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Euler and Logistic Methods · Tutorial 594 of 999

How Carrying Capacity and Growth Rate Shape a Logistic Curve

Use the explicit logistic solution to interpret carrying capacity as the limiting population and growth rate as a parameter that changes the curve’s timing.

Advanced 12 min read

What You'll Learn

  • Interpret the parameters \(K\) and \(r\) in the logistic differential equation.
  • Use the explicit solution to compare populations with different carrying capacities.
  • Explain how changing \(r\) rescales time while preserving the limiting level.
  • Distinguish the time to reach a fixed population from the time to reach a fixed fraction of capacity.
  • Check logistic formulas against the initial value and differential equation.

Why Two Logistic Curves Can Differ

A logistic model describes growth that slows as a population approaches an upper level. Two parameters have especially clear roles: \(K\), the carrying capacity, sets that limiting level, while \(r\), the growth-rate parameter, controls how quickly the population changes relative to its distance from capacity.

These roles are related but not identical. Changing \(K\) changes the population scale and can also change when the curve reaches a chosen target. Changing \(r\), with the other conditions held fixed, changes the timing without changing the limiting population. To compare curves correctly, we must say which quantities are being held fixed and what kind of target we mean.

Model and conditions: We consider the logistic initial-value problem \(P'=rP(1-P/K)\), \(P(0)=P_0\), with \(K>0\), \(r>0\), and \(0<P_0<K\). Here \(P(t)\) is population, \(K\) is carrying capacity, and \(r\) has units of inverse time.

The Solution and the Meaning of Its Parameters

For these conditions, the explicit solution is

$$ P(t)=\frac{K}{1+A e^{-rt}}, \qquad A=\frac{K-P_0}{P_0}=\frac{K}{P_0}-1. $$

The constant \(A\) is determined by the initial value. In particular, changing \(K\) while keeping \(P_0\) fixed also changes \(A\); it is not correct to compare such models while silently treating \(A\) as unchanged. The exponential factor \(e^{-rt}\) shows how \(r\) affects the rate at which the denominator approaches \(1\).

When \(t=0\), the formula gives \(P(0)=K/(1+A)=P_0\). As \(t\) increases, \(e^{-rt}\) approaches zero, so the denominator approaches \(1\) and \(P(t)\) approaches \(K\). Thus \(K\) is the long-run limiting population for the stated positive-growth model.

A useful rearrangement describes the population relative to the capacity it has not yet reached:

$$ \frac{P(t)}{K-P(t)}=\frac{e^{rt}}{A}, \qquad \ln\!\left(\frac{P(t)}{K-P(t)}\right)=rt-\ln A. $$

The ratio on the left compares the current population with the remaining gap \(K-P(t)\). The logarithm of that ratio changes linearly with time, with slope \(r\). This makes the two parameters’ roles especially visible: \(K\) sets the reference level, and \(r\) sets the time scale of the change.

Why the Growth-Rate Parameter Rescales Time

Hold \(K\) and \(P_0\) fixed. Then \(A=K/P_0-1\) is fixed as well. If \(P_r(t)\) denotes the solution using parameter \(r\), then

$$ P_r(t)=\frac{K}{1+A e^{-rt}}=P_1(rt). $$

This identity means that the curve with parameter \(r\) reaches a given population at \(1/r\) times the corresponding time for the curve with parameter \(1\), provided the same time unit is used. More generally, replacing \(r\) by \(cr\), for \(c>0\), compresses the time scale by a factor of \(c\): the population that the original curve reaches at time \(t\) is reached by the faster curve at time \(t/c\).

The same conclusion follows directly from the differential equation. At a population \(P\) between \(0\) and \(K\), the rate is \(rP(1-P/K)\). Increasing \(r\) multiplies that rate by the same factor at every such population. The initial value and limiting value are unchanged, but the model moves between them on a different time scale.

How Carrying Capacity Changes the Curve

Now hold \(P_0\) and \(r\) fixed and change \(K\). The limiting level changes because \(P(t)\) approaches \(K\). The entire trajectory also changes: for any fixed \(t>0\), a larger capacity gives a larger predicted population under these conditions.

