Curve sketching, question 14

Topic: Curve sketching
Difficulty: Standard
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §7 CLP-1 §3.6Key results

The number of fruit flies in a jar, in hundreds, $t$ days after the jar is set up, is modelled by the logistic function below. At what time is the population growing fastest?

P(t) = \frac{10}{1 + 9e^{-t}}, \qquad t \ge 0

Give the answer in days.

Your answer

days

An exact expression, or a decimal to two decimal places.

How sure are you?

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Theory UBC Math 100 §7 CLP-1 §3.6Key results

Worked solution

Strategy

The growth rate is P′, and the largest value of P′ is where P″ changes sign: the inflection point of the S-shaped curve. Differentiate twice and find where P″ changes sign.

Working

$P(t) = 10\left(1 + 9e^{-t}\right)^{-1} \quad\Longrightarrow\quad P'(t) = \frac{90e^{-t}}{(1 + 9e^{-t})^{2}} > 0$
Chain rule; the derivative of 9e^−t is −9e^−t. P always increases, so the question is where it increases fastest.
$P''(t) = \frac{\begin{gathered} -90e^{-t}(1 + 9e^{-t})^{2} \\ {} - 90e^{-t}\cdot 2(1 + 9e^{-t})(-9e^{-t}) \end{gathered}}{(1 + 9e^{-t})^{4}} = \frac{90e^{-t}\left(9e^{-t} - 1\right)}{(1 + 9e^{-t})^{3}}$
Quotient rule, then cancel one factor of 1 + 9e^−t and collect −1 − 9e^−t + 18e^−t.
$P''(t) = 0 \iff 9e^{-t} = 1 \iff t = \log 9$
Every other factor is positive.
$t < \log 9:\ P'' > 0; \qquad t > \log 9:\ P'' < 0$
P′ rises and then falls, so it is largest at t = log 9 ≈ 2.20 days, when P = 5: half the limiting value 10.

Answer

t = \log 9 \approx 2.20 \text{ days}

Check

P(log 9) = 10/(1 + 1) = 5, exactly half the limiting population of 10, which is where a logistic curve grows fastest.

Where this goes wrong

Setting P′ = 0 to find the fastest growth. P′ is never 0, because the population always grows; the fastest growth is the largest value of P′, where P″ changes sign.