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Pure · Differentiation

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Connected rates of change

Linking the rate of change of one quantity to another through a shared variable using the chain rule — dV/dt = (dV/dr)(dr/dt) — for problems like an expanding balloon, a draining cone or a sliding ladder.

Pure · Differentiation

Connected rates of change

Linking the rate of change of one quantity to another through a shared variable using the chain rule — dV/dt = (dV/dr)(dr/dt) — for problems like an expanding balloon, a draining cone or a sliding ladder.

Why it works

Many real situations have several quantities changing together over time: as a balloon inflates, its radius, surface area and volume all grow at once, locked together by geometry. You are usually told one rate and asked for another. The chain rule connects them through the variable they share.

Suppose a spherical balloon is inflated so that air is pumped in at dVdt=50 cm3s−1\frac{dV}{dt} = 50\ \text{cm}^3\text{s}^{-1}, and you want how fast the radius grows, drdt\frac{dr}{dt}, when r=5r = 5. Volume and radius are linked by V=43πr3V = \frac43\pi r^3, so dVdr=4πr2\frac{dV}{dr} = 4\pi r^2. The chain rule strings the rates together: dVdt=dVdr×drdt.\frac{dV}{dt} = \frac{dV}{dr}\times\frac{dr}{dt}. You know dVdt\frac{dV}{dt} and dVdr\frac{dV}{dr}; solve for the unknown: drdt=dV/dtdV/dr=504πr2=504π(25)=12π cm s−1.\frac{dr}{dt} = \frac{dV/dt}{dV/dr} = \frac{50}{4\pi r^2} = \frac{50}{4\pi(25)} = \frac{1}{2\pi}\ \text{cm s}^{-1}.

The reliable recipe:
  1. Name the rates. Write each "how fast …" as a derivative with respect to
tt — the one you're given and the one you want. ("Increasing" is positive, "decreasing" or "leaking" is negative.)
  1. Find the link. Write the equation connecting the two quantities (a volume,
area or Pythagoras formula), and differentiate it to get the bridging derivative.
  1. Chain them. Write d wantdt=d wantd link×d linkdt\frac{d\,\text{want}}{dt} = \frac{d\,\text{want}}{d\,\text{link}}\times\frac{d\,\text{link}}{dt},
substitute the known values, and solve. Put numbers in last, after differentiating — substituting r=5r = 5 too early would differentiate a constant and give zero.

If three quantities are involved, just chain through more factors — e.g. dAdt=dAdr⋅drdt\frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt} once you have drdt\frac{dr}{dt}.