Connected rates of change

Linking rates

Use the chain rule to connect what you know to what you want:

dr/dt = dr/dV × dV/dt = (dV/dt) ÷ (dV/dr).

Find the formula linking the quantities (such as V = (4/3)πr³), differentiate it, then substitute.

Forming differential equations

• 'The rate of change of N is proportional to N': dN/dt = kN.
• 'decreases' or 'cools': use a minus sign with k > 0, for example dθ/dt = -kθ.
• 'inversely proportional to A': dA/dt = k/A.

The derivative is always with respect to time t unless stated.

Units

Rates keep the units of the quantity per unit time, such as cm³ s^-1.

Worked example

A sphere's volume increases at 10 cm³ s^-1. Find the rate of increase of the radius when r = 5 cm.

1. V = (4/3)πr³, so dV/dr = 4πr² = 100π
2. dr/dt = (dV/dt) ÷ (dV/dr) = 10/(100π)

Answer: 0.0318 cm s^-1

Check yourself

1. The radius of a circle increases at 0.2 cm s^-1. Find the rate of increase of its area when r = 10 cm, to 3 significant figures.
   Show answer

   12.6 cm² s^-1. dA/dr = 2πr = 20π, so dA/dt = 20π × 0.2 = 4π = 12.6 cm² s^-1.

2. The side of a cube increases at 0.5 cm s^-1. Find the rate of increase of the volume when the side is 4 cm.
   Show answer

   24 cm³ s^-1. V = x³, dV/dx = 3x² = 48, so dV/dt = 48 × 0.5 = 24 cm³ s^-1.

3. Air is pumped into a spherical balloon at 20 cm³ s^-1. How fast is the radius growing when r = 2 cm? Give 3 significant figures.
   Show answer

   0.398 cm s^-1. dV/dr = 4πr² = 16π, so dr/dt = 20/(16π) = 0.398 cm s^-1.

4. A liquid cools at a rate proportional to its temperature θ above the room. Which equation models this (k > 0)?
   • dθ/dt = -kθ
   • dθ/dt = kθ
   • dθ/dt = -k/θ
   • θ = -kt
   Show answer

   dθ/dt = -kθ. Proportional to θ gives kθ; cooling means θ decreases, so the rate is negative.

5. The rate of increase of a population P is proportional to the square root of P. Which equation models this?
   • dP/dt = kP²
   • P = k sqrt(t)
   • dP/dt = k sqrt(P)
   • dP/dt = sqrt(kt)
   Show answer

   dP/dt = k sqrt(P). The rate is dP/dt, and it equals a constant times sqrt(P).

6. The area A of a spill increases at a rate inversely proportional to A. Which equation models this?
   • dA/dt = k/A
   • dA/dt = kA
   • dA/dt = -kA
   • A = k/t
   Show answer

   dA/dt = k/A. Inversely proportional to A means k divided by A.

7. Water fills a cone so that V = (1/3)πh³. Water flows in at 5 cm³ s^-1. Find dh/dt when h = 3 cm, to 3 significant figures.
   Show answer

   0.177 cm s^-1. dV/dh = πh² = 9π, so dh/dt = 5/(9π) = 0.177 cm s^-1.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3)
• Pearson Edexcel IAL Pure Mathematics 4 scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.