The radius of a cylinder is increasing at the rate $2 \mathrm{~cm} / \mathrm{sec}$ and its height is…

The radius of a cylinder is increasing at the rate $2 \mathrm{~cm} / \mathrm{sec}$ and its height is decreasing at the rate of $3 \mathrm{~cm} / \mathrm{sec}$, then the rate of change of volume, when radius is $3 \mathrm{~cm}$ and the height is $5 \mathrm{~cm}$, is
  1. $44 \pi \mathrm{cm}^3 / \mathrm{sec}$
  2. $11 \pi \mathrm{cm}^3 / \mathrm{sec}$
  3. $23 \pi \mathrm{cm}^3 / \mathrm{sec}$
  4. $33 \pi \mathrm{cm}^3 / \mathrm{sec}$

Solution

$V=\pi r^2 h$ $\begin{aligned} & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=\pi\left(2 r \frac{\mathrm{d} r}{\mathrm{~d} t} h+r^2 \frac{\mathrm{d} h}{\mathrm{~d} t}\right) \\ & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=\pi\left(2 \times 3 \times 2 \times 5+3^2 \times(-3)\right) \\ & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=33 \pi \mathrm{cm}^3 / \mathrm{sec}\end{aligned}$

Asked in: MHT CET 2022 (10 Aug Shift 2)

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