⇒Application Of Derivative

Correct Option is (D)

Let A, P and X be the area, perimeter and length of side of square respectively at time 't' seconds. Then,

$\begin{array}{rl}\mathrm{A}& ={\mathrm{X}}^2,\mathrm{P}=4\mathrm{X}\\ \therefore \mathrm{P}& =4\sqrt{\mathrm{A}}\\ & \text{ Differentiating w.r.t. }\mathrm{t},\text{ we get }& \\ \frac{\mathrm{dP}}{\mathrm{dt}}& =4\frac{1}{2\sqrt{\mathrm{A}}}\cdot \frac{\mathrm{dA}}{\mathrm{dt}}\\ & =\frac{2}{\mathrm{X}}\cdot \frac{\mathrm{dA}}{\mathrm{dt}}\\ & =\frac{2}{20}\times 4\dots \left[\frac{\mathrm{side}=20\mathrm{cm}}{\frac{\mathrm{dA}}{\mathrm{dt}}=4{\mathrm{cm}}^2/\sec }\right]\\ & =\frac{2}{5}\mathrm{cm}/\sec \end{array}$

Correct Option is (B)

In $\mathrm{△}\mathrm{ABC},\mathrm{AC}$ represents ladder

$\mathrm{AB}\to$ vertical wall

Let $\mathrm{AB}=x,\mathrm{BC}=y$

$\therefore \mathrm{\angle }\mathrm{ABC}={90}^{\circ }$

By Pythagoras theorem,

$\begin{array}{rl}& {\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{AC}}^2\\ & \Rightarrow x^2+y^2={17}^2\\ & \Rightarrow x^2=289-y^2\\ & \Rightarrow x^2=289-64\\ & \Rightarrow x^2=225\\ & \Rightarrow x=15\mathrm{m}\end{array}$

Consider equation (i),

$x^2=289-y^2$

Differentiating w.r.t. t, we get

$\begin{array}{rl}& 2x\frac{\mathrm{d}x}{\mathrm{dt}}=-2y\frac{\mathrm{d}y}{\mathrm{dt}}\\ & \Rightarrow 15\frac{\mathrm{d}x}{\mathrm{dt}}=-8(1)\\ & \Rightarrow \frac{\mathrm{d}x}{\mathrm{dt}}=\frac{-8}{15}\mathrm{m}/\mathrm{s}\end{array}$

Negative sign shows that the ladder is moving down. i.e., vertical length is decreasing

$\therefore$ Upper end is coming down at the rate of $\frac{8}{15}\mathrm{m}/\mathrm{s}$.

Correct Option is (B)

Let '$\mathrm{P}$' be the position of the kite and PR be the string.

Let $\mathrm{QR}=x$ and $\mathrm{PR}=y$

By Pythagoras theorem,

$\begin{array}{rl}& {\mathrm{PR}}^2={\mathrm{PQ}}^2+{\mathrm{QR}}^2\\ & \Rightarrow y^2=(120{)}^2+x^2\text{.... (i)}\end{array}$

Differentiating w.r.t. t, we get

$\begin{array}{rl}& 2y\frac{\mathrm{d}y}{\mathrm{dt}}=2x\frac{\mathrm{d}x}{\mathrm{dt}}\\ & \Rightarrow y\frac{\mathrm{d}y}{\mathrm{dt}}=x\frac{\mathrm{d}x}{\mathrm{dt}}\text{.... (ii)}\end{array}$

Now, kite is moving away horizontally at the rate of $39\mathrm{m}/\sec$.

$\begin{array}{rl}\therefore & \frac{\mathrm{d}x}{\mathrm{dt}}=39\mathrm{m}/\sec \\ & \text{ From (i) }\\ & (130{)}^2=(120{)}^2+x^2\\ \Rightarrow & x^2=16900-14400\\ \Rightarrow & x^2=2500\\ \Rightarrow & x=50\end{array}$

From (ii),

$\begin{array}{rl}& 130\frac{\mathrm{d}y}{\mathrm{dt}}=50\times 39\\ \therefore & \frac{\mathrm{d}y}{\mathrm{dt}}=\frac{50\times 39}{130}=15\mathrm{m}/\sec \end{array}$

Correct Option is (A)

Semi-vertical angle $={\tan }^{-1}\left(\frac{1}{2}\right)$

Let $\alpha ={\tan }^{-1}\left(\frac{1}{2}\right)$

$\begin{array}{rl}& \tan \alpha =\frac{1}{2}\\ & \therefore \frac{\mathrm{r}}{\mathrm{h}}=\frac{1}{2}\\ & \mathrm{r}=\frac{\mathrm{h}}{2}\end{array}$

Given, $\frac{\mathrm{dV}}{\mathrm{dt}}=5{\mathrm{m}}^3/\min$.

$\mathrm{V}=$ Volume of cone

Volume of cone $=\frac{1}{3}\pi {\mathrm{r}}^2\mathrm{h}$

$\begin{array}{rl}& \mathrm{V}=\frac{1}{3}\pi {\left(\frac{\mathrm{h}}{2}\right)}^2\times \mathrm{h}\\ & \mathrm{V}=\frac{1}{12}\pi {\mathrm{h}}^3\end{array}$

Differentiating w. r. t. t, we get $\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{1}{12}\times \pi \times 3{\mathrm{h}}^2\times \frac{\mathrm{dh}}{\mathrm{dt}}$

$5=\frac{1}{4}\pi {\mathrm{h}}^2\frac{\mathrm{dh}}{\mathrm{dt}}$

$\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{20}{\pi {\mathrm{h}}^2}$

Now, $\mathrm{h}=10$ .... [Given]

$\begin{array}{rl}\therefore \frac{\mathrm{dh}}{\mathrm{dt}}& =\frac{20}{\pi \times (10{)}^2}\\ \frac{\mathrm{dh}}{\mathrm{dt}}& =\frac{1}{5\pi }\end{array}$

$\therefore$ Rate of change of water level is $\frac{1}{5\pi }\mathrm{m}/\min$.

Correct Option is (B)

$\begin{array}{rl}& 2x\frac{\mathrm{d}x}{\mathrm{dt}}+2y\frac{\mathrm{d}y}{\mathrm{dt}}=0\\ & 2x\frac{\mathrm{d}x}{\mathrm{dt}}+2y(-3)=0\\ & \frac{\mathrm{d}x}{\mathrm{dt}}=\frac{6y}{2x}\\ & \frac{\mathrm{d}x}{\mathrm{dt}}=\frac{3y}{x}\\ & {\frac{\mathrm{d}x}{\mathrm{dt}}|}_{(\frac{1}{2},\frac{\sqrt{3}}{2})}=\frac{3\times \frac{\sqrt{3}}{2}}{\frac{1}{2}}\\ & =3\sqrt{3}\text{ units }/\sec \end{array}$