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Question
The radius of a cylinder is increasing at the rate 2 cm/sec. and its altitude is decreasing at the rate of 3 cm/sec. Find the rate of change of volume when radius is 3 cm and altitude 5 cm.
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Solution
\[\text { Let r e the radius,h be the height and V be the volume of the cylinder at any time t. Then },\]
\[V = \pi r^2 h\]
\[ \Rightarrow \frac{dV}{dt} = 2\pi r h\frac{dr}{dt} + \pi r^2 \frac{dh}{dt}\]
\[ \Rightarrow \frac{dV}{dt} = \pi r\left( 2h\frac{dr}{dt} + r\frac{dh}{dt} \right)\]
\[ \Rightarrow \frac{dV}{dt} = \pi \times 3\left( 2 \times 5 \times 2 + 3 \times - 3 \right)\]
\[ \Rightarrow \frac{dV}{dt} = 3\pi\left( 20 - 9 \right)\]
\[ \Rightarrow \frac{dV}{dt} = 33\pi \text{ cm}^3 /\sec\]
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