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
- A \(44 \pi \mathrm{cm}^3 / \mathrm{sec}\)
- B \(11 \pi \mathrm{cm}^3 / \mathrm{sec}\)
- C \(23 \pi \mathrm{cm}^3 / \mathrm{sec}\)
- D \(33 \pi \mathrm{cm}^3 / \mathrm{sec}\)
Answer & Solution
Correct Answer
(D) \(33 \pi \mathrm{cm}^3 / \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
\(V=\pi r^2 h\)
\(\begin{aligned} & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=\pi\left(2 r \frac{\mathrm{d} r}{\mathrm{~d} t} h+r^2 \frac{\mathrm{d} h}{\mathrm{~d} t}\right) \\ & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=\pi\left(2 \times 3 \times 2 \times 5+3^2 \times(-3)\right) \\ & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=33 \pi \mathrm{cm}^3 / \mathrm{sec}\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=\pi\left(2 r \frac{\mathrm{d} r}{\mathrm{~d} t} h+r^2 \frac{\mathrm{d} h}{\mathrm{~d} t}\right) \\ & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=\pi\left(2 \times 3 \times 2 \times 5+3^2 \times(-3)\right) \\ & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=33 \pi \mathrm{cm}^3 / \mathrm{sec}\end{aligned}\)
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