Related Rates
Sometimes two quantities are tied together by an equation, and both are changing
with time. If you know how fast one changes, the equation tells you how fast the other must
change. These are related rates problems, and the engine that powers them is
the chain rule.
The trick: differentiate the relationship with respect to time
t. Every variable picks up a \frac{d}{dt}
factor through the chain rule, turning a static equation into a relationship between rates.
The expanding balloon
Air is pumped into a spherical balloon so its radius grows. Volume and radius are related by
V = \tfrac{4}{3}\pi r^3. Differentiate both sides with respect to
t — the chain rule gives the 3r^2 and a
\frac{dr}{dt}:
\frac{dV}{dt} = 4\pi r^2 \,\frac{dr}{dt}
Read it: the bigger the balloon, the more air it takes to nudge the radius — the same
\frac{dr}{dt} needs a far larger \frac{dV}{dt}.
Drag the radius and watch the surface factor 4\pi r^2 climb.
The sliding ladder
A 10 m ladder leans against a wall; its foot is pulled out along
the ground. The top and bottom are tied by Pythagoras, x^2 + y^2 = 100.
Differentiate with respect to t:
2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \;\Rightarrow\; \frac{dy}{dt} = -\frac{x}{y}\,\frac{dx}{dt}
Watch the animation: as the foot slides out at a steady pace, the top slides
down — and notice it speeds up dramatically as the ladder nears flat, because the
x/y factor blows up when y\to 0.
The four-step method
- Relate the variables with an equation (geometry, area, Pythagoras…).
- Differentiate both sides with respect to t (chain rule everywhere).
- Substitute the known values and known rate.
- Solve for the unknown rate.
Example: when the balloon's radius is r = 5 cm and air flows at
\frac{dV}{dt} = 100 cm³/s, then
100 = 4\pi(25)\frac{dr}{dt}, so
\frac{dr}{dt} = \frac{1}{\pi} \approx 0.32 cm/s.
Watch it on Khan Academy