5.2 Growth and Decay
- Use separation of variables to solve a simple differential equation.
- Use exponential functions to model growth and decay in applied problems.
Assignment
- Vocabulary and teal boxes
- p384 1, 6, 9, 11, 12, 14, 15, 19, 23, 25, 31, 38, 39 42, 45, 48, 54, 59â61
Additional Resources
- AP Topics: 7.6, 7.7, 7.8
- Khan Academy
Differential Equations
Now that weâve introduced differential equations involving both $x$ and $y$, itâs time to start solving them. The technique weâll use is called separation of variables, and it involves $\frac{dy}{dx}$ behaving like a fraction, even though it isnât.
\[\begin{align} \frac{dy}{dx} &= \frac{2x}{y} \\ y\, dy &= 2x \, dx & \text{All $y$'s on one side, and all $x$'s on the other.}\\ \int y\, dy &= \int 2x \, dx \\ \frac{1}{2}y^2 &= x^2 + C \\ y^2 - 2x^2 &= C \end{align}\]I prefer Leibniz notation, which uses $\frac{dy}{dx}$, while the book introduces the concept using Lagrange notation, which uses $yâ$. Either is fine.
Weâll eventually get into how to solve the more complicated ones, like what we saw when looking at slope fields.
Growth and Decay Models
Youâve likely worked with exponential models before, but now weâll look at them from the perspective of their rate of change, which is often proportional to the amount in question.
\[\begin{align} \frac{dy}{dt} = ky \end{align}\]We can integrate this using separation of variables.
\[\begin{align} \frac{dy}{dt} &= ky \\ \frac{dy}{y} &= k\, dt \\ \int \frac{1}{y} \, dy &= \int k \, dt \\ \ln y &= kt + C \\ y &= e^{kt + c} \\ y &= e^{kt}e^C \label{exp-prop}\\ y &= Ce^{kt} \label{const} \end{align}\]Steps \ref{exp-prop} and \ref{const} are common with separation of variables, and explain why you started seeing $C$ as a factor rather than an addend. Multiplying bases results in addition of powers, so we applied that to step \ref{exp-prop}, and in step \ref{const} $e^C$ is just a constant, so we can just drop the base.
Exponential Growth and Decay
\[\begin{align} y = Ce^{kt} \end{align}\]where $C$ is the initial value, $k$ the constant of proportionality.
A Simple Example Stretched Out
When $t=0$, $y=2$. And when $t=2$, $y=4$. If the rate of change is proportional to $y$, then we can model this easily.
First off, since we get $t=0$, we get the initial condition $C$ for free. Just in case you forget that $C$ is the initial condition, you can always find it yourself.
\[\begin{align} y &= Ce^{kt} \\ 2 &= Ce^{k\cdot0} \\ 2 &= Ce^{0} \\ 2 &= C \end{align}\]We still need $k$, but weâll need to use our other point to find that.
\[\begin{align} y &= 2e^{kt} \\ 4 &= 2e^{k\cdot2} \\ \ln 2 &= 2k \\ \frac{1}{2}\ln 2 &= k \\[1em] y&\approx 2e^{0.3466t} \end{align}\]Though, if you know your logarithm properties you can write this using the exact value of $k$.
\[\begin{align} y &= 2e^{\frac{1}{2}\ln 2 \cdot t} \\ y &= 2e^{\ln \sqrt{2} \cdot t} \\ y &= 2\sqrt{2}^t \\ t&= 2\cdot2^{\frac{t}{2}} \end{align}\]Oh, we could have done this with our old exponential modeling rules. We knew the initial amount, $C=2$, and we technically knew the rate of growth, though it wasnât called out by name. We went from $2$ to $4$, so it doubled, but it took two ticks of $t$ leading to $\frac{t}{2}$ as the exponent.
Population Growth
In this problem, you wonât have the initial condition. Instead, you know there are 100 flies after day two of an experiment and 300 after four days. Weâll use these to create what is essentially a system of equations, solving one for $C$ and then substituting into the other.
\[\begin{align} y &= Ce^{kt} \\ 100 &= Ce^{k\cdot 2} \\ \frac{100}{e^{2k}} &= C \\ 100e^{-2k} &= C \\[1em] 300 &= Ce^{4k} \\ 300 &= \left(100e^{-2k}\right)e^{4k} \\ 300 &= 100e^{2k} \\ \ln3 &= 2k \\ \frac{1}{2}\ln3 &= k \\[1em] k &\approx 0.5493 \end{align}\]We can solve for $C$ now that we have $k$.
\[\begin{align} 100 &= Ce^{0.5493(2)} \\ 100e^{-0.5493(2)} &= C \\[1em] C&\approx 33 \end{align}\]So our model is $y=33e^{0.5493t}$.
But I want to write this using the exact value again.
\[\begin{align} y &= 33 e^{\frac{1}{2}\ln3\cdot t} \\ y &= 33 \sqrt{3}^t \\ y &= 33 \cdot 3^\frac{t}{2} \end{align}\]That just means it tripled every two days, and we knew that from the beginning: 100 flies on day 2, and 300 on day 4. We could have set up ${y=C\cdot 3^{\frac{t}{2}}}$ and solved for $C$ using one of the points.
Newtonâs Law of Cooling
An object cools proportional to the ambient temperature, so how can we model an object that is initially 100ÂșF, but cools to 90ÂșF in 10 minutes in a 60ÂșF room? Weâll have to start with our initial definition of the proportional rate of change, but modified to account for the difference in temperature of the room.
\[\begin{align} \frac{dy}{dt} &= k(y-60) \\ \int \frac{dy}{y-60} &= \int k \, dt \\ \ln(y-60) &= kt + C \\ y &= Ce^{kt}+60 \end{align}\]And now, like before, we can use our two points to calculate $C$ and $k$.
\[\begin{align} 100 &= Ce^{0k} + 60 \\ 40 &= C \\[1em] 90 &= 40e^{10k} + 60 \\ \ln\frac{3}{4} &= \ln e^{10k} \\ \frac{1}{10}\ln\frac{3}{4} &=k \\ -0.0288 &\approx k \\[1em] y&= 40e^{-0.0288t}+60 \end{align}\]And if you go through with the exact value of $k$ and rewrite the base
\[\begin{align} y&= 40e^{\frac{t}{10}\ln\frac{3}{4}}+60 \\ y&= 40\cdot\left(\frac{3}{4}\right)^\frac{t}{10}+60 \\ \end{align}\]You can pull each part of the original problem out of this version, to the point where you can skip the calculus just like the others. I donât advise that, since the exam will find ways to trip you up, but being able to see the relationship between the problem and the model help to spot check common errors.