AP Calculus ABmediummcq1 pt

A spherical balloon is being inflated. If the radius r is increasing at a constant rate of 2 cm/s, at what instantaneous rate is the surface area S = 4πr² changing when r = 5 cm?

Explanation

Core Concept

We need dS/dt = dS/dr · dr/dt. Since S = 4πr², we have dS/dr = 8πr. Therefore dS/dt = 8πr · dr/dt = 8π(5)(2) = 80π cm²/s.

Correct Answer

80pi

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