Carrying capacity and the factor that switches growth off, the direction field and its two equilibria, solving by separation and partial fractions, the S-shaped solution and its inflection at half capacity, and fitting the model to data.
Subject: Calculus II · 68 slides · symbolic lesson
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Title
Calculus II · Section 4.4
Growth with a ceiling: carrying capacity, equilibria and the S-curve
Objectives
Section 4.1 met the exponential model, where a population grows in proportion to its size forever. Real populations run out of food, space or water. This lesson adds one factor to that equation, the logistic brake, and follows it all the way to a formula.
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, pp. 393-404 — learning objectives 4.4.1 to 4.4.3
The exponential model from Section 4.1 is a good description of a population with unlimited food and space, and a terrible one of anything that has been growing for a while. Left alone it predicts that a herd of deer would outweigh the planet within a few centuries.
This lesson repairs it with one extra factor. You will see what that factor does before you solve anything, using only the sign of the rate: which populations grow, which shrink, and where they all end up. Then you will solve the equation exactly, which takes separation of variables and the partial fractions of Section 3.4.
The payoff is a formula with three inputs, the growth rate, the carrying capacity and the starting population, and the last part of the lesson asks the practical question: if you only have data, which of those three can you actually pin down?
Warm-up
Discussion prompt
Solve the initial-value problem below, then find the time it takes the population to double. Do it without looking back at Section 4.1.
Commit to an answer on paper before revealing. This is the equation the whole lesson modifies, so it is worth being fluent with it.
Separating the variables gives the natural logarithm of P equal to r t plus a constant, and exponentiating gives the starting population times e to the r t. Setting that equal to twice the starting population, the starting population cancels, which is the important point: the doubling time is the logarithm of 2 divided by r, whatever size you start from.
Hold on to that idea of a fixed doubling time. It says a herd of a million doubles exactly as fast as a herd of ten. That is the assumption that fails in the real world, and the logistic equation is built to remove it.
Section
Part 1
Concept
Figure (svg): The curve P equals 100 e to the 0.03 t for t from 0 to 100, starting at 100 and bending ever more steeply upward past 2000, with dots at the doubling times 23.1, 46.2 and 69.3.
\[ P(t) = 100e^{0.03t} \quad\Longrightarrow\quad \frac{dP}{dt} = 3e^{0.03t} = 0.03P \]
The rate of growth is a fixed multiple of the population, so the bigger the population, the faster it grows, with no limit. At t equal to 100 this model already stands at 2009, twenty times where it began, and it keeps accelerating.
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 393 — Figure 4.18
This is Figure 4.18 of the book, a population of 100 growing at 3 percent per unit time. The dots mark the doublings, and they are evenly spaced in time: every 23.1 units the population doubles again.
Look at what equal spacing in time does to the height. The first doubling adds 100, the second 200, the fourth 800. The curve keeps getting steeper, because the rate of growth is 0.03 times the current population, and the population keeps rising.
Nothing in the equation dP by dt equals r P can ever slow this down. The factor r is a constant, so the growth per organism never changes. For bacteria in a fresh dish, or a species arriving on an empty island, that is a fine model for a while. For anything that shares a finite environment, it eventually becomes absurd, and the question is how to build in a limit.
Concept
carrying capacity — The maximum population of an organism that a given environment can sustain indefinitely. It is written K.
Keep the exponential rate, and multiply it by a factor that equals one when the population is tiny and zero when it reaches K:
\[ \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) \qquad (4.8) \]
Together with a starting value this is an initial-value problem. Pierre Verhulst published it in 1845.
\[ P(0) = P_0 \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 394 — definitions of carrying capacity and the logistic equation
The carrying capacity is an ecological quantity: how many individuals the food, water and space can support indefinitely. It is written K, and in Example 4.14 it comes from multiplying a sustainable density of deer per square mile by the area of Kentucky.
The logistic equation keeps the exponential rate r P and multiplies it by one minus P over K. Think of that factor as the fraction of the capacity still unused. When the population is tiny almost all of it is unused, the factor is nearly 1, and you get exponential growth back. When the population reaches K, none is unused, the factor is 0, and growth stops.
Notice that you have not solved anything yet. The equation is a statement about rates, and a surprising amount can be read off it directly, which is what the next few slides do.
Notation
Annotate
On: \( \frac{dP}{dt} = r\,P\left(1 - \frac{P}{K}\right) \)
Step through the annotations and connect each piece to something physical. The left side is a rate, measured in organisms per unit of time, so every term on the right must be too.