To justify that comparison, write \(B=e^{-rt}\), which satisfies \(0<B<1\) for \(t>0\). Substituting \(A=K/P_0-1\) gives

$$ P(t)=\frac{K}{1+B(K/P_0-1)} =\frac{K}{1-B+BK/P_0}. $$

Differentiate this expression with respect to \(K\), keeping \(P_0\), \(r\), and \(t\) fixed. The denominator is \(D=1-B+BK/P_0\), so the quotient rule gives

$$ \frac{\partial P}{\partial K} = \frac{D-KB/P_0}{D^2} = \frac{1-B}{D^2}>0. $$

Therefore, at every positive time, increasing \(K\) increases the predicted population as well as its eventual limiting level. This comparison assumes the same initial population and growth-rate parameter; changing those conditions too would be a different comparison.

Targets: An Absolute Population or a Fraction of Capacity?

Timing depends on how the target is specified. To find the time to reach a fraction \(p\) of capacity, where \(P(t)=pK\) and \(0<p<1\), substitute into the solution:

$$ p=\frac{1}{1+A e^{-rt}} \quad\Longrightarrow\quad t=\frac{1}{r}\ln\!\left(A\frac{p}{1-p}\right). $$

This time depends on both parameters through \(A=K/P_0-1\) and the factor \(1/r\). If \(P_0\) is fixed and \(K\) increases, then \(A\) increases. For a fixed fraction \(p\) that is above the initial fraction, the curve takes longer to reach that fraction of its now-higher capacity.

By contrast, for a fixed absolute target \(Y\), with \(P_0<Y<K\), solving \(P(t)=Y\) gives

$$ t=\frac{1}{r} \ln\!\left( \frac{Y(K-P_0)}{P_0(K-Y)} \right). $$

For a target \(Y\) below both capacities being compared, increasing \(K\) makes the time to reach \(Y\) shorter. A higher capacity means less crowding at that same absolute population in the model. These different timing comparisons are not contradictory: “half of capacity” is a moving target when \(K\) changes, but a fixed population \(Y\) is not.

Worked Example: A Population with Capacity 1,000

Worked Example: Evaluating a Logistic Population

A population starts at \(P_0=100\) and follows the logistic model with \(K=1000\) individuals and \(r=0.4\) per year. Find the population after \(5\) years, rounded to the nearest whole individual, and check the model’s initial rate.

First compute the constant in the solution:

$$ A=\frac{K}{P_0}-1=\frac{1000}{100}-1=10-1=9. $$

Substitute \(t=5\) years:

$$ P(5)=\frac{1000}{1+9e^{-(0.4)(5)}} =\frac{1000}{1+9e^{-2}} \approx\frac{1000}{1+9(0.135335)} =\frac{1000}{2.218015} \approx 450.85. $$

Rounded to the nearest whole individual, the model predicts \(451\) individuals. To check the initial value, \(P(0)=1000/(1+9)=100\), as required. The differential equation gives the initial rate \(P'(0)=0.4(100)(1-100/1000)=36\) individuals per year. Differentiating the explicit solution gives \(P'(t)=rP(t)(1-P(t)/K)\), so it satisfies the model and has the same initial rate.

Worked Example: Changing Capacity at a Fixed Time

Worked Example: Comparing Two Carrying Capacities

Two populations each start at \(80\) individuals and have \(r=0.3\) per year. Compare their model predictions after \(4\) years if their capacities are \(800\) and \(1600\) individuals. Round the final populations to the nearest tenth.

For \(K=800\), the constant is \(A=800/80-1=9\). Since \((0.3)(4)=1.2\),

$$ P_{800}(4)=\frac{800}{1+9e^{-1.2}} \approx\frac{800}{1+9(0.301194)} =\frac{800}{3.710746} \approx 215.6. $$

For \(K=1600\), the constant is \(A=1600/80-1=19\). Therefore,

$$ P_{1600}(4)=\frac{1600}{1+19e^{-1.2}} \approx\frac{1600}{1+19(0.301194)} =\frac{1600}{6.722686} \approx 238.0. $$

Thus the larger-capacity model predicts \(238.0\) individuals rather than \(215.6\) after \(4\) years. Both formulas check at \(t=0\): \(800/(1+9)=80\), and \(1600/(1+19)=80\). The comparison changes \(K\) while keeping the initial population and \(r\) fixed, so the higher prediction agrees with the parameter comparison derived above.

