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Calculus 2

Logistic growth

A population grows at a rate proportional to its size while it is small, but the rate falls to zero as the population nears what its surroundings can support. What does that one equation force the population to do, and can it be solved exactly?

You need
Exponential growth and decay modelsCalculus 1Integration by partial fractionsCalculus 2Concavity and inflection pointsCalculus 1
Where you are 7 in Differential equations
Calculus8
Differential equations7
▴ 2 aboveSlope fieldsEuler's methodSeparable differential equationsExponential growth and decay models
Logistic growth

Why this exists

The model \dfrac{dP}{dt} = kP of Exponential growth and decay models predicts that a population grows without bound, and no pond can hold infinitely many fish. Food, oxygen and space run short, and each fish then adds fewer young than before. Logistic growth: when growth levels off read the resulting S-shaped curve from a given formula. This lesson starts one step earlier, from a statement about the rate alone: growth is proportional to size when the population is small and falls off as it nears a ceiling. From that equation, without being handed any formula, calculus finds the ceiling, shows that it attracts every positive solution, locates the moment of fastest growth, and then produces the formula itself.

Definition

Day 1: the logistic equation

A population P(t) follows the logistic differential equation with growth constant k > 0 and carrying capacity K > 0 if

\frac{dP}{dt} = kP\left(1 - \frac{P}{K}\right).

Dividing by P gives the per-capita growth rate, the rate of growth per individual:

\frac{1}{P}\,\frac{dP}{dt} = k\left(1 - \frac{P}{K}\right).

It falls in a straight line from k when P is near 0 to 0 when P = K, and is negative above K. For a small population the equation is close to \dfrac{dP}{dt} = kP, exponential growth; near K the growth stops.

The equation is autonomous: the rate depends on P only, not on t. As in Slope fields, a value c with zero rate gives an equilibrium solution P = c, and the equilibrium is stable if solutions starting near it on both sides approach it, unstable if they move away.

Multiplying out, kP\left(1 - \dfrac{P}{K}\right) = kP - \dfrac{k}{K}P^2. So an equation written \dfrac{dP}{dt} = aP - bP^2 with a, b > 0 is logistic with k = a and K = \dfrac{a}{b}.

Day 2: the solution formula

A solution of the logistic equation with P(0) = P_0 > 0 is a function P(t) that satisfies it for all t \ge 0 and starts at P_0. Day 2 finds every such solution by separating the variables and integrating with partial fractions.

Intuition

The rate kP\left(1 - \dfrac{P}{K}\right) is a downward parabola in P (Figure 1). It is zero at P = 0 and at P = K, positive in between, and negative above K. A population below K grows, and one above K shrinks; either way it moves toward K. The parabola is highest halfway between its zeros, at P = \dfrac{K}{2}, so that is where the population grows fastest.

100 200 300 400 500 600 700 −100 −50 0 50 P dP/dt
Figure 1The rate \dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{500}\right) as a function of P. It is zero at the two equilibria, positive between them, negative above 500, and largest at P = 250.

Read along a solution curve (Figure 2). Starting small, the rate climbs as P climbs, and the curve bends upward. Past \dfrac{K}{2} the rate starts to fall, and the curve bends downward. Close to K the rate is near 0 and the curve flattens onto the line P = K. A population that starts above K falls fastest at the start and slows as it nears K, so its curve bends upward the whole way.

4 8 12 16 250 500 750 t P
Figure 2Solutions of \dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{500}\right) from P(0) = 50, 150 and 800. Each approaches K = 500 without reaching it. The S-curve from 50 bends at P = 250, at t = \dfrac{\ln 9}{0.4} \approx 5.49 months.

Results

Equilibria. The logistic equation has exactly two equilibrium solutions, P = 0, which is unstable, and P = K, which is stable.

Concavity and the fastest growth. Along any solution,

\frac{d^2P}{dt^2} = k\left(1 - \frac{2P}{K}\right)\frac{dP}{dt}.

