A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is propotional to the surface. Prove that the radius is decreasing at a constant rate.
We have, rate of decrease of the volume of spherical ball of salt at any instant is $\propto$ surface. Let the radius of the spherical ball of the salt be $r$.
$$\begin{array}{lr} \therefore & \text { Volume of the ball }(V)=\frac{4}{3} \pi r^3 \\ \text { and } & \text { surface area }(S)=4 \pi r^2 \end{array}$$
$\because\quad \frac{d V}{d t} \propto S \quad \Rightarrow \frac{d}{d t}\left(\frac{4}{3} \pi r^3\right) \propto 4 \pi r^2$
$\Rightarrow \quad \frac{4}{3} \pi \cdot 3 r^2 \cdot \frac{d r}{d t} \propto 4 \pi r^2 \Rightarrow \frac{d r}{d t} \propto \frac{4 \pi r^2}{4 \pi r^2}$
$$\begin{aligned} &\begin{array}{ll} \Rightarrow & \frac{d r}{d t}=k \cdot 1 \quad \text { [where, } k \text { is the proportionality constant] } \\ \Rightarrow & \frac{d r}{d t}=k \end{array}\\ &\text { Hence, the radius of ball is decreasing at a constant rate. } \end{aligned}$$
If the area of a circle increases at a uniform rate, then prove that perimeter varies inversely as the radius.
Let the radius of circle $=r$ And area of the circle, $A=\pi r^2$
$$\begin{array}{ll} \therefore & \frac{d}{d t} A=\frac{d}{d t} \pi r^2 \\ \Rightarrow & \frac{d A}{d t}=2 \pi r \cdot \frac{d r}{d t}\quad\text{.... (i)} \end{array}$$
Since, the area of a circle increases at a uniform rate, then
$$\frac{d A}{d t}=k\quad\text{.... (ii)}$$
where, $k$ is a constant.
From Eqs. (i) and (ii),
$$2 \pi r \cdot \frac{d r}{d t}=k$$
$\Rightarrow \quad \frac{d r}{d t}=\frac{k}{2 \pi r}=\frac{k}{2 \pi} \cdot\left(\frac{1}{r}\right)\quad\text{.... (iii)}$
$$\begin{aligned} &\begin{aligned} \text { Let the perimeter, }\quad P & =2 \pi r \\ \therefore\quad\frac{d P}{d t} & =\frac{d}{d t} \cdot 2 \pi r \quad \Rightarrow \quad \frac{d P}{d t}=2 \pi \cdot \frac{d r}{d t} \end{aligned} \end{aligned}$$
$$=2 \pi \cdot \frac{k}{2 \pi} \cdot \frac{1}{r}=\frac{k}{r}\quad$$ [using Eq. (iii)]
$\Rightarrow \quad \frac{d P}{d t} \propto \frac{1}{r}\quad$ Hence proved.
A kite is moving horizontally at a height of 151.5 m. If the speed of kite is $10 \mathrm{~m} / \mathrm{s}$, how fast is the string being let out, when the kite is 250 m away from the boy who is flying the kite, if the height of boy is 1.5 m ?
We have, height $(h)=151.5 \mathrm{~m}$, speed of kite $(v)=10 \mathrm{~m} / \mathrm{s}$
Let $C D$ be the height of kite and $A B$ be the height of boy.
Let $$\quad D B=x \mathrm{~m}=E A \text { and } A C=250 \mathrm{~m}$$
$$\therefore\quad \frac{d x}{d t}=10 \mathrm{~m} / \mathrm{s}$$
From the figure, we see that
$$\begin{aligned} & E C=151.5-1.5=150 \mathrm{~m} \\ \text{and}\quad& A E=x \\ \text{Also,}\quad& A C=250 \mathrm{~m} \end{aligned}$$
In right angled $\triangle C E A$,
$$\begin{array}{ll} & A E^2+E C^2=A C^2 \\ \Rightarrow & x^2+(150)^2=y^2 \quad\text{..... (i)}\\ \Rightarrow & x^2+(150)^2=(250)^2 \end{array}$$
$\Rightarrow \quad x^2=(250)^2-(150)^2$
$$\begin{aligned} & =(250+150)(250-150) \\ & =400 \times 100 \\ \therefore\quad x & =20 \times 10=200 \end{aligned}$$
$$\begin{aligned} &\text { From Eq. (i), on differentiating w.r.t. } t \text {, we get }\\ &\begin{array}{rlrl} 2 x \cdot \frac{d x}{d t}+0 & =2 y \frac{d y}{d t} \\ \Rightarrow \quad 2 y \frac{d y}{d t} & =2 x \frac{d x}{d t} \\ \therefore \quad \frac{d y}{d t} & =\frac{x}{y} \cdot \frac{d x}{d t} \\ & =\frac{200}{250} \cdot 10=8 \mathrm{~m} / \mathrm{s} & {\left[\because \frac{d x}{d t}=10 \mathrm{~m} / \mathrm{s}\right]} \end{array} \end{aligned}$$
So, the required rate at which the string is being let out is $8 \mathrm{~m} / \mathrm{s}$.
Two men $A$ and $B$ start with velocities $v$ at the same time from the junction of two roads inclined at $45^{\circ}$ to each other. If they travel by different roads, then find the rate at which they are being separated.
Let two men start from the point $C$ with velocity $v$ each at the same time.
Also, $\quad \angle B C A=45^{\circ}$
Since, $A$ and $B$ are moving with same velocity $v$, so they will cover same distance in same time.
Therefore, $\triangle A B C$ is an isosceles triangle with $A C=B C$.
Now, draw $C D \perp A B$.
Let at any instant $t$, the distance between them is $A B$.
Let $$A C=B C=x \quad \text { and } \quad A B=y$$
In $\triangle A C D$ and $\triangle D C B$,
$$\begin{array}{l} \angle C A D =\angle C B D & {[\because A C=B C]} \\ \angle C D A =\angle C D B=90^{\circ} \\ \therefore \quad \angle A C D =\angle D C B \end{array}$$
$$\begin{array}{ll} \text { or } & \angle A C D=\frac{1}{2} \times \angle A C B \\ \Rightarrow & \angle A C D=\frac{1}{2} \times 45^{\circ} \\ \Rightarrow & \angle A C D=\frac{\pi}{8} \\ \therefore & \sin \frac{\pi}{8}=\frac{A D}{A C} \\ \Rightarrow & \sin \frac{\pi}{8}=\frac{y / 2}{x} \quad [\because A D=y / 2]\\ \Rightarrow & \frac{y}{2}=x \sin \frac{\pi}{8} \\ \Rightarrow & y=2 x \cdot \sin \frac{\pi}{8} \end{array}$$
$$\begin{aligned} &\text { Now, differentiating both sides w.r.t. } t \text {, we get }\\ &\begin{aligned} \frac{d y}{d t} & =2 \cdot \sin \frac{\pi}{8} \cdot \frac{d x}{d t} \\ & =2 \cdot \sin \frac{\pi}{8} \cdot v \quad \left[\because v=\frac{d x}{d t}\right]\\ & =2 v \cdot \frac{\sqrt{2-\sqrt{2}}}{2} \quad \left[\because \sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}\right]\\ & =\sqrt{2-\sqrt{2}} v \text { unit/s } \end{aligned} \end{aligned}$$
which is the rate at which A and B are being separated.
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 $\frac{\pi}{3}$.