Mathematics

Problem

Air is pumped into a spherical balloon at a constant rate of

0.040\text{ m}^{3}\text{ s}^{-1}

. Let

r

be the radius (in meters) of the balloon,

V

its volume (in cubic meters), and

S

its surface area (in square meters). The formulas

V=\dfrac{4}{3}\pi r^{3}

and

S=4\pi r^{2}

apply. (a) Derive an expression for the rate of change

\dfrac{dS}{dt}

of the surface area in terms of the volume–inflation rate

\dfrac{dV}{dt}

and the instantaneous radius

r

. (b) The balloon material can safely withstand surface–area expansion rates up to

6.0\text{ m}^{2}\text{ s}^{-1}

. Using your result from part (a) and the given pumping rate, determine the radius (to the nearest centimeter) at which this safety limit is reached. (c) Explain, in context, what your answer in part (b) implies about the continued inflation of the balloon once its radius exceeds the computed value.

Intuitive explanation

This is a related rates problem where you connect how fast the volume changes to how fast the radius grows. Since you know $\frac{dV}{dt}$ (the pumping rate), you'll differentiate the volume formula with respect to time to get a relationship between $\frac{dr}{dt}$ and $\frac{dV}{dt}$ , then solve for how quickly the radius is expanding at any given moment. The key is treating $r$ as a function of time and using the chain rule.

Intuitive explanation

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