Air is being pumped into a spherical balloon so that its volume increases at a rate of 80cm^3/s. How fast is the surface area of the balloon increasing when its radius is 7cm?

Question

Air is being pumped into a spherical balloon so that its volume increases at a rate of 80cm^3/s. How fast is the surface area of the balloon increasing when its radius is 7cm? Recall that a ball of radius r has volume V=4/3 πr^3 and surface area S=4πr^2.

Answer 1

No where is it mentioned the rate of change of the radius dr/dt, but that value appears in the derivative with respect to time of both volume and surface area.

Let;s see what happens if we do take derivatives with respect to time.

dV/dt = 4πr²(dr/dt) = 80cm³/s -- think about solving this for (dr/dt) and substituting into the next.

dS/dt = 8πr (dr/dt) = 8πr (80/4πr²)

Simplify and then substitute 7 for r (given). dS/dt will have units cm²/s.