William was given this problem: $\newline$ The radius $r(t)$ of a sphere is increasing at a rate of $7$ . $5$ meters per minute. At a certain instant $t_{0}$ , the radius is $5$ meters. What is the rate of change of the surface area $S(t)$ of the sphere at that instant? $\newline$ Which equation should William use to solve the problem? $\newline$ Choose $1$ answer: $\newline$ (A) $S(t)=2 \pi \cdot r(t)$ $\newline$ (B) $S(t)=\pi[r(t)]^{2}$ $\newline$ (C) $S(t)=4 \pi[r(t)]^{2}$ $\newline$ (D) $S(t)=\frac{4}{3} \pi[r(t)]^{3}$

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Q. William was given this problem:

\newline

The radius

r(t)

of a sphere is increasing at a rate of

7

5

meters per minute. At a certain instant

t_{0}

, the radius is

5

meters. What is the rate of change of the surface area

S(t)

of the sphere at that instant?

\newline

Which equation should William use to solve the problem?

\newline

Choose

1

answer:

\newline

(A)

S(t)=2 \pi \cdot r(t)

\newline

(B)

S(t)=\pi[r(t)]^{2}

\newline

(C)

S(t)=4 \pi[r(t)]^{2}

\newline

(D)

S(t)=\frac{4}{3} \pi[r(t)]^{3}

Identify Formula: Identify the correct formula for the surface area of a sphere. $\newline$ The surface area $S(t)$ of a sphere with radius $r(t)$ is given by the formula $S(t) = 4\pi[r(t)]^2$ . This is because the surface area of a sphere is $4$ times $\pi$ times the square of the radius.
Determine Correct Option: Determine which of the given options corresponds to the correct formula for the surface area of a sphere. $\newline$ (A) $S(t) = 2\pi\cdot r(t)$ is the formula for the circumference of a circle, not the surface area of a sphere. $\newline$ (B) $S(t) = \pi[r(t)]^2$ is the formula for the area of a circle, not the surface area of a sphere. $\newline$ (C) $S(t) = 4\pi[r(t)]^2$ is the correct formula for the surface area of a sphere. $\newline$ (D) $S(t) = \frac{4}{3}\pi[r(t)]^3$ is the formula for the volume of a sphere, not the surface area. $\newline$ Therefore, the correct equation that William should use to solve the problem is (C) $S(t) = 4\pi[r(t)]^2$ .
Calculate Rate of Change: Calculate the rate of change of the surface area $S(t)$ with respect to time $t$ . To find the rate of change of the surface area with respect to time, we need to differentiate the surface area function $S(t)$ with respect to time $t$ . This is done using the chain rule of differentiation. $\frac{dS}{dt} = \frac{d}{dt} [4\pi[r(t)]^2] = 4\pi \cdot 2r(t) \cdot \frac{dr}{dt}$
Substitute Given Values: Substitute the given rate of change of the radius and the radius at the instant $t_0$ . We are given that the radius $r(t)$ is increasing at a rate of $\frac{dr}{dt} = 7.5$ meters per minute and that at the instant $t_0$ , the radius $r(t)$ is $5$ meters. $\frac{dS}{dt} = 4\pi \times 2 \times 5$ meters $\times 7.5$ meters/minute $= 4\pi \times 10 \times 7.5$ meters $^2$ /minute
Perform Calculation: Perform the calculation to find the rate of change of the surface area. $\frac{dS}{dt} = 4\pi \times 10 \times 7.5$ meters $^2$ /minute $= 300\pi$ meters $^2$ /minute This is the rate of change of the surface area of the sphere at the instant when the radius is $5$ meters.