Q54 / 120 · VITEEE 2019 · Mathematics
The radius of a right circular cylinder increases at the rate of $0.1 \mathrm{~cm} / \mathrm{min}$, and the height decreases at the rate of $0.2 \mathrm{~cm} / \mathrm{min}$. The rate of change of the volume of the cylinder, in $\mathrm{cm}^{3} / \mathrm{min}$, when the radius is $2 \mathrm{~cm}$ and the height is $3 \mathrm{~cm}$ is
- A$-2 \pi$
- B$-\frac{8 \pi}{5}$
- C$-\frac{3 \pi}{5}$
- Correct answerD$\frac{2 \pi}{5}$
Solution
Given $V=\pi r^{2} h$.
Differentiating both sides, we get
$$
\begin{array}{l}
\frac{d V}{d t}=\pi\left(r^{2} \frac{d h}{d t}+2 r \frac{d r}{d t} h\right)=\pi r\left(r \frac{d h}{d t}+2 h \frac{d r}{d t}\right) \\
\frac{d r}{d t}=\frac{1}{10} \text { and } \frac{d h}{d t}=-\frac{2}{10}
\end{array}
$$
$$
\frac{d V}{d t}=\pi r\left(r\left(-\frac{2}{10}\right)+2 h\left(\frac{1}{10}\right)\right)=\frac{\pi r}{5}(-r+h)
$$
Thus, when $r=2$ and $h=3$,
$$
\frac{d V}{d t}=\frac{\pi(2)}{5}(-2+3)=\frac{2 \pi}{5}
$$
Hence, the correct answer is option: 'D'.