For 0 < P < K a solution is concave up while P < \dfrac{K}{2} and concave down while P > \dfrac{K}{2}. The rate \dfrac{dP}{dt} is largest at P = \dfrac{K}{2}, where it equals \dfrac{kK}{4}. A solution above K is decreasing and concave up.

The solution formula. For P_0 > 0 the solution with P(0) = P_0 is

P(t) = \frac{K}{1 + Ae^{-kt}}, \quad A = \frac{K - P_0}{P_0},

and \lim_{t \to \infty} P(t) = K. If 0 < P_0 < \dfrac{K}{2}, the fastest growth comes at t = \dfrac{\ln A}{k}.

No solution crosses K. A solution that starts below K stays below K for all t \ge 0, and one that starts above K stays above it. Neither ever equals K.

Why it is true

Equilibria and stability. A constant P = c has derivative 0, so it is a solution exactly when kc\left(1 - \dfrac{c}{K}\right) = 0, that is, c = 0 or c = K. Between 0 and K both factors kP and 1 - \dfrac{P}{K} are positive, so solutions rise; above K the second factor is negative, so solutions fall. Solutions on both sides of K head toward it, so it is stable. Just above 0 solutions rise away from 0 (and just below, where P has no meaning as a count, the rate is negative and they fall away), so 0 is unstable.

The second derivative. Write f(P) = kP - \dfrac{k}{K}P^2, so \dfrac{dP}{dt} = f(P). By the chain rule,

\frac{d^2P}{dt^2} = f'(P)\,\frac{dP}{dt} = k\left(1 - \frac{2P}{K}\right)\frac{dP}{dt}.

For 0 < P < K, \dfrac{dP}{dt} > 0, so the sign of the second derivative is the sign of 1 - \dfrac{2P}{K}: positive below \dfrac{K}{2} and negative above, as Concavity and inflection points reads it. For P > K, both factors are negative and the product is positive. The rate itself is the quadratic f(P), whose vertex is where f'(P) = 0, at P = \dfrac{K}{2}, with f\left(\dfrac{K}{2}\right) = k \cdot \dfrac{K}{2} \cdot \dfrac{1}{2} = \dfrac{kK}{4}.

The formula, by partial fractions. Separate, writing kP\left(1 - \dfrac{P}{K}\right) = \dfrac{kP(K - P)}{K}:

\int \frac{K}{P(K - P)}\,dP = \int k\,dt.

The fraction splits as \dfrac{K}{P(K - P)} = \dfrac{1}{P} + \dfrac{1}{K - P}, which can be checked by adding the right side over the common denominator P(K - P). Integrating,

\begin{aligned} \ln|P| - \ln|K - P| &= kt + C, \\ \left\lvert\frac{P}{K - P}\right\rvert &= e^C e^{kt}. \end{aligned}

Drop the absolute value with a constant B of either sign: \dfrac{P}{K - P} = Be^{kt}, and t = 0 gives B = \dfrac{P_0}{K - P_0}. Solving for P, P = (K - P)Be^{kt}, so P\left(1 + Be^{kt}\right) = KBe^{kt} and

\begin{aligned} P &= \frac{K}{1 + \frac{1}{B}e^{-kt}} = \frac{K}{1 + Ae^{-kt}}, \\ A &= \frac{1}{B} = \frac{K - P_0}{P_0}. \end{aligned}

The division by P(K - P) assumed P_0 \ne K; the case P_0 = K is the equilibrium, which the formula also gives, with A = 0. Differentiating back confirms the formula: with u = 1 + Ae^{-kt}, P' = \dfrac{KAke^{-kt}}{u^2}, while kP\left(1 - \dfrac{P}{K}\right) is \dfrac{kK}{u}\cdot\dfrac{u - 1}{u}, which is \dfrac{kKAe^{-kt}}{u^2}, the same. As t \to \infty, e^{-kt} \to 0 and P \to K. When A > 1, P = \dfrac{K}{2} exactly when Ae^{-kt} = 1, at t = \dfrac{\ln A}{k}.

