Exponential Models (College Board AP® Calculus AB): Revision Note

If an exam question is based on a real world example, be sure that your answers are given in the context of the question.

At any point in time, the rate of growth of a colony of bacteria is proportional to the current population size, $P$.

(a) Write a differential equation to model the size of the population of bacteria.

At time $t=0$ hours, the population size is 5000.

(b) Write down the particular solution of the differential equation from part (a).

After 1 hour, the population has grown to 7000.

(c) Determine how long it will take from time $t=0$, according to the model, for the population of bacteria to grow to 100 000.

Answer:

(a)

This is a description of an exponential growth model

$\frac{dP}{dt}=kP$

(b)

For $\frac{dy}{dt}=ky$, with the initial condition $y=y_0$ when $t=0$, the particular solution is$y=y_0{e}^{kt}$

$P=5000e^{kt}$

(c)

The question doesn't say what units of time to use, but looking at the information given it will be easiest to use hours

Substitute the values for $t=1$ hour into the particular solution, and solve for $k$

$\begin{array}{rcl}7000& =& 5000e^{k\left(1\right)}\\ e^k& =& \frac{7000}{5000}=\frac{7}{5}=1.4\\ k& =& \ln \left(1.4\right)\end{array}$

Using that value of $k$, substitute $P=100000$ into the particular solution and solve for $t$

$\begin{array}{rcl}100000& =& 5000e^{\ln \left(1.4\right)t}\\ e^{\ln \left(1.4\right)t}& =& \frac{100000}{5000}=20\\ \ln \left(1.4\right)t& =& \ln \left(20\right)\\ t& =& \frac{\ln \left(20\right)}{\ln \left(1.4\right)}=8.903356...\end{array}$

8.903 hours (to 3 decimal places)

Notice that the doubling-time and half-life are independent of the initial value.

The mass of a radioactive isotope decays at a rate proportional to the mass present, modeled by the differential equation $\frac{dM}{dt}=kM$, where $M$ is the mass in grams and $t$ is the time in years.

If the half-life of the isotope is 25 years, which of the following is the particular solution to the differential equation given an initial mass of 100 grams?

(A) $M\left(t\right)=100e^{-25\ln \left(2\right)t}$

(B) $M\left(t\right)=100e^{-\frac{\ln 2}{25}t}$

(C) $M\left(t\right)=100e^{\frac{\ln 2}{25}t}$

(D) $M\left(t\right)=100e^{-0.5t}$

Answer:

Identify the general solution

• Note here that the value of $k$ is negative

$M\left(t\right)=100e^{kt}$

Apply the half-life condition and solve for $k$

$\begin{array}{rcl}M\left(25\right)& =& 50\\ 100e^{25k}& =& 50\\ e^{25k}& =& \frac{1}{2}\\ 25k& =& \ln \left(\frac{1}{2}\right)\\ 25k& =& -\ln 2\\ k& =& -\frac{\ln 2}{25}\end{array}$

Substitute the value of $k$ into the general solution

$M\left(t\right)=100e^{-\frac{\ln 2}{25}t}$

Option (B)