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Amy Ferguson Moncure

Calculus BC · Unit 7

Logistic Models

Find the carrying capacity, the fastest growth, and the explicit solution from the initial population.

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Key Ideas Easy Medium Hard

\(\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{L}\right)\)

What to Remember

The logistic equation

\(\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{L}\right)\)

\(L\) is the carrying capacity. The same model is sometimes written \(\dfrac{dP}{dt}=cP(L-P)\).

Read the rate without solving

The rate is positive when \(00\). It is zero at \(P=0\) and at \(P=L\).

The fastest growth is at \(P=L/2\).

The explicit solution

\(P(t)=\dfrac{L}{1+Ae^{-kt}}\)

with \(A=\dfrac{L-P_0}{P_0}\).

The long-run value is L

If \(k>0\) and \(P_0>0\), then \(\lim_{t\to\infty} P(t)=L\).

Level 1

Easy

Identify L, or evaluate the rate at one population.

Example
\(\dfrac{dP}{dt}=0.2P\left(1-\dfrac{P}{500}\right)\). Find the carrying capacity and the growth rate when \(P=250\).

The carrying capacity is the value that makes the second factor zero: \(L=500\).

Half of 500 is 250, so this is the fastest point.

\(0.2\cdot 250\cdot\left(1-\dfrac{250}{500}\right)\)

\(50\cdot \dfrac{1}{2}\)

\(25\)

Answer: \(L=500\), rate 25

1
State the carrying capacity of \(dP/dt=0.04P\left(1-P/2000\right)\).
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\(2000\)

Answer: \(2000\)

2
\(dP/dt=3P\left(1-\dfrac{P}{12}\right)\). Find the rate when \(P=4\).
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\(3\cdot 4\cdot\left(1-\dfrac{4}{12}\right)\)

\(12\cdot \dfrac{2}{3}\)

\(8\)

Answer: \(8\)

3
For the same equation, find the rate when \(P=12\).
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\(1-\dfrac{12}{12}=0\)

The rate is 0.

Answer: \(0\)

4
A logistic population starts between 0 and its carrying capacity. What happens to \(P(t)\) as \(t\to\infty\)?
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It approaches the carrying capacity.

It does not grow without bound.

Answer: approaches L

5
Where is \(dP/dt=kP(1-P/L)\) largest for \(00\)?
A. \(P=L/2\)
B. \(P=L\)
C. \(P=0\)
D. \(P=L/4\)
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The quadratic \(P(1-P/L)\) opens downward and its zeros are 0 and \(L\).

The vertex is halfway between the zeros.

Answer: A. \(P=L/2\)

Level 2

Medium

Build the solution, or compare two populations with the same rate.

Example
\(dP/dt=0.2P(1-P/500)\) and \(P(0)=100\). Write \(P(t)\), then find \(t\) when \(P=250\).

\(A=\dfrac{500-100}{100}=4\)

\(P=\dfrac{500}{1+4e^{-0.2t}}\)

\(250=\dfrac{500}{1+4e^{-0.2t}}\)

\(1+4e^{-0.2t}=2\)

\(e^{-0.2t}=\dfrac{1}{4}\)

\(t=\dfrac{\ln 4}{0.2}\)

\(t=10\ln 2\)

Answer: \(t=10\ln 2\)

1
Write \(P(t)\) if \(L=800\), \(k=0.1\), and \(P(0)=200\).
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\(A=\dfrac{800-200}{200}=3\)

\(P=\dfrac{800}{1+3e^{-0.1t}}\)

Answer: \(P=\dfrac{800}{1+3e^{-0.1t}}\)

2
\(dP/dt=0.1P\left(1-\dfrac{P}{800}\right)\). Compute the rate at \(P=200\) and at \(P=600\).
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At 200: \(0.1\cdot 200\cdot(1-0.25)=15\).

At 600: \(0.1\cdot 600\cdot(1-0.75)=15\).

The two populations are the same distance from the midpoint 400, so the rates match.

Answer: both rates are 15

3
The carrying capacity is 800 and the population is 900. Is it increasing?
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\(1-\dfrac{900}{800}<0\)

With \(k>0\), the rate is negative.

The population is decreasing toward 800.

Answer: no, it is decreasing

4
Find the maximum value of \(dP/dt=0.4P\left(1-\dfrac{P}{100}\right)\).
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The maximum is at \(P=50\).

\(0.4\cdot 50\cdot \dfrac{1}{2}\)

\(10\)

Answer: \(10\)

5
Which model has a carrying capacity?
A. \(dP/dt=0.3P(1-P/40)\)
B. \(dP/dt=0.3P\)
C. \(dP/dt=0.3\)
D. \(dP/dt=-0.3P\)
Show Answer

A carrying capacity appears as a second factor that is zero at a positive population.

The plain exponential models do not level off.

Answer: A. \(dP/dt=0.3P(1-P/40)\)

Level 3

Hard

Rewrite cP(L − P), and describe concavity from the fastest point.

Example
Rewrite \(dP/dt=0.02P(800-P)\) in the form \(kP(1-P/L)\). Then find the rate at \(P=400\).

\(0.02P(800-P)=16P\left(1-\dfrac{P}{800}\right)\)

Because \(0.02\cdot 800=16\) and \(800-P=800(1-P/800)\).

So \(L=800\) and \(k=16\).

At half capacity: \(0.02\cdot 400\cdot 400\).

\(3200\)

Answer: \(k=16\), \(L=800\), rate 3200

1
\(P=\dfrac{400}{1+3e^{-kt}}\). This came from \(P(0)=100\) and \(L=400\). Find \(t\) when \(P=200\), in terms of \(k\).
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\(200=\dfrac{400}{1+3e^{-kt}}\)

\(1+3e^{-kt}=2\)

\(e^{-kt}=\dfrac{1}{3}\)

\(t=\dfrac{\ln 3}{k}\)

Answer: \(t=\dfrac{\ln 3}{k}\)

2
\(dP/dt=0.5P\left(1-\dfrac{P}{1000}\right)\) and \(P(0)=200\). Compare the starting rate with the fastest rate.
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Starting rate: \(0.5\cdot 200\cdot(1-0.2)=80\).

Fastest population: \(P=500\).

Fastest rate: \(0.5\cdot 500\cdot 0.5=125\).

The population is still speeding up at \(t=0\).

Answer: 80 now, 125 at the fastest

3
\(P=\dfrac{100}{1+2^{-t}}\). Find \(P(0)\), \(P(1)\), and \(P(2)\).
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\(P(0)=\dfrac{100}{1+1}=50\)

\(P(1)=\dfrac{100}{1+1/2}=\dfrac{200}{3}\)

\(P(2)=\dfrac{100}{1+1/4}=80\)

Answer: \(50,\ \dfrac{200}{3},\ 80\)

4
A logistic population has \(k>0\) and \(0
Show Answer

The rate \(dP/dt\) increases while \(P\) climbs from \(P_0\) to \(L/2\).

An increasing rate means \(P\) is concave up on that part.

After \(L/2\), the rate decreases while \(P\) is still increasing, so \(P\) is concave down.

Answer: concave up, then concave down

5
Find \(\lim_{t\to\infty}\dfrac{800}{1+3e^{-0.1t}}\).
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\(e^{-0.1t}\to 0\)

The denominator approaches 1.

The limit is 800, the carrying capacity.

Answer: \(800\)

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