Never crossing K. From the formula, K - P = \dfrac{KAe^{-kt}}{1 + Ae^{-kt}}. For P_0 < K, A > 0 and this is positive. For P_0 > K, -1 < A < 0, so 1 + Ae^{-kt} \ge 1 + A > 0 for t \ge 0 and K - P is negative. The same fact follows without the formula from a theorem stated here without proof, the uniqueness theorem: if f and f' are continuous, then through each point (t_0, P_0) passes exactly one solution of \dfrac{dP}{dt} = f(P). The constant P = K is a solution. If another solution reached K at some time t_0, two solutions would pass through (t_0, K), so they would be the same solution, and the other one would have been K all along.

Worked examples

Example 1

Day 1: Nigora’s pond and its equilibria

Nigora stocks a fish farm pond with 50 trout. She models the population, t in months, by \dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{500}\right). Find the equilibria and classify them, find the rate at P = 50, 250, 450 and 800, and describe the long run.

Solution

Here k = 0.4 per month and K = 500. The rate is zero when 0.4P = 0 or 1 - \dfrac{P}{500} = 0, so the equilibria are P = 0 and P = 500. The sign is fixed on each interval between them, so one value per interval decides it:

P\dfrac{dP}{dt}The population
500.4 \cdot 50 \cdot 0.9 = 18rises
2500.4 \cdot 250 \cdot 0.5 = 50rises
4500.4 \cdot 450 \cdot 0.1 = 18rises
8000.4 \cdot 800 \cdot (-0.6) = -192falls

Solutions rise below 500 and fall above it, so 500 is stable and 0 is unstable. Nigora’s 50 trout grow toward 500. The rates at 50 and 450 are equal, 18 trout per month, because the parabola of Figure 1 is symmetric about 250: the factor P is small at one end and the factor 1 - \dfrac{P}{500} is small at the other.

Example 2

Day 1: where the S bends

For Nigora’s pond, find the second derivative of P along a solution, the value of P where the growth is fastest and the rate there, and the concavity of her population’s graph at P = 100 and at P = 400.

Solution

By the Results, with \dfrac{2P}{K} = \dfrac{P}{250},

\frac{d^2P}{dt^2} = 0.4\left(1 - \frac{P}{250}\right)\frac{dP}{dt}.

While 0 < P < 500 the rate is positive, so the second derivative is positive for P < 250 and negative for P > 250. The growth is fastest at P = 250, at \dfrac{kK}{4} = \dfrac{0.4 \cdot 500}{4} = 50 trout per month, the peak in Figure 1.

At P = 100, \dfrac{dP}{dt} = 0.4 \cdot 100 \cdot 0.8 = 32 and \dfrac{d^2P}{dt^2} = 0.4 \cdot 0.6 \cdot 32 = 7.68: concave up, growth speeding up. At P = 400, the rate is again 32, and \dfrac{d^2P}{dt^2} = 0.4 \cdot (-0.6) \cdot 32 = -7.68: concave down, growth slowing. The graph has its inflection point where P = 250; Day 2 finds the time.

Example 3

Day 1: Tahmina’s pond, multiplied out

Tahmina’s pond follows \dfrac{dP}{dt} = 0.3P - 0.0004P^2, t in months. Find k, K and the per-trout growth rate. Her pond now holds 450 trout: how fast is it growing, and is that growth speeding up or slowing down?

Solution

Factor out 0.3P:

\begin{aligned} 0.3P - 0.0004P^2 &= 0.3P\left(1 - \frac{0.0004}{0.3}P\right) \\ &= 0.3P\left(1 - \frac{P}{750}\right), \end{aligned}

since \dfrac{0.3}{0.0004} = 750. So k = 0.3 per month and K = \dfrac{a}{b} = 750 trout.

The per-trout rate is \dfrac{1}{P}\,\dfrac{dP}{dt} = 0.3 - 0.0004P, which falls by 0.0004 per month for each extra trout and reaches 0 at 750.

