(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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$33 \pi \text{ cm}^3/\text{sec}$

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

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

Medium

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 $5 \text{ cm}$.

MHT CET 2022 All MHT CET

A
$44 \pi \text{ cm}^3/\text{sec}$
B
$11 \pi \text{ cm}^3/\text{sec}$
C
$23 \pi \text{ cm}^3/\text{sec}$
D
$33 \pi \text{ cm}^3/\text{sec}$

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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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