Wemazi

Topic 7.9 · BC

Find the fastest growth

Builds on deciding whether a logistic population grows or declines. 6 minutes.

Where is the rate greatest?

Bacteria grow in a dish. The curve shows the rate \(\cy{\frac{dP}{dt}}\), in thousands of bacteria per hour, against the number \(P\) of bacteria, in thousands. It is green where the number increases and red where it decreases.

Move the orange marker from 0 to 60. The rate rises to a highest point and then falls back to 0.

Press the first button to mark the highest point. It is at \(P = 30\), where the rate is 7.5 thousand per hour.

Press the second button to compare two points. At \(P = 15\) the rate is 5.625, and at \(P = 45\) it is again 5.625, because 15 and 45 lie the same distance from 30.

The curve crosses the axis at 0 and at 60. The highest point, at 30, lies halfway between these two crossings.

The peak is halfway to K

The same graph without numbers shows the rate of any logistic population, with the crossings at 0 and at \(K\).

Multiplied out, the rate \(kP\left(1 - \frac{P}{K}\right)\) is \(kP - \frac{k}{K}P^2\). Its graph is a parabola that opens downward, because the coefficient of \(P^2\) is negative.

A parabola is symmetric about the vertical line through the midpoint of its crossings, and its highest point lies on that line.

The midpoint of 0 and \(K\) is \(\cy{\frac{K}{2}}\), so the population grows fastest when it is half of the carrying capacity.

Move the slider for \(K\) or \(k\). The peak stays above \(\frac{K}{2}\), and its height changes.

Notation

  1. Call the rate \(R\) and write it as a function of the population \(P\).

    \[\begin{aligned} R(P) &= kP\left(1 - \frac{P}{K}\right) \\ &= kP - \frac{k}{K}P^{2} \end{aligned}\]

  2. At a maximum, the derivative of \(R\) with respect to \(P\) is 0.

    \[R'(P) = k - \frac{2k}{K}P = 0\]

  3. Solve for \(P\). The population of fastest growth is half of the carrying capacity.

    \[P = \cy{\frac{K}{2}}\]

  4. This point is a maximum, not a minimum, because the parabola opens downward.

  5. To find the greatest rate, substitute \(P = \frac{K}{2}\) into the rate.

    \[\begin{aligned} R\!\left(\frac{K}{2}\right) &= k\cdot\frac{K}{2}\left(1 - \frac12\right) \\ &= \cy{\frac{kK}{4}} \end{aligned}\]

When it fails

Press the first button to start the population to the right of \(\frac{K}{2}\). Then press the second button. The population increases toward \(K\), so the marker moves away from the peak and the rate falls.

The rate is greatest at \(\frac{K}{2}\), but a population passes through \(\frac{K}{2}\) only if it starts below it. If it starts above \(\frac{K}{2}\), its fastest growth is at time 0.

Example 1: A rumor

A rumor spreads through a school of 1000 students. The number \(P\) who have heard it satisfies \(\dfrac{dP}{dt} = 0.3P\left(1 - \dfrac{P}{1000}\right)\), where \(t\) is in days. For what value of \(P\) does the rumor spread fastest, and what is the greatest rate?

  1. The curve crosses the axis at 0 and at \(K = 1000\). The peak is halfway between them.

    \[P = \frac{1000}{2} = \cy{500}\]

  2. The rumor spreads fastest when 500 students have heard it. Substitute 500 into the rate to find how fast.

    \[0.3(500)\left(1 - \frac{500}{1000}\right) = \cy{75}\]

  3. The greatest rate is 75 students per day.

Example 2: A town

The population of a town satisfies \(\dfrac{dP}{dt} = 0.004P(700 - P)\). Find the population at which the town grows fastest and the greatest rate.

  1. The rate is 0 at \(P = 0\) and at \(P = 700\), so \(K = 700\).

  2. A tempting wrong move: The population is largest at 700, so the town grows fastest there.

    \[P = 700\]

    At \(P = 700\) the curve crosses the axis, so the rate there is 0. The peak of the curve lies above a smaller population.

  3. The peak of the parabola is halfway between its crossings.

    \[P = \frac{0 + 700}{2} = \cy{350}\]

  4. Substitute 350 into the rate. The greatest rate is 490 people per unit of time.

    \[0.004(350)(700 - 350) = \cy{490}\]

    The rate is \(0.004P(700 - P) = 2.8P\left(1 - \frac{P}{700}\right)\), so \(k = 2.8\) and \(\frac{kK}{4} = \frac{2.8\cdot700}{4} = 490\) as well.

Practice

A seal colony satisfies \(\dfrac{dP}{dt} = 0.25P\left(1 - \dfrac{P}{320}\right)\). Find the population at which the colony grows fastest and the greatest rate.

The greatest rate is the height of the highest point of the curve, halfway between the crossings. Move the marker there to see it.

Solution

  1. The curve crosses the axis at 0 and at \(K = 320\).

  2. The population of fastest growth is halfway between them.

    \[P = \frac{320}{2} = \cy{160}\]

  3. Substitute 160 into the rate. The greatest rate is 20 seals per unit of time.

    \[0.25(160)\left(1 - \frac{160}{320}\right) = \cy{20}\]

Questions

The graph shows the rate \(\dfrac{dP}{dt}\) of the number of fish \(P\) on a reef. The equation is \(\dfrac{dP}{dt} = 0.2P\left(1 - \dfrac{P}{800}\right)\).

  1. At which population is the rate of growth the greatest?

    • 400
    • 800
    • 40
    • 200
    Answer

    400

  2. How does the rate at \(P = 200\) compare with the rate at \(P = 600\)?

    • They are equal.
    • The rate at 600 is greater, because 600 is a larger population.
    • The rate at 200 is greater, because it is closer to 0.
    • The rate at 600 is 0.
    Answer

    They are equal.

  3. A population satisfies \(\dfrac{dP}{dt} = 0.05P\left(1 - \dfrac{P}{1200}\right)\). What is the greatest rate?

    • 15
    • 30
    • 600
    • 60
    Answer

    15

Summary

You find the crossings of the rate curve at 0 and \(K\), take the midpoint \(\frac{K}{2}\), and substitute it into the rate to find \(\frac{kK}{4}\).

This works because the rate is a downward parabola, and its highest point lies halfway between its crossings.

You use it to find the size at which a logistic population grows fastest, and how fast it grows there.

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