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A-level Maths revision · Pure, year 2

Connected rates of change

The idea

Link the rate you want to the rate you know with the chain rule: dAdt=dAdr×drdt\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}. A sentence such as 'decreases at a rate proportional to T−20T - 20' becomes the differential equation dTdt=−k(T−20)\frac{dT}{dt} = -k(T - 20).

  1. Write a formula linking the two quantities, such as A=πr2A = \pi r^2.
  2. Differentiate it: dAdr=2πr\frac{dA}{dr} = 2\pi r.
  3. Multiply by the given rate and substitute the moment's values.

Watch out A decreasing quantity has a negative rate: include the minus sign.

See it worked

Worked example

A sphere's volume increases at 10 cm3 s−110\text{ cm}^3\text{ s}^{-1}. Find the rate its radius increases when r=5 cmr = 5\text{ cm}.

  1. V=43πr3V = \frac{4}{3}\pi r^3, so dVdr=4πr2=100π\frac{dV}{dr} = 4\pi r^2 = 100\pi.
  2. drdt=dVdt÷dVdr=10100π\frac{dr}{dt} = \frac{dV}{dt} \div \frac{dV}{dr} = \frac{10}{100\pi}.
  3. =110π=0.0318 cm s−1= \frac{1}{10\pi} = 0.0318\text{ cm s}^{-1}.

Answer 0.0318 cm s−10.0318\text{ cm s}^{-1}

Your turn

Question 1Non-calculator3 marks

A circle's radius increases at 0.5 cm s−10.5\text{ cm s}^{-1}. Find the exact rate its area increases when r=4 cmr = 4\text{ cm}.

Question 2Non-calculator3 marks

A cube's side xx increases at 0.2 cm s−10.2\text{ cm s}^{-1}. Find the rate its volume increases when x=3 cmx = 3\text{ cm}.

Question 3Non-calculator3 marks

A drink cools according to dTdt=−k(T−20)\frac{dT}{dt} = -k(T - 20). When T=80T = 80, dTdt=−3\frac{dT}{dt} = -3. Find kk.

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