At P = 450, \dfrac{dP}{dt} = 0.3 \cdot 450 \cdot 0.4 = 54 trout per month. The fastest growth is at \dfrac{K}{2} = 375, at \dfrac{0.3 \cdot 750}{4} = 56.25 trout per month. Since 450 > 375, the factor 1 - \dfrac{2P}{K} is negative, so the second derivative is negative: growth is slowing, and it will keep slowing as the pond fills toward 750.

Example 4

Day 2: solving Nigora’s equation

Solve \dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{500}\right) with P(0) = 50. Find when the pond holds 400 trout, when it is growing fastest, and P(12).

Solution

Separate, with 0.4P\left(1 - \dfrac{P}{500}\right) = \dfrac{0.4P(500 - P)}{500}:

\int \frac{500}{P(500 - P)}\,dP = \int 0.4\,dt.

Partial fractions give \dfrac{500}{P(500 - P)} = \dfrac{1}{P} + \dfrac{1}{500 - P}, so

\ln|P| - \ln|500 - P| = 0.4t + C.

The solution starts between 0 and 500 and, by the Results, stays there, so both absolute values can be dropped: \ln\dfrac{P}{500 - P} = 0.4t + C. At t = 0, C = \ln\dfrac{50}{450} = -\ln 9, so \dfrac{P}{500 - P} = \dfrac{e^{0.4t}}{9}. Then 9P = (500 - P)e^{0.4t}, and solving for P,

P(t) = \frac{500}{1 + 9e^{-0.4t}}.

This is the general formula with A = \dfrac{500 - 50}{50} = 9.

For 400 trout: 1 + 9e^{-0.4t} = \dfrac{500}{400} = 1.25, so e^{-0.4t} = \dfrac{1}{36} and t = \dfrac{\ln 36}{0.4} \approx 8.96 months.

The growth is fastest at P = 250: 9e^{-0.4t} = 1, so t = \dfrac{\ln 9}{0.4} \approx 5.49 months, the dot in Figure 2.

After a year, P(12) = \dfrac{500}{1 + 9e^{-4.8}} \approx 465.5, about 466 trout, and the formula never reaches 500.

Example 5

Day 2: Lutfullo’s overstocked pond

Lutfullo puts 800 trout into a pond with the same model, \dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{500}\right). Solve for P(t), find when the population is down to 600, and describe its graph.

Solution

The separation and the partial fractions are those of the previous example. Now P > 500, and the solution stays above 500, so |500 - P| = P - 500:

\begin{aligned} \ln\frac{P}{P - 500} &= 0.4t + C, \\ C &= \ln\frac{800}{300} = \ln\frac{8}{3}. \end{aligned}

So \dfrac{P}{P - 500} = \dfrac{8}{3}e^{0.4t}. Then P = \dfrac{8}{3}e^{0.4t}(P - 500), so P\left(\dfrac{8}{3}e^{0.4t} - 1\right) = \dfrac{4000}{3}e^{0.4t}, and dividing through by \dfrac{8}{3}e^{0.4t},

P(t) = \frac{500}{1 - \frac{3}{8}e^{-0.4t}}.

This is the general formula with A = \dfrac{500 - 800}{800} = -\dfrac{3}{8}. For t \ge 0, \dfrac{3}{8}e^{-0.4t} lies in \left(0, \dfrac{3}{8}\right], so the denominator lies in \left[\dfrac{5}{8}, 1\right) and P > 500 always.

For 600 trout: 1 - \dfrac{3}{8}e^{-0.4t} = \dfrac{5}{6}, so e^{-0.4t} = \dfrac{1}{6} \cdot \dfrac{8}{3} = \dfrac{4}{9} and t = \dfrac{\ln(9/4)}{0.4} \approx 2.03 months.

At the start the rate is -192 trout per month (Example 1); at 600 it is 0.4 \cdot 600 \cdot (-0.2) = -48. The population falls more and more slowly, the graph is decreasing and concave up throughout, and it levels off onto P = 500 from above (Figure 2).

