Related rates of change

The idea

• A rate of change is a derivative with respect to time: $\dfrac{dV}{dt}$ is how fast a volume is changing, in units per second.
• In a related rates problem two quantities are linked by a formula, and you know one rate and want the other.
• The chain rule is what connects them.

The linking chain rule

$$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$$

• Read it as: the rate you want = (how $A$ depends on $r$) × (the rate you know).

• The $dr$ terms "cancel" in the notation, which is a useful check that you have written the chain the right way round.

• Longer chains work the same way:

$$\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$$

The method

• Identify the rate you are given and the rate you want, and write both in derivative notation.
• Find a formula connecting the two quantities.
• Differentiate that formula with respect to the variable it contains.
• Build the chain that links the wanted rate to the given rate.
• Substitute the known values — at the very end, never before differentiating.
• State the answer with units, and say whether the quantity is increasing or decreasing.

Writing rates in symbols

• Turning words into notation is half the problem:

  • "the radius increases at 2 cm/s" — $\dfrac{dr}{dt} = 2$
  • "the volume decreases at 5 cm³/s" — $\dfrac{dV}{dt} = -5$ (negative for a decrease)
  • "how fast is the area growing?" — find $\dfrac{dA}{dt}$
  • "when the radius is 3 cm" — substitute $r = 3$ after differentiating

• A decreasing quantity has a negative rate. Missing that sign flips the whole answer.

A worked pattern — the expanding sphere

• A balloon's radius grows at 0.5 cm/s. How fast is the volume growing when $r = 10$ cm?

  • Given: $\dfrac{dr}{dt} = 0.5$. Want: $\dfrac{dV}{dt}$ at $r = 10$.
  • Formula: $V = \frac{4}{3}\pi r^3$
  • Differentiate: $\dfrac{dV}{dr} = 4\pi r^2$
  • Chain: $\dfrac{dV}{dt} = 4\pi r^2 \times 0.5 = 2\pi r^2$
  • Substitute: at $r = 10$, $\dfrac{dV}{dt} = 200\pi \approx 628$ cm³/s

• The 2025 report lists "solved a related rates problem, involving the volume of a sphere" as a Merit behaviour.

When there are two variables in the formula

• A cone's volume $V = \frac{1}{3}\pi r^2 h$ depends on both $r$ and $h$, which cannot be differentiated as it stands.
• Use the shape's fixed proportions to eliminate one variable — exactly as in an optimisation constraint.
• If a cone is always twice as tall as it is wide, then $h = 4r$, and substituting gives $V = \frac{4}{3}\pi r^3$, a one-variable formula.

Reversing the chain

• Sometimes the rate you want is the inner one:

$$\frac{dr}{dt} = \frac{dV}{dt} \div \frac{dV}{dr}$$

• Note that $\dfrac{dr}{dV} = \dfrac{1}{\;dV/dr\;}$ — derivatives can be inverted like fractions in this notation, which makes rearranging safe.

Units

• The units of a rate are always (units of the quantity) per (unit of time):
  • $\dfrac{dV}{dt}$ — cm³ s⁻¹
  • $\dfrac{dA}{dt}$ — cm² s⁻¹
  • $\dfrac{dh}{dt}$ — cm s⁻¹
• The specification notes that inappropriate use of units "may count as a minor error", so give them.