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The radius of a cylinder is increasing at the rate of $2$ cm/sec and its altitude is decreasing at the rate of $3$cm/sec . The rate of change of volume when the radius is $3$cm and the altitude is $5$cm is

\[\begin{array}{1 1}(1)23\pi&(2)33\pi\\(3)43\pi&(4)53\pi\end{array}\]

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Let V be the volume, r be the radius h be the height of the cylinder for the time 't'.
$V= \pi r^2 h$
Given $\large\frac{dr}{st}$$=-3 ;cm sec$
$r=3 \;cm$
$h=5 \;cm$
$V= \pi r^2 h$
$\large\frac{dV}{dt}$$=\pi \bigg[ r^2 \large\frac{dh}{dt} $$+h 2 r \large\frac{dr}{dt}\bigg]$
$\qquad= \pi [9 \times -3 +2 \times 2 \times 5 \times 3 \times 2 ]$
$\qquad =33 \pi$
Hence 2 is the correct answer
answered May 19, 2014 by meena.p

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