GCSE-ready- Home
- Lessons
- IGCSE Further Pure Mathematics
- Rates of change and small changes
Rates of change and small changes
🎬 The doodle video for this lesson is coming soon. Subscribe on YouTube to see it first.
When one quantity changes with time, everything linked to it changes too. The chain rule connects the rates.
Connected rates
dAdt = dAdr × drdt
A circle's radius grows at 0.5 cm/s. A = πr2, so dAdr = 2πr.
When r = 4: dAdt = 8π × 0.5 = 4π ≈ 12.6 cm2/s
A circle's radius grows at 0.5 cm/s. A = πr2, so dAdr = 2πr.
When r = 4: dAdt = 8π × 0.5 = 4π ≈ 12.6 cm2/s
Working backwards
A sphere has V = 43πr3 and dVdt = 100 cm3/s.
dVdr = 4πr2, so drdt = dVdt ÷ dVdr
When r = 5: drdt = 100100π = 1π ≈ 0.318 cm/s
dVdr = 4πr2, so drdt = dVdt ÷ dVdr
When r = 5: drdt = 100100π = 1π ≈ 0.318 cm/s
Small changes
For a small change δx in x,
δy ≈ dydx × δx
y = x2, x changes from 3 to 3.01: δy ≈ 6 × 0.01 = 0.06
δy ≈ dydx × δx
y = x2, x changes from 3 to 3.01: δy ≈ 6 × 0.01 = 0.06
A cube has side x cm, increasing at 0.2 cm/s. Find the rate of increase of the volume when x = 5.
- V = x3, dVdx = 3x2 = 75
- dVdt = 75 × 0.2
Answer: 15 cm3/s
Chain the rates: dydt = dydx × dxdt. For small changes, δy ≈ dydx δx.
Check you have got it
Answer 6 quick questions with instant marking. If you get one wrong, GCSE-ready shows you why and gives you another go. It is free, and you do not need an account.