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The radius of a sphere is increasing at a rate of 7.5 meters per minute.
At a certain instant, the radius is 5 meters.
What is the rate of change of the surface area of the sphere at that instant (in square meters per minute)?
Choose 1 answer:
(A)
330 pi
(B)
100 pi
(C)
300 pi
(D)
225 pi
The surface area of a sphere with radius
r is
4pir^(2).

The radius of a sphere is increasing at a rate of $7$ . $5$ meters per minute. $\newline$ At a certain instant, the radius is $5$ meters. $\newline$ What is the rate of change of the surface area of the sphere at that instant (in square meters per minute)? $\newline$ Choose $1$ answer: $\newline$ (A) $330 \pi$ $\newline$ (B) $100 \pi$ $\newline$ (C) $300 \pi$ $\newline$ (D) $225 \pi$ $\newline$ The surface area of a sphere with radius $r$ is $4 \pi r^{2}$ .

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Area of quadrilaterals and triangles: word problems

Full solution

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

7

.

5

meters per minute.

\newline

At a certain instant, the radius is

5

meters.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

330 \pi

\newline

(B)

100 \pi

\newline

(C)

300 \pi

\newline

(D)

225 \pi

\newline

The surface area of a sphere with radius

r

is

4 \pi r^{2}

.

Formula Differentiation: The formula for the surface area of a sphere is $S = 4 \pi r^2$ . To find the rate of change of the surface area, we need to differentiate this formula with respect to time ( $t$ ).
Rate of Change Calculation: $\frac{dS}{dt} = \frac{d}{dt} (4 \pi r^2) = 8 \pi r \frac{dr}{dt}$ , where $\frac{dr}{dt}$ is the rate of change of the radius.
Substitute Values: We know that $\frac{dr}{dt} = 7.5$ meters per minute and $r = 5$ meters. Plug these values into the differentiated formula to find $\frac{dS}{dt}$ .
Final Result: $\frac{dS}{dt} = 8 \times \pi \times 5 \times 7.5 = 300 \times \pi$ square meters per minute.

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