The radius of a right circular cylinder incre | vedclass

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

Vedclass · Accepted Answer

$\frac{2\pi}{5} 	ext{ cm}^3	ext{/min}$

Vedclass · Answer

(D) The volume $V$ of a right circular cylinder is given by $V = \pi r^2 h$. Differentiating both sides with respect to time $t$,we get: $\frac{dV}{dt} = \pi \left( 2r \cdot \frac{dr}{dt} \cdot h + r^2 \cdot \frac{dh}{dt} ight)$. Given: $r = 2 	ext{ cm}$,$h = 3 	ext{ cm}$,$\frac{dr}{dt} = 0.1 	ext{ cm/min}$,and $\frac{dh}{dt} = -0.2 	ext{ cm/min}$. Substituting these values: $\frac{dV}{dt} = \pi \left( 2 	imes 2 	imes 0.1 	imes 3 + 2^2 	imes (-0.2) ight)$ $= \pi \left( 1.2 - 0.8 ight) = 0.4 \pi = \frac{2\pi}{5} 	ext{ cm}^3	ext{/min}$.

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

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