The factor r P is exactly the exponential model: births minus deaths, proportional to how many organisms there are. The bracket is the new part, and the thing to notice is that P over K has no units: it is the fraction of capacity in use, 0.84 for the Kentucky deer in 2004.
Because the bracket is a pure number, it can switch sign. Below the capacity it is positive and the population grows; above it, it is negative and the population shrinks. One formula covers both situations, which is why the logistic equation is so widely used.
Intuition
Divide both sides by P to get the growth per organism, the relative growth rate:
\[ \frac{1}{P}\frac{dP}{dt} = r\left(1 - \frac{P}{K}\right) = r - \frac{r}{K}P \]
Figure (svg): The relative growth rate plotted against the deer population in thousands. The exponential model is a flat line at 0.2311. The logistic model is a straight line falling from 0.2311 at zero to zero at the carrying capacity 1072.8 thousand, and negative beyond it. A dot marks 900 thousand deer, where the logistic rate is 0.037.
For the exponential model this is the constant r. For the logistic model it is a straight line in P: close to r while P is small, so early growth looks exponential, and falling to zero at the capacity.
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 394 — the discussion after equation 4.8
Dividing by P turns the rate into a rate per organism, the relative growth rate. For the exponential model it is the flat dashed line: every deer contributes the same 23 percent a year no matter how crowded the land is.
For the logistic model it is the green line, falling steadily as the herd grows. Near the left end it almost touches the dashed line, which is the precise meaning of the book's remark that small populations grow nearly exponentially.
Now look at the dot. With 900 thousand deer, the relative growth rate is only about 0.037, under 4 percent a year. The model says the Kentucky herd of 2004 was already heavily braked. That single number explains everything in Example 4.14, including why the herd will come nowhere near doubling in three years.
Picture it
Figure (svg): The growth rate dP/dt of the deer herd, in thousands per year, plotted against the population in thousands: a downward parabola through zero at P equal 0 and at P equal 1072.8, peaking at 62.0 when P is 536.4, and negative for larger P. Dots mark 900 thousand (rate 33.5) and 1200 thousand (rate minus 32.9).
Multiply out the right side of 4.8 and it is a downward parabola in P, with roots 0 and K. For the Kentucky deer of Example 4.14, the herd grows fastest, about 62 thousand deer a year, when it is half the capacity.
\[ \frac{dP}{dt} = rP - \frac{r}{K}P^2 \]
Here the vertical axis is the actual rate, deer per year, not the rate per deer. Multiplying the falling line of the previous slide by P turns it into a parabola that opens downward.
Read the parabola from left to right. With few deer the rate is small because there are few parents. With many deer it is small again because the brake is nearly on. In between it peaks, and the peak sits exactly halfway between the two roots, at half the capacity: about 536 thousand deer, growing by about 62 thousand a year.
The two dots are the two starting herds of Example 4.14. The herd of 900 thousand sits on the falling side, still growing but slowly. The herd of 1.2 million is past the root at K, where the rate is negative. Keep this picture; the phase line in Part 2 is just its sign pattern.
Worked example
Show that the logistic rate is largest when the population is half the carrying capacity, and find that largest rate.
\[ g(P) = rP\left(1 - \frac{P}{K}\right) = rP - \frac{r}{K}P^2 \]
Differentiate with respect to P
Why: The rate is a quadratic in P, so its derivative is linear.
\[ g'(P) = r - \frac{2r}{K}P \]
Set the derivative to zero
Why: Divide by r, which is positive.
\[ r - \frac{2r}{K}P = 0 \;\Longrightarrow\; P = \frac{K}{2} \]
Confirm it is a maximum
Why: The second derivative is negative.
\[ g''(P) = -\frac{2r}{K} < 0 \]
Evaluate the rate there
Why: Half times one minus a half is a quarter.
\[ g\left(\frac K2\right) = r\cdot\frac{K}{2}\left(1 - \frac12\right) = \frac{rK}{4} \]
Check with the deer
Why: With r equal to 0.2311 and K equal to 1,072,764, the peak rate beats the rate at the 2004 herd of 900,000.
\[ \frac{rK}{4} \approx 61{,}979 > 33{,}496 \approx g(900{,}000) \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 406 — Exercise 200
The rate g is a function of P, so finding its maximum is an ordinary first-derivative problem, the kind you did in Calculus I. The only care needed is to remember that r and K are constants here.
