The radius of a right circular cylinder increases at the rate of 0.1 cm/min, and the height decreases at the rate of 0.2 cm/min. The rate of change of the volume of the cylinder, in ${cm}^{3} / \min$ , when the radius is 2 cm and the height is 3 cm is

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$- 2 π$

$- \frac{3 π}{5}$

$- \frac{8 π}{5}$

$\frac{2 π}{5}$

answer is D.

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Detailed Solution

Given $v = π r^{2} h . \frac{d V}{d t} = π (r^{2} \frac{d h}{d t} + 2 r \frac{d r}{d t} h)$

$put \frac{d r}{d t} = \frac{1}{10}, \frac{d h}{d t} = \frac{2}{10}; r = 2, h = 3$

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