Example 6

Day 2: Lutfullo finds K from three counts

In another pond Lutfullo counts 60, 150 and 240 trout at 0, 4 and 8 months. Assuming a logistic model, find K and k, the time of fastest growth, and the population at 12 months.

Solution

The reciprocal of the formula is \dfrac{1}{P} = \dfrac{1 + Ae^{-kt}}{K} = \dfrac{1}{K} + \dfrac{A}{K}e^{-kt}. So u = \dfrac{1}{P} - \dfrac{1}{K} decays exponentially, and over each 4 months it is multiplied by the same factor e^{-4k}. Write u_0 = \tfrac{1}{60}, u_1 = \tfrac{1}{150}, u_2 = \tfrac{1}{240} for the reciprocals of the counts and c = \dfrac{1}{K}. Equal factors mean

\frac{u_1 - c}{u_0 - c} = \frac{u_2 - c}{u_1 - c}.

Cross-multiplying, u_1^2 - 2u_1c = u_0u_2 - (u_0 + u_2)c (the c^2 terms cancel), so

\begin{aligned} c &= \frac{u_0u_2 - u_1^2}{u_0 + u_2 - 2u_1} = \frac{\frac{1}{14400} - \frac{1}{22500}}{\frac{9}{1200}} \\ &= \frac{\frac{1}{40000}}{\frac{3}{400}} = \frac{1}{300}. \end{aligned}

The method needs the three counts at equally spaced times, so that the same factor applies twice, and a nonzero denominator u_0 + u_2 - 2u_1. The fit describes a growing logistic population only if the factor lies strictly between 0 and 1, which means 0 < c < u_2; here \dfrac{1}{300} < \dfrac{1}{240}.

So K = 300. The factor is \dfrac{u_1 - c}{u_0 - c} = \dfrac{1/300}{4/300} = \dfrac{1}{4}, so e^{-4k} = \dfrac{1}{4} and k = \dfrac{\ln 4}{4} \approx 0.3466 per month. With A = \dfrac{300 - 60}{60} = 4, P(t) = \dfrac{300}{1 + 4e^{-kt}}; as a check, \displaystyle P(8) = \dfrac{300}{1 + 4 \cdot \frac{1}{16}} = 240.

The fastest growth comes at t = \dfrac{\ln A}{k} = \dfrac{\ln 4}{(\ln 4)/4} = 4 months, which is the second count: 150 = \dfrac{K}{2}. At 12 months, \displaystyle P(12) = \dfrac{300}{1 + 4 \cdot \frac{1}{64}} \approx 282.4 trout.

Boundaries

  • The constants must stay constant. The model assumes the pond’s capacity K and the growth constant k do not change. A drought, a change of feed or stocking more fish partway through changes the equation, and the formula no longer applies.
  • Only P_0 > 0 means anything. The formula’s A = \dfrac{K - P_0}{P_0} needs P_0 \ne 0; the start P_0 = 0 is the equilibrium. A negative start solves the equation but describes no population.
  • A start above K has a past that ends. For P_0 > K, A < 0 and the denominator 1 + Ae^{-kt} is zero at the negative time t = \dfrac{\ln(-A)}{k}. Run backward, the model’s population grows without bound as t approaches that time, so the formula describes the population only after it.
  • Fastest growth at \dfrac{K}{2} belongs to this equation. A population whose per-capita rate does not fall in a straight line with P is not logistic, and its fastest growth can come at another level. A start with P_0 \ge \dfrac{K}{2} never has an inflection point at a time t > 0.
  • Uniqueness needs a smooth rate. The no-crossing argument uses the uniqueness theorem, whose hypothesis is that the rate function and its derivative are continuous. The logistic rate, a polynomial in P, meets it; an equation such as \displaystyle \dfrac{dy}{dt} = \sqrt{|y|} does not, and there the solution y = 0 is left by other solutions.
  • Counts are integers. The model is a smooth approximation to a whole-fish count and is most reliable for large populations.