The derivative is r minus 2 r over K times P, which vanishes at P equal to K over 2. The second derivative is a negative constant, so this is a maximum, as the downward parabola already told you. Putting half the capacity back in gives r times K over 4.
The check uses the deer. Their peak rate would be about 62 thousand a year, and their actual rate in 2004 was about 33.5 thousand, smaller, as it must be. Remember the number r K over 4: it will return as the largest sustainable fishing catch in Part 5.
Trap
A student writes the deer model as:
\[ \frac{dP}{dt} = 0.2311P(1 - P) \]
Wrong. With P equal to 900,000 the bracket is about minus 900,000.
The brake must compare the population with the capacity. One minus P only works if P is measured in units of K, as a fraction of capacity. Written with head counts, the bracket needs P over K, a pure number between zero and one.
\[ 1 - \frac{P}{K} = 1 - \frac{900{,}000}{1{,}072{,}764} \approx 0.161 \]
This mistake usually comes from textbooks that write the logistic equation for a population measured as a fraction of its capacity, where K is 1 and the bracket really is one minus P. Copy that form into a problem with head counts and the bracket becomes enormous and negative.
The fix is to ask what the bracket means. It must compare the population with the capacity, and the only way to compare two counts is to divide one by the other. With 900 thousand deer out of a possible 1,072,764, the bracket is about 0.161: sixteen percent of the capacity is still free.
A quick sanity check catches this every time. Put in the starting population: if the bracket is not a number between zero and one for a population below capacity, something is wrong.
Prediction
\[ \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right), \qquad P_0 > K \]
Predict first
A population starts above its carrying capacity. What does the logistic equation say happens next?
Correct: It declines toward K, never quite reaching it
Why: Above K the bracket is negative, so the rate is negative and the population falls. As it approaches K the bracket shrinks toward zero, so the decline slows and K is approached only as t goes to infinity. It can never cross K, because K itself is a constant solution.
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 394 — the case P greater than K
Make a prediction before revealing, and base it on the sign of the bracket, not on intuition about animals.
Above the capacity, P over K is bigger than 1, so the bracket is negative and the rate is negative: the population falls. As it falls toward K, the bracket shrinks toward zero, so the fall slows. The population approaches K from above but never reaches it in finite time.
Why can it never cross K? Because the constant function equal to K is itself a solution, and two solutions of an equation like this cannot cross: at a crossing point they would share the same value and the same slope, and the solution through that point is unique. You will see this in the direction field shortly: no curve ever passes through the dashed line at K.
Sorting
Sort into buckets
Use dP/dt = 0.2311P(1 − P/1,072,764). Sort each herd size by what the model says it does next.
Work out the bracket for each herd size, one minus the herd divided by 1,072,764, and let its size tell you which bucket. A bracket near 1 means nearly exponential; a small positive bracket means heavily braked; zero means no change; negative means decline.
The two that surprise people are 900 thousand and 1,050,000. Both are growing, but at a sixth and a fiftieth of the exponential rate respectively. The closer the herd is to capacity, the harder the brake.
The empty herd and the herd at exactly the capacity both sit still, but for very different reasons, and the next part of the lesson is about that difference: one of them is a balance the population returns to, and the other is one it runs away from.
Section
Part 2
Concept
The logistic equation is autonomous: the right side depends on P alone, not on t. A constant population is a solution exactly when the rate vanishes there.
\[ rP\left(1 - \frac{P}{K}\right) = 0 \]
\[ P = 0 \quad\text{or}\quad 1 - \frac{P}{K} = 0 \;\Longrightarrow\; P = K \]
No organisms means no births, so an empty habitat stays empty. A population sitting exactly at capacity has births and deaths in balance, so it never changes.
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 399 — Step 1 of the solution method
An autonomous equation is one whose right side does not mention time. For such an equation, a constant is a solution precisely when the right side is zero at that constant, because a constant function has derivative zero.
The logistic rate is a product of r, P and the bracket, and a product is zero only when a factor is zero. So the equilibria are P equal to 0 and P equal to K, and there are no others.
Both make biological sense. No animals means no births; a population at capacity has births and deaths exactly balanced. The book lists this as Step 1 of solving the equation, and the reason will become clear in Part 3: the separation step divides by P and by K minus P, so these two solutions must be recorded before they are lost.
