Air is pumped into a spherical balloon so that its volume increases at a constant cubic centimeters per second. At the instant when the balloon's surface area is square centimeters, at what rate is the surface area increasing, in square centimeters per second? (For a sphere, and .)
Answer and explanation
Correct answer: A.
Step 3. Differentiate surface area : . Substitute and : .
- Step 1: Determine the radius when surface area : .
- Step 2: Relate to : differentiating gives . Substitute and : $36pi = 4pi(36)dfrac{dr}{dt} implies dfrac{dr}{dt} = dfrac{1}{4} ext{cm/s}$.
Answer
.
Why not the others: Other values result from misapplying the chain rule on the volume formula or failing to solve for the radius from the surface area.
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