# related rates problem-calculus

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A spherical balloon is being filled with air in such a way that its radius is increasing at a rate of 2 centimeters per second. At what rate is the volume of the balloon changing at the instant when its surface has an area of 4 pi square centimeters?

• related rates problem-calculus -

The balloon is spherical, so
V = (4/3)(pi)r^3
SA = 4(pi)r^2

Differentiate both with respect to time.

dV/dt = 4(pi)r^2(dr/dt)
dSA/dt = 8(pi)r(dr/dt)

We're given the rate of change in SA and need to find the rate of change in volume. Let's write an equation for it.

(dV/dt)/r = 4(pi)r(dr/dt)
(dSA/dt)/2 = 4(pi)r(dr/dt)

(dV/dt)/r = (dSA/dt)/2
(dV/dt) = (r/2)(dSA/dt)

• related rates problem-calculus -

Edit: actually we're given dr/dt and SA.

Since dV/dt = 4(pi)r^2(dr/dt) and SA = 4(pi)r^2, dV/dt = SA(dr/dt).

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