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Rate of change of radius

A spherical balloon is being inflated so that air is pumped into it at a constant rate of 100 cm3/s100 \text{ cm}^3/\text{s}. The balloon's volume is V=43πr3V = \frac{4}{3}\pi r^3 and its surface area is S=4πr2S = 4\pi r^2, where rr is the radius.

At the moment when the radius of the balloon is r=5r = 5 cm, find the rate at which the radius is increasing, drdt\dfrac{dr}{dt}, in cm/s.

Hint

Differentiate V=43πr3V = \frac{4}{3}\pi r^3 implicitly with respect to tt to relate dVdt\frac{dV}{dt} and drdt\frac{dr}{dt}, then substitute the known values dVdt=100\frac{dV}{dt}=100 and r=5r=5.

Step-by-step solution

Start with the volume formula V=43πr3V = \frac{4}{3}\pi r^3. Differentiating both sides with respect to tt using the chain rule gives dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}, since rr is a function of time.

Substitute the known values dVdt=100\frac{dV}{dt} = 100 and r=5r = 5: 100=4π(5)2drdt=100πdrdt100 = 4\pi (5)^2 \frac{dr}{dt} = 100\pi \frac{dr}{dt}. Solving gives drdt=1π≈0.318\frac{dr}{dt} = \frac{1}{\pi} \approx 0.318 cm/s.

Rate of change of surface area

At the same moment, when r=5r = 5 cm, find the rate at which the surface area of the balloon is increasing, dSdt\dfrac{dS}{dt}, in cm2^2/s.

Hint

Differentiate S=4πr2S = 4\pi r^2 with respect to tt to get dSdt\frac{dS}{dt} in terms of rr and drdt\frac{dr}{dt}, then use the value of drdt\frac{dr}{dt} found in part (a).

Step-by-step solution

Differentiate S=4πr2S = 4\pi r^2 with respect to tt: dSdt=8πrdrdt\frac{dS}{dt} = 8\pi r \frac{dr}{dt}. This relates the rate of change of surface area to the rate of change of radius.

Substitute r=5r = 5 and drdt=1π\frac{dr}{dt} = \frac{1}{\pi} from part (a): dSdt=8π(5)(1π)=40\frac{dS}{dt} = 8\pi (5)\left(\frac{1}{\pi}\right) = 40 cm2^2/s.

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