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Calculus BC · Unit 7 · 7.8
AP Calculus · Differential Equations
Exponential Models with Differential Equations
One sentence — rate proportional to amount — and the whole model follows.
key idea$\frac{dy}{dt}=ky\ \Longleftrightarrow\ y=y_0e^{kt}$
Setting up the model
Exponential behaviour is defined by a single sentence: the rate of change is proportional to the amount present. Every population, isotope, investment and cooling problem of this type reduces to $\dfrac{dy}{dt}=ky$, whose solution you should recognise on sight rather than re-derive.
Building the model from a word problem
- Name the quantity $y$ and the time variable $t$, with units for both. Write down what $t=0$ means in the story.
- Write $\dfrac{dy}{dt}=ky$. The constant $k$ has units of per unit time and is positive for growth, negative for decay.
- Record $y_0=y(0)$ — the initial amount — separately from the equation.
- Use the second piece of data (a later measurement, a doubling time, a half-life) to solve for $k$. This always means taking a logarithm.
- Answer the actual question with the finished model $y=y_0e^{kt}$, and check that the units of your answer match what was asked.
growth: $k\gt 0$, doubling time $\dfrac{\ln2}{k}$
decay: $k\lt 0$, half-life $\dfrac{\ln2}{\lvert k\rvert}$
| Given | What to do | Result |
|---|
| $y$ doubles in $T$ | $2y_0=y_0e^{kT}$ | $k=\dfrac{\ln2}{T}$ |
| half-life is $T$ | $\tfrac12y_0=y_0e^{kT}$ | $k=-\dfrac{\ln2}{T}$ |
| $y(5)=3y(0)$ | $3=e^{5k}$ | $k=\dfrac{\ln3}{5}$ |
| $y$ falls $10\%$ per year | $0.9=e^{k}$ | $k=\ln(0.9)\approx-0.105$ |
Every row is the same move: divide out $y_0$, then take $\ln$. The initial amount always cancels, which is why a half-life can be quoted without saying how much you started with.
Worked examples
EX 1
Find $k$ from a doubling time
bacteriaA culture doubles every $3$ hours. Write the model and find how long it takes to triple.
1
Set up
$\dfrac{dP}{dt}=kP$ with $P$ in cells and $t$ in hours, so $P=P_0e^{kt}$.
2
Use the doubling time
$2P_0=P_0e^{3k}$, so $2=e^{3k}$.
3
Solve for $k$
$k=\dfrac{\ln2}{3}\approx0.231$ per hour.
4
Set up the tripling question
$3=e^{kt}$, so $t=\dfrac{\ln3}{k}=\dfrac{3\ln3}{\ln2}$.
5
Evaluate
$t\approx4.75$ hours. Note that tripling takes less than twice as long as doubling — growth compounds.
EX 2
Radioactive decay
half-lifeCarbon-14 has a half-life of $5730$ years. A sample retains $62\%$ of its original carbon-14. How old is it?
1
Find $k$ from the half-life
$\tfrac12=e^{5730k}$, so $k=-\dfrac{\ln2}{5730}\approx-1.21\times10^{-4}$ per year.
2
Write the equation to solve
$0.62y_0=y_0e^{kt}$, so $0.62=e^{kt}$.
3
Take logarithms
$t=\dfrac{\ln(0.62)}{k}$.
4
Evaluate
$t=\dfrac{-0.478}{-1.21\times10^{-4}}\approx3950$ years.
5
Sanity-check
$62\%$ is a little above half, so the age should be a little under one half-life. $3950\lt 5730$ ✓
EX 3
Interpreting $k$ in context
units and meaningA model gives $\dfrac{dM}{dt}=-0.04M$ with $M$ in grams and $t$ in days. Interpret $-0.04$, and find the half-life.
1
Read the sign
Negative, so $M$ is decreasing — this is decay.
2
Read the units
$k$ has units of $\text{day}^{-1}$: it is a rate per gram present, per day.
3
State the plain-English meaning
The mass falls by about $4\%$ of whatever is currently there, each day.
4
Find the half-life
$\tfrac12=e^{-0.04T}$, so $T=\dfrac{\ln2}{0.04}\approx17.3$ days.
5
Note the distinction
‘Falls by $4\%$ of the current amount’ is not the same as falling by a fixed number of grams per day — the absolute loss shrinks as $M$ does.
EX 4
Derive the solution rather than quote it
show the workStarting from $\dfrac{dy}{dt}=ky$ and $y(0)=y_0$, derive $y=y_0e^{kt}$.
1
Separate
$\dfrac{dy}{y}=k\,dt$.
2
Integrate
$\ln\lvert y\rvert=kt+C$.
3
Apply the initial condition first
At $t=0$: $\ln\lvert y_0\rvert=C$.
4
Combine and exponentiate
$\ln\left\lvert\dfrac{y}{y_0}\right\rvert=kt$, so $\dfrac{y}{y_0}=e^{kt}$ taking the sign of $y_0$.
5
Finish
$y=y_0e^{kt}$. Free-response scoring often wants this derivation, so it is worth being able to produce it in four lines.
Units & interpretation
reading k
- $k\gt 0$ is growth, $k\lt 0$ is decay, and $\lvert k\rvert$ sets how fast either happens.
- Doubling time and half-life are independent of $y_0$ — the starting amount always cancels.
- $\dfrac{dy}{dt}$ carries the units of $y$ per unit of $t$; $k$ carries units of $\text{time}^{-1}$.
- An exponential model never reaches zero and never levels off. If the story has a ceiling, this is the wrong model.
| Phrase in the problem | Symbolic form | Interpretation |
|---|
| “grows at a rate proportional to its size” | $y'=ky$, $k\gt 0$ | unbounded growth |
| “decays at a rate proportional to the amount present” | $y'=ky$, $k\lt 0$ | approaches zero, never reaches it |
| “half-life $T$” | $k=-\dfrac{\ln2}{T}$ | a fixed fraction lost per unit time |
| “continuously compounded at $r$” | $k=r$ | $$1$ becomes $$e^{r}$ in one year |
| “increases by $8\%$ per year” | $e^{k}=1.08$ | $k=\ln(1.08)$, slightly less than $0.08$ |
The last row is the one that trips people: an $8\%$ annual increase corresponds to a continuous rate of $\ln(1.08)\approx0.077$, not $0.08$. Discrete percentages and continuous rates are related by a logarithm, not by equality.
Watch out
$y_0$ cancels — do not hunt for it
Doubling-time and half-life questions never need the starting amount. If you find yourself needing it, the equation was set up wrong.
A percentage rate is not $k$
“Rises $8\%$ a year” means $e^{k}=1.08$. Setting $k=0.08$ is close but wrong, and the gap grows with the rate.
Keep $k$ symbolic until the end
Rounding $k$ early and then exponentiating magnifies the error. Carry $\dfrac{\ln2}{T}$ through the algebra where you can.
Exponential models have no ceiling
If the situation described has a maximum sustainable level, $y'=ky$ cannot represent it — the model will overshoot without limit.