefore\quad \Delta y & =\frac{d y}{d x} \times \Delta x=3 x^2 \times 0.0005 \\ & =3 \times 3^2 \times 0.0005 \\ & =27 \times 0.0005=0.0135 \end{aligned} \end{aligned}$$
$$ \begin{aligned} &\begin{aligned} \text{Also,}\quad(3.0005)^3 & =y+\Delta y \\ & =3^3+0.0135=27.0135 \\ \therefore\quad V & =\frac{4}{3} \pi[27.0135-27.000] \quad \text { [using Eq. (i)] }\\ & =\frac{4}{3} \pi[0.0135]=4 \pi \times(0.0045) \\ & =0.0180 \pi \mathrm{~cm}^3 \end{aligned}\\ \end{aligned}$$
A man, 2 m tall, walks at the rate of $1 \frac{2}{3} \mathrm{~m} / \mathrm{s}$ towards a street light which is $5 \frac{1}{3} \mathrm{~m}$ above the ground. At what rate is the tip of his shadow moving and at what rate is the length of the shadow changing when he is $3 \frac{1}{3} \mathrm{~m}$ from the base of the light?
Let $A B$ be the street light post and $C D$ be the height of man i.e., $C D=2 \mathrm{~m}$.
Let $B C=x \mathrm{~m}, C E=y \mathrm{~m}$ and $\frac{d x}{d t}=\frac{-5}{3} \mathrm{~m} / \mathrm{s}$
From $\triangle A B E$ and $\triangle D C E$, we see that
$\triangle A B E \sim \triangle D C E\quad$ [by AAA similarity]
$$\begin{array}{ll} \therefore & \frac{A B}{D C}=\frac{B E}{C E} \Rightarrow \frac{\frac{16}{3}}{2}=\frac{x+y}{y} \\ \Rightarrow & \frac{16}{6}=\frac{x+y}{y} \\ \Rightarrow & 16 y=6 x+6 y \Rightarrow 10 y=6 x \\ \Rightarrow & y=\frac{3}{5} x \end{array}$$
On differentiating both sides w.r.t. $t$, we get
$$\frac{d y}{d t}=\frac{3}{5} \cdot \frac{d x}{d t}=\frac{3}{5} \cdot\left(-1 \frac{2}{3}\right)\quad$$ [since, man is moving towards the light post]
$$=\frac{3}{5} \cdot\left(\frac{-5}{3}\right)=-1 \mathrm{~m} / \mathrm{s}$$
Let $$z=x+y$$
Now, differentiating both sides w.r.t. $t$, we get
$$\begin{aligned} \frac{d z}{d t} & =\frac{d x}{d t}+\frac{d y}{d t}=-\left(\frac{5}{3}+1\right) \\ & =-\frac{8}{3}=-2 \frac{2}{3} \mathrm{~m} / \mathrm{s} \end{aligned}$$
Hence, the tip of shadow is moving at the rate of $2 \frac{2}{3} \mathrm{~m} / \mathrm{s}$ towards the light source and length of the shadow is decreasing at the rate of $1 \mathrm{~m} / \mathrm{s}$.
A swimming pool is to be drained for cleaning. If $L$ represents the number of litres of water in the pool $t$ seconds after the pool has been plugged off to drain and $L=200(10-t)^2$. How fast is the water running out at the end of 5 s and what is the average rate at which the water flows out during the first 5 s?
Let $L$ represents the number of litres of water in the pool $t$ seconds after the pool has been plugged off to drain, then
$$L=200(10-t)^2$$
$\therefore$ Rate at which the water is running out $=-\frac{d L}{d t}$
$$\begin{aligned} \frac{d L}{d t} & =-200 \cdot 2(10-t) \cdot(-1) \\ & =400(10-t) \end{aligned}$$
$$\begin{aligned} &\text { Rate at which the water is running out at the end of } 5 \mathrm{~s}\\ &\begin{aligned} & =400(10-5) \\ & =2000 \mathrm{~L} / \mathrm{s}=\text { Final rate } \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { Since, } \quad \text { initial rate }=-\left(\frac{d L}{d t}\right)_{t=0}=4000 \mathrm{~L} / \mathrm{s}\\ &\begin{aligned} \therefore \quad \text { Average rate during } 5 \mathrm{~s} & =\frac{\text { Initial rate }+ \text { Final rate }}{2} \\ & =\frac{4000+2000}{2} \\ & =3000 \mathrm{~L} / \mathrm{s} \end{aligned} \end{aligned}$$