Concept preview

What you will be able to do

1. 01Recognise that any quantity whose rate of change is proportional to its amount, $\dfrac{dy}{dt}=ky$, has the exponential solution $y=y_0e^{kt}$, and identify $y_0$ and $k$ from context.
2. 02Compute the growth constant $k$ from two data points, and convert between $k$, the doubling time $\dfrac{\ln 2}{k}$, and the half-life $\dfrac{\ln 2}{|k|}$.
3. 03Set up and solve exponential word problems (population, radioactive decay, continuously compounded interest, Newton cooling), interpreting $k$ and avoiding the error of modelling exponential change as linear.

Start from the statement "the rate of change of $y$ is proportional to $y$." Proportional means "a constant times," so $\dfrac{dy}{dt}=ky$ for some constant $k$. This is separable: $\dfrac{dy}{y}=k\,dt$, and integrating gives $\ln|y|=kt+C$. Exponentiating, $y=e^{kt+C}=Ae^{kt}$. Applying the initial amount $y(0)=y_0$ forces $A=y_0$, so the particular solution is $y=y_0e^{kt}$. Memorise this pairing: whenever you see $\dfrac{dy}{dt}=ky$, the answer is $y=y_0e^{kt}$ with $y_0$ the starting amount. You should be able to reproduce the derivation, but on the exam you may quote the result directly.