Exponential growth and decay

The differential equation dN/dt = kN and its solution N = N0 e^kt, finding the growth constant, solving growth and decay problems, rates of growth, and half-life.

The model

Growth

Example 1●●●

A culture of $500$ bacteria grows to $1500$ in $4$ hours, where $N = 500e^{kt}$.

$1500 = 500e^{4k}$, so $e^{4k} = 3$ and $k = \dfrac{\ln 3}{4} \approx 0.2747$.

Find k

At $t = 10$: $N = 500e^{10k} = 500 \times 3^{2.5} \approx 7794$.

Predict

$e^{kt} = 20$, so $t = \dfrac{\ln 20}{k} = \dfrac{4\ln 20}{\ln 3} \approx 10.9$ hours.

Time to reach 10 000

When $N = 1500$, $\dfrac{dN}{dt} = kN \approx 412$ bacteria per hour.

Rate

Decay and half-life

The half-life is the time for a decaying quantity to halve. From $e^{kT} = \tfrac12$, the half-life is $T = \dfrac{\ln 2}{|k|}$.

Example 2●●

Carbon-14 has a half-life of $5730$ years. What percentage of a sample remains after $1000$ years?

$k = -\dfrac{\ln 2}{5730}$.

$\dfrac{N}{N_0} = e^{1000k} = 2^{-\frac{1000}{5730}} \approx 0.886$, so about $88.6\%$ remains.

Flashcards

Solution of $dfrac{dN}{dt} = kN$

$N = N_0e^{kt}$

Finding $k$

Substitute a known value, then take $\ln$.

Half-life

$\dfrac{\ln 2}{|k|}$

Practice

The solution of $\dfrac{dN}{dt} = kN$ is:
$N = 500e^{kt}$ and $N = 1500$ when $t = 4$. Find $k$ to 4 decimal places.
For the same culture, find $N$ when $t = 10$, to the nearest whole number.
When does the culture reach $10\,000$? Answer in hours to 1 decimal place.
Find the growth rate when $N = 1500$, to the nearest whole number.
With a half-life of $5730$ years, what percentage of carbon-14 remains after $1000$ years, to 1 decimal place?
In $N = N_0e^{kt}$, a negative $k$ means:
The half-life of a quantity with decay constant $k$ is:
$N = 200e^{0.1t}$. Find $\dfrac{dN}{dt}$ at the moment when $N = 600$.