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MHT CET · Maths · Application of Derivatives

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. A \(44 \pi \mathrm{cm}^3 / \mathrm{sec}\)
  2. B \(11 \pi \mathrm{cm}^3 / \mathrm{sec}\)
  3. C \(23 \pi \mathrm{cm}^3 / \mathrm{sec}\)
  4. D \(33 \pi \mathrm{cm}^3 / \mathrm{sec}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(33 \pi \mathrm{cm}^3 / \mathrm{sec}\)

Step-by-step Solution

Detailed explanation

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