Where this goes next

Euler’s method can be tested against the exact logistic solution, and the change of concavity at \dfrac{K}{2} is exactly what decides whether its estimate runs high or low. A constant harvest, as in the fishing model of Writing a differential equation from a description, subtracts from the rate and moves the equilibria, and Exercise 22 shows that the largest catch a pond can sustain is the fastest growth rate \dfrac{kK}{4}. The same equation, with other letters, describes the spread of a rumor or of an infection and the progress of some chemical reactions, and in later courses it is the starting point for models of two interacting species.

Exercises

Day 1: equilibria and the sign of the rate

  1. For \dfrac{dP}{dt} = 0.25P\left(1 - \dfrac{P}{1200}\right), find and classify the equilibria, and find the rate at P = 300 and at P = 1500.
  2. For \dfrac{dy}{dt} = 2y\left(1 - \dfrac{y}{8}\right), find the rate at y = 2, y = 6 and y = 10. Why are the first two equal?
  3. For Nigora’s pond, \dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{500}\right), give \lim_{t \to \infty} P(t) for the starts P(0) = 0, 120, 500 and 650.
  4. For \dfrac{dP}{dt} = 0.6P\left(1 - \dfrac{P}{90}\right), find the per-capita growth rate at P = 30 and P = 60, and the population at which it is half of k.
  5. A pond follows \dfrac{dP}{dt} = 0.35P\left(1 - \dfrac{P}{700}\right) with P(0) = 900. Find the rate at the start, and say whether the graph is concave up or down there.

Day 1: concavity and the fastest growth

  1. For \dfrac{dP}{dt} = 0.25P\left(1 - \dfrac{P}{1200}\right), write \dfrac{d^2P}{dt^2} in terms of P, and find where the growth is fastest and the rate there.
  2. For \dfrac{dy}{dt} = 3y\left(1 - \dfrac{y}{40}\right), find \dfrac{dy}{dt} and \dfrac{d^2y}{dt^2} at y = 10 and at y = 30, and the value of y at the inflection point of a solution starting at y = 5.
  3. A population follows \dfrac{dP}{dt} = 0.5P\left(1 - \dfrac{P}{2000}\right), t in years. What is the largest rate of growth it ever has, and at what size?
  4. A logistic trout population grows fastest, at 60 trout per month, when the pond holds 400 trout. Find K and k and write the equation.

Day 1: equations multiplied out

  1. Find k, K, the level of fastest growth and the rate there for \dfrac{dP}{dt} = 0.08P - 0.0002P^2.
  2. Find k, K and the largest growth rate for \dfrac{dN}{dt} = 1.2N - 0.004N^2.
  3. Tahmina measures the per-trout growth rate of a population: 0.28 per month at 100 trout and 0.08 per month at 600 trout. Assuming it is linear in P, write the equation in the form aP - bP^2, then in logistic form, and give the largest growth rate.

Day 2: solving by partial fractions

  1. Solve \dfrac{dP}{dt} = 0.25P\left(1 - \dfrac{P}{1200}\right) with P(0) = 200. Find P(4) and the time of fastest growth.
  2. Solve \dfrac{dy}{dt} = 2y\left(1 - \dfrac{y}{8}\right) with y(0) = 1, and find when y = 6.
  3. In Nigora’s pond, how many months does the population take to grow from 100 to 300 trout?
  4. For \dfrac{dP}{dt} = 0.5P\left(1 - \dfrac{P}{2000}\right), use \displaystyle t = \displaystyle\int_{P_1}^{P_2} \frac{dP}{dP/dt} to find how long a population takes to grow from 100 to 1{,}500.

Day 2: above the ceiling, and constants from data

  1. Solve the equation of Exercise 5 with P(0) = 900, and find when the population is down to 750.
  2. A logistic population has K = 1000, P(0) = 200 and P(4) = 400. Find k and P(8).
  3. Counts of 100, 250 and 500 at 0, 5 and 10 months follow a logistic model. Find K and k.

