The radius of the base of a cylinder is decreasing at a rate of $12$ kilometers per second. $\newline$ The height of the cylinder is fixed at $2$ . $5$ kilometers. $\newline$ At a certain instant, the radius is $40$ kilometers. $\newline$ What is the rate of change of the volume of the cylinder at that instant (in cubic kilometers per second)? $\newline$ Choose $1$ answer: $\newline$ (A) $-2400 \pi$ $\newline$ (B) $-4000 \pi$ $\newline$ (C) $-1600 \pi$ $\newline$ (D) $-360 \pi$ $\newline$ 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 decreasing at a rate of

12

kilometers per second.

\newline

The height of the cylinder is fixed at

2

5

kilometers.

\newline

At a certain instant, the radius is

40

kilometers.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

-2400 \pi

\newline

(B)

-4000 \pi

\newline

(C)

-1600 \pi

\newline

(D)

-360 \pi

\newline

The volume of a cylinder with radius

r

and height

h

\pi r^{2} h

Differentiate Volume Formula: First, let's differentiate the volume formula with respect to time $t$ to find $\frac{dV}{dt}$ . $\newline$ $\frac{dV}{dt} = \frac{d(\pi r^2 h)}{dt}$ $\newline$ Since $h$ is constant, we can take it outside the differentiation. $\newline$ $\frac{dV}{dt} = \pi h \cdot \frac{d(r^2)}{dt}$
Differentiate $r^2$ : Now, we differentiate $r^2$ with respect to time $t$ . $\newline$ $\frac{d(r^2)}{dt} = 2r \cdot \frac{dr}{dt}$
Calculate $\frac{dV}{dt}$ : We plug in the values for $\frac{dr}{dt}$ , $r$ , and $h$ into the differentiated volume formula. $\newline$ $\frac{dV}{dt} = \pi h \cdot (2r \cdot \frac{dr}{dt})$ $\newline$ $\frac{dV}{dt} = \pi \cdot 2.5 \cdot (2 \cdot 40 \cdot -12)$
Calculate final value: Now, we calculate the value. $\newline$ $\frac{dV}{dt} = \pi \times 2.5 \times (2 \times 40 \times -12)$ $\newline$ $\frac{dV}{dt} = \pi \times 2.5 \times (-960)$ $\newline$ $\frac{dV}{dt} = -2400\pi \, \text{km}^3/s$