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(D) The volume of a cylinder is given by $V = \pi r^2 h$.Differentiating both sides with respect to time $t$,we get:$\frac{dV}{dt} = \pi \left( 2rh \f

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$33 \pi 	ext{ cm}^3/	ext{sec}$

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(D) The volume of a cylinder is given by $V = \pi r^2 h$.Differentiating both sides with respect to time $t$,we get:$\frac{dV}{dt} = \pi \left( 2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} ight)$Given: $r = 3 	ext{ cm}$,$h = 5 	ext{ cm}$,$\frac{dr}{dt} = 2 	ext{ cm/sec}$,and $\frac{dh}{dt} = -3 	ext{ cm/sec}$ (since height is decreasing).Substituting these values into the derivative:$\frac{dV}{dt} = \pi \left( 2 	imes 3 	imes 5 	imes 2 + 3^2 	imes (-3) ight)$$\frac{dV}{dt} = \pi \left( 60 - 27 ight)$$\frac{dV}{dt} = 33 \pi 	ext{ cm}^3/	ext{sec}$.

The radius of a cylinder is increasing at the rate of $2 \text{ cm/sec}$ and its height is decreasing at the rate of $3 \text{ cm/sec}$. Find the rate of change of volume when the radius is $3 \text{ cm}$ and the height is $5 \text{ cm}$.

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