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William was given this problem: 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? Which equation should William use to solve the problem? Choose 1 answer: (A) S(t)=2pi*r(t) (B) S(t)=pi[r(t)]^(2) (C) S(t)=4pi[r(t)]^(2) (D) S(t)=(4)/(3)pi[r(t)]^(3)

William was given this problem:\newlineThe radius r(t) r(t) of a sphere is increasing at a rate of 77.55 meters per minute. At a certain instant t0 t_{0} , the radius is 55 meters. What is the rate of change of the surface area S(t) S(t) of the sphere at that instant?\newlineWhich equation should William use to solve the problem?\newlineChoose 11 answer:\newline(A) S(t)=2π⋅r(t) S(t)=2 \pi \cdot r(t) \newline(B) S(t)=π[r(t)]2 S(t)=\pi[r(t)]^{2} \newline(C) S(t)=4π[r(t)]2 S(t)=4 \pi[r(t)]^{2} \newline(D) S(t)=43π[r(t)]3 S(t)=\frac{4}{3} \pi[r(t)]^{3}

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Q. William was given this problem:\newlineThe radius r(t) r(t) of a sphere is increasing at a rate of 77.55 meters per minute. At a certain instant t0 t_{0} , the radius is 55 meters. What is the rate of change of the surface area S(t) S(t) of the sphere at that instant?\newlineWhich equation should William use to solve the problem?\newlineChoose 11 answer:\newline(A) S(t)=2π⋅r(t) S(t)=2 \pi \cdot r(t) \newline(B) S(t)=π[r(t)]2 S(t)=\pi[r(t)]^{2} \newline(C) S(t)=4π[r(t)]2 S(t)=4 \pi[r(t)]^{2} \newline(D) S(t)=43π[r(t)]3 S(t)=\frac{4}{3} \pi[r(t)]^{3}
  1. Identify Formula: Identify the correct formula for the surface area of a sphere.\newlineThe surface area S(t)S(t) of a sphere with radius r(t)r(t) is given by the formula S(t)=4π[r(t)]2S(t) = 4\pi[r(t)]^2. This is because the surface area of a sphere is 44 times π\pi times the square of the radius.
  2. 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)=2π⋅r(t)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)=π[r(t)]2S(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)=4π[r(t)]2S(t) = 4\pi[r(t)]^2 is the correct formula for the surface area of a sphere.\newline(D) S(t)=43π[r(t)]3S(t) = \frac{4}{3}\pi[r(t)]^3 is the formula for the volume of a sphere, not the surface area.\newlineTherefore, the correct equation that William should use to solve the problem is (C) S(t)=4π[r(t)]2S(t) = 4\pi[r(t)]^2.
  3. Calculate Rate of Change: Calculate the rate of change of the surface area S(t)S(t) with respect to time tt. To find the rate of change of the surface area with respect to time, we need to differentiate the surface area function S(t)S(t) with respect to time tt. This is done using the chain rule of differentiation. dSdt=ddt[4π[r(t)]2]=4π⋅2r(t)⋅drdt\frac{dS}{dt} = \frac{d}{dt} [4\pi[r(t)]^2] = 4\pi \cdot 2r(t) \cdot \frac{dr}{dt}
  4. Substitute Given Values: Substitute the given rate of change of the radius and the radius at the instant t0t_0. We are given that the radius r(t)r(t) is increasing at a rate of drdt=7.5\frac{dr}{dt} = 7.5 meters per minute and that at the instant t0t_0, the radius r(t)r(t) is 55 meters. dSdt=4π×2×5\frac{dS}{dt} = 4\pi \times 2 \times 5 meters ×7.5\times 7.5 meters/minute =4π×10×7.5= 4\pi \times 10 \times 7.5 meters2^2/minute
  5. Perform Calculation: Perform the calculation to find the rate of change of the surface area. dSdt=4π×10×7.5\frac{dS}{dt} = 4\pi \times 10 \times 7.5 meters2^2/minute =300π= 300\pi meters2^2/minute This is the rate of change of the surface area of the sphere at the instant when the radius is 55 meters.

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