⇒Application Of Derivative

Correct Option is (D)

According to the figure, $x^2+y^2=25$ ..... (i)

Note that $\cos \theta =\frac{\mathrm{OB}}{\mathrm{AB}}=\frac{x}{5}$

$\begin{array}{rl}& \therefore x=5\cos \theta \\ & \therefore \text{ (i) }\Rightarrow 25{\cos }^2\theta +y^2=25\end{array}$

Differentiating w.r.t. 't', we get

$\begin{array}{rl}& -50\cos \theta \sin \theta \frac{\mathrm{d}\theta }{\mathrm{dt}}+2y\frac{\mathrm{d}y}{\mathrm{dt}}=0\\ & 25\sin \theta \cos \theta \frac{\mathrm{d}\theta }{\mathrm{dt}}=y\frac{\mathrm{d}y}{\mathrm{dt}}\\ & \therefore 25\sin \theta \cos \theta \frac{\mathrm{d}\theta }{\mathrm{dt}}=y(-0.1)\\ & \dots [\because \frac{\mathrm{d}y}{\mathrm{d}x}=-10\mathrm{cm}/\mathrm{s}=-0.1\mathrm{m}/\mathrm{s}]\\ & \therefore 25\sin \theta \cos \theta \frac{\mathrm{d}\theta }{\mathrm{dt}}=-(0.1)y\dots \text{ (ii) }\\ & \text{ at }x=4,\cos \theta =\frac{4}{5},\sin \theta =\frac{3}{5}\text{ and }y=3\\ & \therefore \text{ (ii) }\Rightarrow 25\times \frac{3}{5}\times \frac{4}{5}\times \frac{\mathrm{d}\theta }{\mathrm{dt}}=-0.3\\ & \Rightarrow \frac{\mathrm{d}\theta }{\mathrm{dt}}=-0.025\end{array}$

i.e., the angle is decreasing at the rate of $0.025\mathrm{rad}/\mathrm{s}$

Correct Option is (C)

Note that $\mathrm{△}\mathrm{OAB}$ is a right angled triangle.

Let $\mathrm{OA}=x\mathrm{ft}$ and $\mathrm{OB}=y\mathrm{ft}$.

$y^2=169-x^2$

Now, differentiating above function w.r.t. time '$\mathrm{t}$', we get

$2y\frac{\mathrm{d}y}{\mathrm{dt}}=-2x\frac{\mathrm{d}x}{\mathrm{dt}}$ ..... (i)

At $x=5,\frac{\mathrm{d}x}{\mathrm{dt}}=3\mathrm{ft}/\sec$ ..... [Given]

Also, at $x=5,y=12$

$\therefore$ (i) $\Rightarrow 2(12)\frac{\mathrm{d}y}{\mathrm{dt}}=-2(5)(3)$

$\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{-5}{4}$

Negative sign indicates that B is moving downwards.

$\therefore$ B is moving at the rate $\frac{5}{4}\mathrm{ft}/\sec$ downwards.