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) $252 \pi$ $\newline$ (B) $216 \pi$ $\newline$ (C) $1512 \pi$ $\newline$ (D) $126 \pi$ $\newline$ 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)

252 \pi

\newline

(B)

216 \pi

\newline

(C)

1512 \pi

\newline

(D)

126 \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 volume with respect to time.
Given Information: Given that $\frac{dr}{dt} = 7 \, \text{mm/hour}$ (rate at which radius is increasing) and $h = 1.5 \, \text{mm}$ (height of the cylinder is constant).
Differentiate Volume Formula: Differentiate the volume formula with respect to time: $\frac{dV}{dt} = \frac{d(\pi r^2 h)}{dt} = \pi h \cdot \frac{d(r^2)}{dt}$ .
Differentiate $r^2$ with respect to time: Now differentiate $r^2$ with respect to time: $\frac{d(r^2)}{dt} = 2r \cdot \frac{dr}{dt}$ .
Substitute Values: Substitute the values of $\frac{dr}{dt}$ and $r$ into the equation: $\frac{d(r^2)}{dt} = 2 \times 12 \, \text{mm} \times 7 \, \text{mm/hour} = 168 \, \text{mm}^2/\text{hour}.$
Calculate $dV/dt$ : Now substitute $d(r^2)/dt$ and $h$ into the $dV/dt$ equation: $dV/dt = \pi \times 1.5$ mm $\times 168$ mm $^2$ /hour.
Calculate $dV/dt$ : Now substitute $d(r^2)/dt$ and $h$ into the $dV/dt$ equation: $dV/dt = \pi \times 1.5 \, \text{mm} \times 168 \, \text{mm}^2/\text{hour}$ .Calculate $dV/dt$ : $dV/dt = \pi \times 1.5 \times 168 = 252\pi \, \text{mm}^3/\text{hour}$ .