Reasoning

  1. For the solution of Exercise 17, find the time t < 0 at which the formula’s denominator is zero, and say what this means for running the model backward.
  2. Using the formula, prove that when 0 < P_0 < K the solution is increasing, stays strictly between 0 and K, and tends to K.
  3. Lutfullo removes h trout per month from Nigora’s pond, so \dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{500}\right) - h. Show that there is an equilibrium exactly when h \le 50, and explain why the cutoff is the fastest growth rate of the pond.
  4. Without the formula, explain why a solution of Nigora’s equation with P(0) = 450 never reaches 500, and one with P(0) = 800 never drops below 500.
  5. With 0 < P_0 < \dfrac{K}{2} and t^* = \dfrac{\ln A}{k}, prove that P(t^* + s) + P(t^* - s) = K for every s. What does this say about the shape of the S-curve?
Answers
  1. P = 0, unstable; P = 1200, stable. The rate is 0.25 \cdot 300 \cdot 0.75 = 56.25 at 300 and 0.25 \cdot 1500 \cdot (-0.25) = -93.75 at 1500.
  2. 3, 3 and -5. The rate 2y\left(1 - \dfrac{y}{8}\right) is a parabola symmetric about y = 4, and 2 and 6 are both 2 away from 4.
  3. 0, 500, 500, 500. The start 0 is an equilibrium; every positive start tends to the stable equilibrium 500.
  4. 0.6\left(1 - \dfrac{30}{90}\right) = 0.4 and 0.6\left(1 - \dfrac{60}{90}\right) = 0.2. Half of k = 0.6 is 0.3, reached at P = 45 = \dfrac{K}{2}.
  5. 0.35 \cdot 900 \cdot \left(1 - \dfrac{9}{7}\right) = -90 trout per month. Concave up: \dfrac{d^2P}{dt^2} = 0.35\left(1 - \dfrac{1800}{700}\right)(-90), which is 49.5 > 0.
  6. \dfrac{d^2P}{dt^2} = 0.25\left(1 - \dfrac{P}{600}\right)\dfrac{dP}{dt}, which is 0.0625P\left(1 - \dfrac{P}{1200}\right) times \left(1 - \dfrac{P}{600}\right). Fastest at P = 600, rate \dfrac{0.25 \cdot 1200}{4} = 75.
  7. At y = 10: \dfrac{dy}{dt} = 22.5 and \dfrac{d^2y}{dt^2} = 3\left(1 - \dfrac{10}{20}\right)(22.5) = 33.75. At y = 30: \dfrac{dy}{dt} = 22.5 and \dfrac{d^2y}{dt^2} = -33.75. The inflection point is at y = 20.
  8. \dfrac{0.5 \cdot 2000}{4} = 250 per year, at P = 1000.
  9. K = 2 \cdot 400 = 800, and \dfrac{k \cdot 800}{4} = 60 gives k = 0.3: \dfrac{dP}{dt} = 0.3P\left(1 - \dfrac{P}{800}\right).
  10. k = 0.08, K = \dfrac{0.08}{0.0002} = 400; fastest at P = 200, rate \dfrac{0.08 \cdot 400}{4} = 8.
  11. k = 1.2, K = \dfrac{1.2}{0.004} = 300; largest rate \dfrac{1.2 \cdot 300}{4} = 90, at N = 150.
  12. From a - 100b = 0.28 and a - 600b = 0.08: b = 0.0004, a = 0.32. So \dfrac{dP}{dt} = 0.32P - 0.0004P^2, which is 0.32P\left(1 - \dfrac{P}{800}\right), and the largest rate is \dfrac{0.32 \cdot 800}{4} = 64 trout per month.
  13. A = \dfrac{1200 - 200}{200} = 5, so P = \dfrac{1200}{1 + 5e^{-0.25t}}; P(4) = \dfrac{1200}{1 + 5e^{-1}} \approx 422.6; fastest at t = \dfrac{\ln 5}{0.25} \approx 6.44.
