Unit 3 · Lesson 4 of 6

Growth changes when resources become limited.

An exponential model grows forever, but a carrying capacity makes growth slow and can pull an oversized population downward.

Change the input. Follow the mathematics.

What you'll be able to do

Set up the model and calculate its population.
Connect the equation to the changing visual.
Explain the assumptions and identify a misleading interpretation.

Predict before exploring

What will change if you increase time? Predict the direction, then explain the response.

A model you can changeMove one control at a time.

64.879 · illustrative relative amount

PopulationCarrying capacity · dashed

Time10 hInitial population20 Growth constant r0.2 1/h

population

P(10)=\frac{100}{1+(100/20-1)e^{-2}}=64.8786

Below carrying capacity, growth slows as the population approaches 100. Above it, the derivative is negative and the population falls toward it.

Assumptions, units, and limits

Carrying capacity 100; positive initial population may start below or above it. r is constant.

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

From the situation to an equation

Separate variables or verify the closed-form logistic solution. Rate is proportional to both population and the remaining capacity fraction. Positive initial values below and above capacity approach the same stable equilibrium.

P'=rP(1-P/100),\quad P(t)=\frac{100}{1+(100/P_0-1)e^{-rt}}

Keep the reasoning visible

P'=0\text{ at }P=0\text{ or }P=K

Below carrying capacity, growth slows as the population approaches 100. Above it, the derivative is negative and the population falls toward it.

What makes this model useful?

Carrying capacity 100; positive initial population may start below or above it. r is constant.

A carrying capacity is a model parameter, not a universal biological constant. At half capacity the absolute growth rate is largest for this model.

A worked example

Calculate the starting scenario

Use time = 10 h; initial population = 20 ; growth constant r = 0.2 1/h.
$P(10)=\frac{100}{1+(100/20-1)e^{-2}}=64.8786$
Below carrying capacity, growth slows as the population approaches 100. Above it, the derivative is negative and the population falls toward it. The checked quantity is population.

Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the population using time 10 h, initial population 20 , growth constant r 0.2 1/h. State the meaning of its sign and units.

Checked quantity: population · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

64.879 · illustrative relative amount

PopulationCarrying capacity · dashed

Time10 hInitial population20 Growth constant r0.2 1/h

population

P(10)=\frac{100}{1+(100/20-1)e^{-2}}=64.8786

Below carrying capacity, growth slows as the population approaches 100. Above it, the derivative is negative and the population falls toward it.

Assumptions, units, and limits

Carrying capacity 100; positive initial population may start below or above it. r is constant.

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

P'=rP(1-P/100),\quad P(t)=\frac{100}{1+(100/P_0-1)e^{-rt}}

Calculate and explain

Apply the idea to a new scenario

For P₀=50, $r=0.1$ h⁻¹, find P at $t=20$ h with capacity 100.

Checked quantity: population · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

88.08 · illustrative relative amount

PopulationCarrying capacity · dashed

Time20 hInitial population50 Growth constant r0.1 1/h

population

P(20)=\frac{100}{1+(100/50-1)e^{-2}}=88.0797

Below carrying capacity, growth slows as the population approaches 100. Above it, the derivative is negative and the population falls toward it.

Assumptions, units, and limits

Carrying capacity 100; positive initial population may start below or above it. r is constant.

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

P'=rP(1-P/100),\quad P(t)=\frac{100}{1+(100/P_0-1)e^{-rt}}

Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual

64.879 · illustrative relative amount

PopulationCarrying capacity · dashed

Time10 hInitial population20 Growth constant r0.2 1/h

population

P(10)=\frac{100}{1+(100/20-1)e^{-2}}=64.8786

Below carrying capacity, growth slows as the population approaches 100. Above it, the derivative is negative and the population falls toward it.

Assumptions, units, and limits

Carrying capacity 100; positive initial population may start below or above it. r is constant.

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Separate variables or verify the closed-form logistic solution. Rate is proportional to both population and the remaining capacity fraction. Positive initial values below and above capacity approach the same stable equilibrium.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

Completion is saved on this browser. This is separate from account practice progress.