The radius of the base of a cylinder is increasing at a rate of $7$ millimeters per hour. $\newline$ The height of the cylinder is fixed at $1$ . $5$ millimeters. $\newline$ At a certain instant, the radius is $12$ millimeters. $\newline$ What is the rate of change of the volume of the cylinder at that instant (in cubic millimeters per hour)? $\newline$ Choose $1$ answer: $\newline$ (A) $216 \pi$ $\newline$ (B) $1512 \pi$ $\newline$ (C) $126 \pi$ $\newline$ (D) $252 \pi$ $\newline$ The volume of a cylinder with radius $r$ and height $h$ is $\pi r^{2} h$ .

Full solution

Q. The radius of the base of a cylinder is increasing at a rate of

7

millimeters per hour.

\newline

The height of the cylinder is fixed at

1

5

millimeters.

\newline

At a certain instant, the radius is

12

millimeters.

\newline

What is the rate of change of the volume of the cylinder at that instant (in cubic millimeters per hour)?

\newline

Choose

1

answer:

\newline

(A)

216 \pi

\newline

(B)

1512 \pi

\newline

(C)

126 \pi

\newline

(D)

252 \pi

\newline

The volume of a cylinder with radius

r

and height

h

\pi r^{2} h

Volume Formula Derivation: The formula for the volume of a cylinder is $V = \pi r^2 h$ . We need to find $\frac{dV}{dt}$ , the rate of change of the volume.
Differentiating Volume Formula: First, let's differentiate $V$ with respect to $t$ : $\frac{dV}{dt} = \frac{d(\pi r^2 h)}{dt}$ . Since $h$ is constant, we can take it outside the differentiation: $\frac{dV}{dt} = h \cdot \frac{d(\pi r^2)}{dt}$ .
Differentiating $\pi r^2$ with respect to $r$ : Now differentiate $\pi r^2$ with respect to $r$ : $\frac{d(\pi r^2)}{dr} = 2\pi r$ . Then multiply by $\frac{dr}{dt}$ to get $\frac{d(\pi r^2)}{dt}$ : $\frac{d(\pi r^2)}{dt} = 2\pi r \cdot \frac{dr}{dt}$ .
Calculating $\frac{d(\pi r^2)}{dt}$ : We know $\frac{dr}{dt} = 7\,\text{mm/hour}$ (the rate at which the radius is increasing) and $r = 12\,\text{mm}$ (the radius at the instant we're interested in).
Substitute Values: Plug in the values: $\frac{d(\pi r^2)}{dt} = 2\pi \times 12 \text{ mm} \times 7 \text{ mm/hour} = 168\pi \text{ mm}^2/\text{hour}.$
Calculate dV/dt: Now, multiply by the height $h = 1.5$ mm to find $dV/dt$ : $dV/dt = 168\pi\, \text{mm}^2/\text{hour} \times 1.5\, \text{mm} = 252\pi\, \text{mm}^3/\text{hour}$ .