(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

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

(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: $\frac{dV}{dt} = \pi \left( 2r \cdot \frac{dr}{dt} \cdot h + r^2 \cdot \frac{dh}{dt} \right)$. Given: $r = 2 \text{ cm}$,$h = 3 \text{ cm}$,$\frac{dr}{dt} = 0.1 \text{ cm/min}$,and $\frac{dh}{dt} = -0.2 \text{ cm/min}$. Substituting these values: $\frac{dV}{dt} = \pi \left( 2 \times 2 \times 0.1 \times 3 + 2^2 \times (-0.2) \right)$ $= \pi \left( 1.2 - 0.8 \right) = 0.4 \pi = \frac{2\pi}{5} \text{ cm}^3\text{/min}$.

Class 12 Mathematics Application of Derivatives Rate of Change of Quantities

Medium

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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A
$-2 \pi \text{ cm}^3\text{/min}$
B
$\frac{-3\pi}{5} \text{ cm}^3\text{/min}$
C
$\frac{-8\pi}{5} \text{ cm}^3\text{/min}$
D
$\frac{2\pi}{5} \text{ cm}^3\text{/min}$

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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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The sides of an equilateral triangle are increasing at the rate of $2 \, cm/sec$. The rate at which the area increases,when the side is $10 \, cm$ is:

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