A-level Maths revision · Pure, year 1

Exponential models

The idea

Growth and decay are modelled by y=Aekty = Ae^{kt}: AA is the starting value and kk the rate (k>0k > 0 grows, k<0k < 0 decays). In y=c+Ae−kty = c + Ae^{-kt}, the value levels off at cc.

  1. Put t=0t = 0 to read the starting value.
  2. Use a known pair of values and ln⁡\ln to find kk.
  3. Interpret: the long-run value, and whether the model stays sensible.

Watch out Check the model in context: no population grows for ever.

See it worked

Worked example

A population is modelled by P=200e0.03tP = 200e^{0.03t}. Find when it doubles.

  1. Doubles: 200e0.03t=400200e^{0.03t} = 400, so e0.03t=2e^{0.03t} = 2.
  2. 0.03t=ln⁡20.03t = \ln 2.
  3. t=ln⁡20.03=23.1t = \frac{\ln 2}{0.03} = 23.1 years.

Answer After 23.123.1 years

Your turn

Question 1Calculator allowed3 marks

A car's value in pounds is V=15 000e−0.2tV = 15\,000e^{-0.2t}, where tt is its age in years. Find its value after 33 years, to the nearest pound.

Question 2Calculator allowed3 marks

N=50ektN = 50e^{kt}, and N=150N = 150 when t=4t = 4. Find kk to 3 significant figures.

Question 3Non-calculator3 marks

A cup of tea cools so that its temperature is T=20+80e−0.05tT = 20 + 80e^{-0.05t} degrees Celsius after tt minutes. What temperature does the model predict in the long run?

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