§3.1 Related rates

Introduction

A spherical balloon has volume $V = \frac{4}{3}\pi r^3$. Differentiating both sides with respect to time $t$ gives

The balloon is inflated at $\frac{dV}{dt} = 20$ cubic inches per second. At the instant the diameter is $12$ inches, the radius satisfies $\frac{dr}{dt} = \frac{20}{4\pi r^2}$, which equals

With constant inflation rate, is the radius changing more rapidly when $d = 12$ or when $d = 16$?

For the balloon with $\frac{dV}{dt} = 20$ cubic inches per second, compute $\frac{dr}{dt}$ when the diameter is $12$ inches, rounded to three decimals.

Suppose the same balloon has radius growing at $\frac{dr}{dt} = 0.5$ inches per second when $r = 2$. Compute $\frac{dV}{dt}$ at that instant, rounded to three decimals.

The equation $\frac{dr}{dt} = \frac{20}{4\pi r^2}$ comes from differentiating $V = \frac{4}{3}\pi r^3$ with respect to $t$ and then solving for $\frac{dr}{dt}$.

Sketching several spheres of different radii suggests the question: as the balloon inflates, which quantities are changing?