  14. y = \dfrac{8}{1 + 7e^{-2t}}. From 1 + 7e^{-2t} = \dfrac{4}{3}: e^{-2t} = \dfrac{1}{21}, t = \dfrac{\ln 21}{2} \approx 1.52.
  15. About 4.48 months. From P = \dfrac{500}{1 + 9e^{-0.4t}}, P = 100 when 9e^{-0.4t} = 4, at t = \dfrac{\ln(9/4)}{0.4}, and P = 300 when 9e^{-0.4t} = \dfrac{2}{3}, at t = \dfrac{\ln(27/2)}{0.4}; the difference is \dfrac{\ln 6}{0.4}.
  16. 2\ln 57 \approx 8.09. The time is \displaystyle \displaystyle\int_{100}^{1500} \frac{2000}{0.5P(2000 - P)}\,dP, which by partial fractions is 2\Big[\ln P - \ln(2000 - P)\Big]_{100}^{1500}, and that equals 2\ln\dfrac{1500 \cdot 1900}{100 \cdot 500}.
  17. A = \dfrac{700 - 900}{900} = -\dfrac{2}{9}, so \displaystyle P = \dfrac{700}{1 - \frac{2}{9}e^{-0.35t}}. From \displaystyle 1 - \frac{2}{9}e^{-0.35t} = \frac{14}{15}: e^{-0.35t} = 0.3, t = \dfrac{\ln(10/3)}{0.35} \approx 3.44.
  18. A = 4. From \dfrac{1000}{1 + 4e^{-4k}} = 400: e^{-4k} = \dfrac{3}{8}, k = \dfrac{\ln(8/3)}{4} \approx 0.2452. Then e^{-8k} = \dfrac{9}{64} and \displaystyle P(8) = \dfrac{1000}{1 + \frac{9}{16}} = 640.
  19. With u_i the reciprocals, c = \dfrac{u_0u_2 - u_1^2}{u_0 + u_2 - 2u_1} = \dfrac{1/250\,000}{1/250} = \dfrac{1}{1000}, so K = 1000. The factor is \dfrac{1/250 - 1/1000}{1/100 - 1/1000} = \dfrac{1}{3} = e^{-5k}, so k = \dfrac{\ln 3}{5} \approx 0.2197.
  20. \displaystyle 1 = \frac{2}{9}e^{-0.35t} gives t = -\dfrac{\ln 4.5}{0.35} \approx -4.297. As t decreases toward -4.297, P \to \infty: the model cannot describe the pond earlier than that.
  21. Here A = \dfrac{K - P_0}{P_0} > 0, so 1 + Ae^{-kt} > 1 and 0 < P < K. As t increases, Ae^{-kt} decreases, so the denominator decreases and P increases. As t \to \infty, Ae^{-kt} \to 0 and P \to K.
  22. An equilibrium needs 0.4P\left(1 - \dfrac{P}{500}\right) = h. The left side is a downward parabola with largest value 50, at P = 250, so it equals h for some P exactly when h \le 50 (equivalently, 0.0008P^2 - 0.4P + h = 0 has discriminant 0.16 - 0.0032h \ge 0). The pond cannot replace more trout per month than its fastest growth rate.
  23. P = 500 is a solution, and the rate is a polynomial, so the uniqueness theorem applies. If a solution reached 500 at some time, two solutions would pass through that point, so it would be the constant 500 all along, which neither is.
  24. Since Ae^{-kt^*} = 1, Ae^{-k(t^* \pm s)} = e^{\mp ks}. So P(t^* + s) = \dfrac{K}{1 + e^{-ks}} and P(t^* - s) = \dfrac{K}{1 + e^{ks}} = \dfrac{Ke^{-ks}}{e^{-ks} + 1}, and the sum is K. The S-curve is symmetric about its inflection point \left(t^*, \dfrac{K}{2}\right): turned half a turn about that point, it lands on itself.
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