The volume of a sphere is increasing at $8\pi$ cm³/s. The rate of change of surface area when the radius is 2 cm is:
A$2\pi$ cm²/s
B$4\pi$ cm²/s
C$8\pi$ cm²/s
D$16\pi$ cm²/s
Answer & Solution
Correct answer: C. $8\pi$ cm²/s
$V = (4/3)\pi r^3$ ⇒ $dV/dt = 4\pi r^2 \cdot dr/dt = 8\pi$.
At $r = 2$: $4\pi(4)\cdot dr/dt = 8\pi$ ⇒ $dr/dt = 1/2$.
$S = 4\pi r^2$ ⇒ $dS/dt = 8\pi r \cdot dr/dt = 8\pi(2)(1/2) = 8\pi$ cm²/s.
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