MISSION · Calculus
Related Rates
When two things are linked, their speeds are linked too
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The idea in one sentence
If two quantities are tied together by an equation, differentiating that equation with respect to time ties their rates together too.
The ladder
A 5 m ladder leans against a wall. Its foot is pulled away at 1 m/s. How fast does the top slide down?
The link is Pythagoras, and it holds at every instant:
\[x^2 + y^2 = 25\]Differentiate both sides with respect to time. Each variable is a function of \(t\), so the chain rule attaches its rate:
\[2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\]Now substitute the moment you care about. When the foot is 3 m out, \(y = 4\), and \(\frac{dx}{dt} = 1\):
\[6(1) + 8\frac{dy}{dt} = 0 \quad \Rightarrow \quad \frac{dy}{dt} = -0.75 \text{ m/s}\]The negative sign is the answer telling you the height is decreasing, which it obviously is. Signs in these problems are information, not bookkeeping.
The surprise worth noticing
The top does not descend at a steady rate, even though the foot moves steadily. As the ladder flattens, \(y\) shrinks and \(\frac{dy}{dt}\) grows without bound — the top is racing down at the moment the ladder goes flat.
That’s the whole point of related rates. A constant rate in one quantity almost never means a constant rate in the other, and intuition is a poor guide to which way the difference goes.
The method
- Draw it, and label every quantity that changes with a letter — never a number.
- Write the equation linking them.
- Differentiate with respect to \(t\), applying the chain rule to every variable.
- Now substitute the values for the instant in question.
- Solve for the rate you want, and check the sign makes sense.
Step 4 is where most attempts fail. If you substitute \(x = 3\) before differentiating, \(x\) becomes a constant and its derivative vanishes — you’ve told the equation the foot isn’t moving, and the answer comes out zero.
Typical setups
- Expanding circle. \(A = \pi r^2\), so \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\). A ripple growing at a constant radius rate gains area faster and faster.
- Draining cone. Volume links to depth, and the surface shrinks as it empties, so the level falls at an accelerating rate.
- Two cars leaving a junction. Pythagoras again, with both legs changing at once.
Common mix-ups
- Substituting before differentiating. Keep the variables as variables.
- Forgetting the chain rule. \(\frac{d}{dt}(y^2) = 2y\frac{dy}{dt}\), not \(2y\).
- Ignoring the sign. A shrinking quantity has a negative rate, and putting it in positive gives an answer that’s wrong in direction.
- Mismatched units. Metres with seconds, consistently, or the answer means nothing.
Check yourself
A circular ripple’s radius grows at 2 cm/s. How fast is its area growing when the radius is 10 cm?
\(\frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 2\pi(10)(2) = 40\pi \approx 126\) cm²/s. At twice the radius it would be twice as fast again.
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