The radius of a right circular cylinder incre | vedclass

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

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$(\frac{2\pi}{5})$

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(D) The volume of a right circular cylinder is given by $V = \pi r^2h$.Differentiating both sides with respect to time $t$,we get:$\frac{dV}{dt} = \pi \left( r^2 \frac{dh}{dt} + 2rh \frac{dr}{dt} ight)$.Given that $\frac{dr}{dt} = 0.1 = \frac{1}{10} 	ext{ cm/min}$ and $\frac{dh}{dt} = -0.2 = -\frac{2}{10} 	ext{ cm/min}$.Substituting these values into the derivative expression:$\frac{dV}{dt} = \pi \left( r^2 \left( -\frac{2}{10} ight) + 2rh \left( \frac{1}{10} ight) ight)$.$\frac{dV}{dt} = \frac{\pi r}{10} (-2r + 2h) = \frac{2\pi r}{10} (h - r) = \frac{\pi r}{5} (h - r)$.At $r = 2 	ext{ cm}$ and $h = 3 	ext{ cm}$:$\frac{dV}{dt} = \frac{\pi (2)}{5} (3 - 2) = \frac{2\pi}{5} 	ext{ cm}^3/	ext{min}$.

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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