Connect rates of change between related quantities
Differentiate a geometric relation, multiply by the changing input's rate and check the direction and units of the resulting rate.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Use an informal understanding of limits; formal general differentiation from first principles and points of inflexion are not required in section 1.7. Written practice is self-checked.
Before you begin
- Differentiate powers and apply the chain rule.
- Use geometric formulae and distinguish a quantity from its rate of change.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which relation connects a circle's area rate to its radius rate?
What you will learn
- Use the chain rule to connect rates such as dr/dt and dA/dt.
- Recover an unknown input rate from a supplied output rate with correct units and signs.
Relate the quantities before relating their rates
If area A depends on radius r and r changes with time t, then dA/dt = (dA/dr)(dr/dt). For a circle A = πr², dA/dr = 2πr, so dA/dt = 2πr dr/dt. The gradient dA/dr alone has units cm²/cm = cm, while multiplying by dr/dt in cm/s gives area-rate units cm²/s.
At r = 5 cm with dr/dt = 2/5 cm/s, the circle's area rate is 2π(5)(2/5) = 4π cm²/s. Radius and radius rate are different inputs: use the actual radius at the instant, together with its stated rate. A negative radius rate would indicate shrinking and give a negative area rate for a positive radius.
Reverse the rate relation or change the geometric quantity
For a sphere V = (4/3)πr³, the relation is dV/dt = 4πr² dr/dt. If r = 3 cm and dr/dt = −1/5 cm/s, then dV/dt = −36π/5 cm³/s, so the volume decreases. The square of the radius is positive, but the signed radius rate still determines the direction of change.
When an area rate is given, rearrange its connected-rate equation instead of assuming it equals the radius rate. If a circle at r = 3 cm has dA/dt = 12π cm²/s, then dr/dt = 12π/[2π(3)] = 2 cm/s. Its circumference C = 2πr changes at dC/dt = 2π dr/dt = 4π cm/s. Constant geometric parameters must actually be fixed by the supplied model before treating them as constants.
Pause and explain
A positive-radius circle has dr/dt < 0. What is the sign of dA/dt?
Worked example
A circle has radius 5 cm at an instant when dr/dt = 2/5 cm/s. Find its area rate at that instant.
Show the worked solution
- Start from A = πr² and differentiate with respect to r to get dA/dr = 2πr.
- Use dA/dt = (dA/dr)(dr/dt) = 2πr dr/dt.
- Substitute r = 5 and dr/dt = 2/5 to obtain 4π cm²/s, positive because the radius is increasing.
Answer dA/dt = 4π cm²/s at the stated instant.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For a cube V = x³, find the volume rate when x = 5 cm and dx/dt = 1/5 cm/s.
Give me a hint
Differentiate volume with respect to side length, then multiply by the side's time rate.
Compare my reasoning
- dV/dx = 3x².
- The chain rule gives dV/dt = 3x² dx/dt.
- At the instant, dV/dt = 3(5²)(1/5) = 15 cm³/s.
15 cm³/s.
Look for these in your work
- Use both the current side and its rate.
- State volume-rate units rather than length-rate units.
A sphere obeys V = (4/3)πr³. Find its volume rate when r = 3 cm and dr/dt = −1/5 cm/s.
Give me a hint
The negative radius rate indicates shrinking; retain its sign after differentiating.
Compare my reasoning
- dV/dr = 4πr².
- Thus dV/dt = 4πr² dr/dt.
- Substituting gives 4π(9)(−1/5) = −36π/5 cm³/s, indicating decreasing volume.
dV/dt = −36π/5 cm³/s; the volume decreases.
Look for these in your work
- Differentiate the supplied sphere-volume formula.
- Keep the signed rate and interpret its direction.
A circle at radius 3 cm has area increasing at 12π cm²/s. Find its radius rate and circumference rate using A = πr² and C = 2πr.
Give me a hint
Reverse the area-rate relation to find dr/dt before using it for circumference.
Compare my reasoning
- From dA/dt = 2πr dr/dt, obtain dr/dt = 12π/[2π(3)] = 2 cm/s.
- Differentiate circumference with respect to time: dC/dt = 2π dr/dt.
- Hence dC/dt = 4π cm/s; the two rates describe different lengths despite sharing length/time units.
dr/dt = 2 cm/s; dC/dt = 4π cm/s.
Look for these in your work
- Rearrange the known output-rate relation to recover the input rate.
- Use that rate in the second supplied geometric relation.
Common mistakes
- A derivative with respect to r is different from a rate with respect to time.
- Substitute both the current quantity and its rate, rather than confusing them.
- Preserve negative rates and use units for the correct changing quantity.
What can you explain now?
For a cube V = x³, x = 5 cm and dx/dt = 1/5 cm/s. Find dV/dt.
Compare with the explanation
15 cm³/s.
Use dV/dt = 3x² dx/dt = 3(25)(1/5) = 15.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Write a connected-rate equation with its variable units before substituting. Explain how you would reverse it when an output rate is supplied.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.