Worked Example: Changing the Growth-Rate Parameter

Worked Example: Comparing Two Growth Rates

Let \(K=600\) individuals and \(P_0=60\) individuals. Compare the populations after \(3\) years for \(r=0.2\) per year and \(r=0.5\) per year. Then find how long each model takes to reach half its capacity. Round populations to the nearest tenth and times to the nearest hundredth of a year.

For both models, \(A=600/60-1=9\). With \(r=0.2\),

$$ P_{0.2}(3) =\frac{600}{1+9e^{-(0.2)(3)}} =\frac{600}{1+9e^{-0.6}} \approx\frac{600}{1+9(0.548812)} =\frac{600}{5.939308} \approx 101.0. $$

With \(r=0.5\),

$$ P_{0.5}(3) =\frac{600}{1+9e^{-(0.5)(3)}} =\frac{600}{1+9e^{-1.5}} \approx\frac{600}{1+9(0.223130)} =\frac{600}{3.008170} \approx 199.5. $$

To reach half the capacity, set \(p=1/2\) in the fraction formula. Since \(p/(1-p)=1\), the time is \(t=\ln(9)/r\). Thus the \(r=0.2\) model takes \(\ln(9)/0.2\approx10.99\) years, while the \(r=0.5\) model takes \(\ln(9)/0.5\approx4.39\) years. Checking the half-capacity times, \(e^{rt}=9\) in either case, so \(P=600/(1+9e^{-rt})=600/(1+1)=300\) individuals. The faster model has the same capacity but reaches that level sooner.

A Common Mistake: Mixing Up the Target

A common error is to say, “Increasing \(K\) always makes the population take longer to reach a target.” That conclusion can arise from thinking about a fixed fraction of capacity, then applying it to a fixed absolute population—or the reverse. The two targets behave differently.

For example, let \(P_0=10\), \(r=1\) per time unit, and compare \(K=100\) with \(K=200\). For a fixed target of \(50\), the first model takes

$$ t_{100} =\ln\!\left(\frac{50(100-10)}{10(100-50)}\right) =\ln(9) \approx 2.197. $$

The second takes

$$ t_{200} =\ln\!\left(\frac{50(200-10)}{10(200-50)}\right) =\ln\!\left(\frac{19}{3}\right) \approx 1.846. $$

So the model with the larger capacity reaches the fixed target \(50\) sooner, not later. But half of capacity means a target of \(50\) for \(K=100\) and \(100\) for \(K=200\); the corresponding times are \(\ln(9)\approx2.197\) and \(\ln(19)\approx2.944\). Always identify whether a comparison concerns a fixed population or a fixed fraction before drawing a conclusion.

Putting the Parameter Roles Together

For a positive logistic model starting between zero and capacity, the solution approaches \(K\), so the carrying capacity determines the limiting population. The parameter \(r\), measured in inverse time, controls the time scale: with \(K\) and \(P_0\) fixed, a larger \(r\) makes the population reach any specified intermediate level sooner.

Changing \(K\) does more than move the horizontal asymptote. With the same \(P_0\) and \(r\), it changes the model’s predictions at positive times and changes \(A\). When describing timing, state the target explicitly. The time to reach a fraction of capacity and the time to reach a fixed number of individuals answer different questions.

Key takeaway: \(K\) sets the logistic curve’s limiting population and scale; \(r\) rescales its timing. Comparisons are meaningful only when the initial population, other parameters, and target definition are specified.

Check Your Understanding

Answer these questions using the logistic model and the conditions stated in the tutorial.

  1. For \(K=500\) and \(P_0=50\), find \(A\) and verify the solution’s value at \(t=0\).
  2. If \(K\) and \(P_0\) stay fixed while \(r\) doubles, what happens to the time required to reach a specified intermediate population?
  3. Why does increasing \(K\), with \(P_0\) and \(r\) fixed, increase the predicted population at a positive time?
  4. Explain why “time to reach half of capacity” is not the same comparison as “time to reach 100 individuals” when \(K\) changes.
  5. What units should \(r\) have if time is measured in months and \(P\) is measured in individuals?

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