Use a logistic solution or observations

Fit the curve to the measurements

The mass \(P\) of algae in a tank, in kilograms, was measured on days 0, 4 and 8. The three dots show 50, 200 and 500 kilograms.

The tank supports at most 800 kilograms, so the dashed line marks this carrying capacity.

The curve is a logistic model with capacity 800 that starts at 50. The slider sets the constant \(k\), and the orange readouts give the curve's values on days 4 and 8.

Move the slider. A small \(k\) gives a curve below the dots at days 4 and 8, and a large \(k\) gives a curve above them.

The curve passes through the dots when \(k\) is about 0.40, so the measurements fix the constant \(k\).

Three numbers fix the solution

The graph now has no numbers. It shows any logistic solution, which three numbers fix: the capacity \(K\), the start \(P_0\) and the constant \(k\).

The solution is a fraction with \(K\) on top and 1 plus a decaying exponential below. As \(t\) grows, the exponential approaches 0, so \(P\) approaches \(K\). At \(t = 0\) the exponential equals 1, and the number in front of it is chosen so that \(P\) equals \(P_0\).

Move the slider for \(P_0\). The start moves, and the curve still ends at \(K\). Move the slider for \(k\), and the curve climbs faster or slower.

The open dot on the orange line at \(K/2\) marks where the curve changes from concave up to concave down.

Notation

1. The solution of the logistic equation \(\frac{dP}{dt} = kP\left(1 - \frac{P}{K}\right)\) is this function, where the number \(A\) depends on the start.

   \[P(t) = \frac{K}{1 + Ae^{-kt}}\]

2. Substitute \(t = 0\). The exponential equals 1, so the start \(P_0\) gives an equation that you solve for \(A\).

   \[P_0 = \frac{K}{1 + A} \;\Rightarrow\; A = \frac{K - P_0}{P_0}\]

3. To find the population at a time, substitute that time into the formula.

4. To find \(k\) or a time from a measurement \(P_1\), set \(P = P_1\) and solve for the exponential. The orange line marks \(P_1\), and the curve meets it at the time \(t\).

   \[\begin{aligned} 1 + Ae^{-kt} &= \frac{K}{P_1} \\ e^{-kt} &= \frac{K/P_1 - 1}{A} \end{aligned}\]

5. Take the natural logarithm of both sides. If \(k\) is known, this gives the time \(t\). If \(t\) is known, it gives \(k\).

   \[\begin{aligned} -kt &= \ln\!\left(\frac{K/P_1 - 1}{A}\right) \\ t &= -\frac{1}{k}\ln\!\left(\frac{K/P_1 - 1}{A}\right) \end{aligned}\]

When it fails

Press the button to raise the orange line, the target \(P_1\), toward \(K\). The curve meets it later and later, and at \(K\) the curve never meets it.

Setting \(P_1 = K\) makes \(K/P_1 - 1 = 0\), so \(e^{-kt} = 0\), which has no solution. The population approaches \(K\) but never reaches it, so the target must lie between \(P_0\) and \(K\).

Example 1: The population at a given time

A beetle population satisfies \(\dfrac{dP}{dt} = 0.3P\left(1 - \dfrac{P}{1200}\right)\) with \(P(0) = 100\), where \(t\) is in months. Find \(P(5)\).

1. Read the constants from the equation.

   \[\begin{aligned} k &= 0.3 \\ K &= 1200 \\ P_0 &= 100 \end{aligned}\]

2. Find \(A\) from the start.

   \[A = \frac{1200 - 100}{100} = 11\]

3. Write the solution and substitute \(t = 5\).

   \[\begin{aligned} P(5) &= \frac{1200}{1 + 11e^{-0.3\cdot5}} \\ &= \frac{1200}{1 + 11e^{-1.5}} \end{aligned}\]

4. The population after 5 months is about 347 beetles, between the start 100 and the capacity 1200.

   \[P(5) \approx \cy{347}\]

   Growth without a limit, \(100e^{0.3t}\), would give about 448 beetles. The capacity slows the growth.

Example 2: The time to reach a population

For the beetles in Example 1, \(P(t) = \dfrac{1200}{1 + 11e^{-0.3t}}\). Find the time at which \(P = 900\).

1. Set \(P = 900\). Multiply both sides by the denominator. Then divide by 900.

   \[1 + 11e^{-0.3t} = \frac{1200}{900} = \frac43\]

2. A tempting wrong move: Divide both sides by 11. This gives \(e^{-0.3t} = \frac{4}{33}\), so \(t \approx 7.0\).

   \[11e^{-0.3t} = \frac43\]

   This step drops the 1 on the left, as if \(1 + 11e^{-0.3t}\) were \(11e^{-0.3t}\). At \(t \approx 7.0\) the curve is at about 514, far below the orange line at 900.

3. Subtract 1 first. Then divide by 11.

   \[\begin{aligned} 11e^{-0.3t} &= \frac43 - 1 = \frac13 \\ e^{-0.3t} &= \frac{1}{33} \end{aligned}\]

4. Take the natural logarithm of both sides. The population reaches 900 after about 11.7 months, where the curve meets the orange line.

   \[\begin{aligned} -0.3t &= \ln\tfrac{1}{33} = -\ln 33 \\ t &= \frac{\ln 33}{0.3} \approx \cy{11.7} \end{aligned}\]

Practice

A culture has carrying capacity 500 and starts at 20. After 3 days its population is 100. Find \(k\).

Move the slider until the curve passes through the dot. For the exact \(k\), substitute \(t = 3\) and \(P = 100\), and solve for the exponential before you take the logarithm.

Questions

The graph shows a logistic solution with carrying capacity 600 and starting population 150.

1. At about what time is the population 450?

   • 8
   • 4
   • 12
   • 450
   Answer

   8

2. Which formula gives this curve?

   • \(P = \dfrac{600}{1 + 3e^{-kt}}\)
   • \(P = \dfrac{600}{1 + 4e^{-kt}}\)
   • \(P = \dfrac{600}{1 + 450e^{-kt}}\)
   • \(P = \dfrac{150}{1 + 3e^{-kt}}\)
   Answer

   \(P = \dfrac{600}{1 + 3e^{-kt}}\)

3. A population has capacity 800, \(P(0) = 200\) and \(k = 0.1\). At what time does it reach 400?

   • \(10\ln 3 \approx 11.0\)
   • \(10\ln\frac32 \approx 4.1\)
   • \(10\ln 4 \approx 13.9\)
   Answer

   \(10\ln 3 \approx 11.0\)

Summary

You find \(A\) from the start, write \(P = \frac{K}{1 + Ae^{-kt}}\), and then solve for the unknown population, time or \(k\).

This works because the exponential equals 1 at \(t = 0\) and approaches 0 as \(t\) grows, so the curve runs from \(P_0\) to \(K\).

Given a logistic model, you use it to find a population, \(k\), or the time to reach a value between \(P_0\) and \(K\).

This is one of five skills in topic 7.9 of Which & Why: AP Calculus.