flowchart LR
S((source)) -->|growth = rP| P[Population P]flowchart LR
S((source)) -->|growth = rP| P[Population P]
Where we are going
By the end of class, you should be able to:
- tell the difference between an amount, a rate, and a change;
- turn a differential equation into an update rule;
- recognize the stock, flow, parameter, and initial condition in a model; and
- build and graph a simple simulation in R.
First: make a prediction
A population begins with 100 individuals and grows continuously at a rate of 10% per hour.
Without calculating, predict the population after 8 hours:
- less than 180;
- between 180 and 240; or
- greater than 240.
Then sketch the graph you expect. Will the population gain the same number of individuals every hour?
Name the pieces
Let \(P(t)\) be the population at time \(t\).
| Symbol | Meaning | Units |
|---|---|---|
| \(P\) | current population | individuals |
| \(P_0\) | initial population | individuals |
| \(r\) | per-capita growth rate | per hour |
| \(t\) | time | hours |
From words to mathematics
The population’s rate of change is proportional to the current population.
On the board, translate that sentence into an equation:
\[ \frac{dP}{dt} = \boxed{\phantom{rP}} \]
For a short time step \(\Delta t\), we approximate the change with
\[ \Delta P \approx \text{rate of change} \times \text{time step}. \]
Complete the update rule:
\[ P_{\text{new}} = P_{\text{old}} + \boxed{\phantom{rP_{\text{old}}\Delta t}} \]
Try one step by hand
Use \(P_0 = 100\), \(r = 0.10\) per hour, and \(\Delta t = 1\) hour.
\[ P_1 = \underline{\hspace{7em}} \]
Before doing a second step, answer this: should the next increase be the same size, larger, or smaller?
See the system
The population is a stock: an amount that accumulates. Growth is a flow: a rate that changes the stock.
The feedback loop matters:
\[ \text{larger population} \longrightarrow \text{more growth} \longrightarrow \text{larger population}. \]
Build the simulation in R
1. Choose the model settings
initial_population <- 100
growth_rate <- 0.10
time_step <- 1
end_time <- 8Which of these is the initial condition? Which is the model parameter?
2. Create the time points
time <- seq(
from = 0,
to = TODO,
by = TODO
)
timeHow many values should time contain? Make a prediction before running it.
3. Make space for the results
population <- numeric(length(time))
population[1] <- TODO
populationWhy does the initial population belong in position 1 even though time begins at 0?
4. Repeat the update
for (i in 2:length(time)) {
old_population <- population[i - 1]
growth <- TODO
population[i] <- TODO
}Read the loop aloud in plain English. What does one trip through the loop represent?
5. Put time and population together
results <- data.frame(
time = time,
population = population
)
resultsCheck the first two rows against the step you calculated by hand.
6. Graph the story
plot(
TODO,
TODO,
type = "TODO",
xlab = "Time (hours)",
ylab = "Population",
main = "Unconstrained growth"
)If we have time: test the simulation
The differential equation also has an exact solution:
\[ P(t) = P_0 e^{rt}. \]
Add the exact solution to the results:
results$exact <- TODOThen add it to the graph:
lines(results$time, results$exact, col = "darkorange", lwd = 3)
legend(
"topleft",
legend = c("simulation", "exact"),
col = c("black", "darkorange"),
lty = 1,
lwd = c(1, 3)
)Do the two curves agree exactly? What part of our simulation introduced the difference?
A smaller time step
Change time_step from 1 to 0.25 and run the model again.
- What else in the code changes automatically?
- Does the simulation move closer to or farther from the exact solution?
- What do we gain—and what does the computer have to do more often?
Growth becomes decay
What single change would turn growth into decay?
growth_rate <- TODOBefore running the code, predict the graph’s shape. Does the model ever produce a negative population?
Before you leave
In one or two sentences, explain the update rule
\[ P_{\text{new}} = P_{\text{old}} + rP_{\text{old}}\Delta t \]
without using mathematical symbols.