The radius of a sphere is decreasing at a rate of $4$ centimeters per second. $\newline$ At a certain instant, the radius is $10$ centimeters. $\newline$ What is the rate of change of the surface area of the sphere at that instant (in square centimeters per second)? $\newline$ Choose $1$ answer: $\newline$ (A) $-64 \pi$ $\newline$ (B) $-160 \pi$ $\newline$ (C) $-320 \pi$ $\newline$ (D) $-400 \pi$ $\newline$ The surface area of a sphere with radius $r$ is $4 \pi r^{2}$ .

Full solution

Q. The radius of a sphere is decreasing at a rate of

4

centimeters per second.

\newline

At a certain instant, the radius is

10

centimeters.

\newline

What is the rate of change of the surface area of the sphere at that instant (in square centimeters per second)?

\newline

Choose

1

answer:

\newline

(A)

-64 \pi

\newline

(B)

-160 \pi

\newline

(C)

-320 \pi

\newline

(D)

-400 \pi

\newline

The surface area of a sphere with radius

r

4 \pi r^{2}

Surface Area Formula: The formula for the surface area of a sphere is $S = 4\pi r^2$ . We need to find the rate of change of the surface area, which is $\frac{dS}{dt}$ .
Chain Rule Application: To find $\frac{dS}{dt}$ , we use the chain rule from calculus: $\frac{dS}{dt} = \frac{dS}{dr} \cdot \frac{dr}{dt}$ . We know $\frac{dr}{dt} = -4 \, \text{cm/s}$ (since the radius is decreasing).
Calculate $dS/dr$ : First, we find $dS/dr$ by differentiating $S = 4\pi r^2$ with respect to $r$ . $dS/dr = 8\pi r$ .
Substitute $r=10$ cm: Now we plug in the value of $r = 10$ cm into $\frac{dS}{dr}$ to get $\frac{dS}{dr} = 8\pi(10) = 80\pi$ .
Find $\frac{dS}{dt}$ : Finally, we multiply $\frac{dS}{dr}$ by $\frac{dr}{dt}$ to find $\frac{dS}{dt}$ : $\frac{dS}{dt} = 80\pi \times (-4) = -320\pi$ square centimeters per second.