Picture it
Figure (svg): Top: a downward parabola for dP/dt against P, positive between P equal 0 and P equal K and negative outside. Bottom, aligned with it: a horizontal phase line with an open dot at 0, a filled dot at K, an arrow pointing right between them and arrows pointing away from 0 on the left and toward K from the right.
Squash the whole direction field onto one vertical line of P values, here drawn lying down under the rate parabola. Each arrow records only which way the population moves from there.
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 395 — Figure 4.19
The phase line throws away the time axis and keeps only direction. For each value of P, ask one question: is the rate positive or negative there? Positive means an arrow toward larger P; negative means an arrow toward smaller P.
The figure puts the parabola directly above the line so you can see where the arrows come from. Between 0 and K the parabola is above the axis, so the arrow points right. Outside that interval it is below, so the arrows point left.
Now read the dots. At K, the arrows on both sides point in: that is a stable equilibrium, marked filled. At 0 they point out: unstable, marked open. The book draws the same line vertically in Figure 4.19; the information is identical. Negative populations are not biology, but the mathematics still has them, and the arrow there shows why 0 repels from both sides.
Concept
An equilibrium is stable when nearby solutions move toward it, and unstable when they move away. The sign of the rate's slope at the equilibrium decides it.
\[ g(P) = rP - \frac{r}{K}P^2, \qquad g'(P) = r - \frac{2r}{K}P \]
\[ g'(0) = r > 0 \;\Longrightarrow\; P = 0 \text{ is unstable} \]
\[ g'(K) = r - 2r = -r < 0 \;\Longrightarrow\; P = K \text{ is stable} \]
A positive slope means the rate is negative just below the equilibrium and positive just above: both sides are pushed away. A negative slope pushes both sides back in.
The phase line gives stability by eye. The derivative of the rate gives it by calculation, which is what you need when the picture is not handy.
Near an equilibrium the rate curve is approximately a straight line through zero. If that line slopes up, the rate is negative just to the left and positive just to the right, so a population displaced either way is pushed further away. If it slopes down, both displacements are pushed back.
For the logistic rate, the slope is r at zero and minus r at the capacity. So zero is unstable and the capacity is stable, for every positive r and K. That is a strong conclusion: any positive population, however small, eventually approaches the carrying capacity. The model never predicts extinction, which is one reason Part 5 adds harvesting and a threshold.
Picture it
Figure (svg): A slope field for dP/dt equals 0.5 P times one minus P over 10, for t from 0 to 14 and P from minus 2 to 16. Solution curves start at P equal 0.5, 2 and 5 and rise in S shapes to the dashed line P equal 10; curves from 14 and 16 fall to it; P equal 0 and P equal 10 are flat lines; a curve from minus 0.5 plunges downward.
Every positive starting population ends at the carrying capacity. Starting below half of it gives an S; starting between half and K gives a curve that only bends down; starting above K gives a curve falling to K.
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 394 — learning objective 4.4.2
A direction field draws a short segment at many points, with slope equal to the rate there. Because the logistic rate depends only on P, every row of segments is identical: the field looks the same at every time, which is what autonomous means in a picture.
Follow the blue curves from the bottom. The one starting at 0.5 rises slowly, steepens, then bends over into the dashed line: that is the S. The one starting at 5, half the capacity, starts at its steepest and only bends down. The orange curves above 10 fall to it.
The red curve starting at minus one half has no biological meaning, but it shows the instability of zero from the other side: it runs away downward. And notice what is missing: no curve crosses the dashed line, and no curve crosses the axis. Equilibria are walls.
Worked example
200 rabbits in a meadow, growth rate 0.04 per month, carrying capacity 750. Write the model and sketch its solution.
Write the initial-value problem
Why: Put r equal to 0.04 and K equal to 750 into equation 4.8.
\[ \frac{dP}{dt} = 0.04P\left(1 - \frac{P}{750}\right), \quad P(0) = 200 \]
Find the equilibria
Why: The rate vanishes at zero and at the capacity.
\[ P = 0 \;(\text{unstable}), \qquad P = 750 \;(\text{stable}) \]
Slope where the solution starts
Why: Evaluate the rate at 200.
\[ 0.04(200)\left(1 - \frac{200}{750}\right) = 8(0.7333) \approx 5.87 \]
Slope at half capacity
Why: The largest slope in the field, from Exercise 200.
\[ \frac{rK}{4} = \frac{0.04(750)}{4} = 7.5 \quad\text{at } P = 375 \]
Figure (svg): The slope field of dP/dt equals 0.04 P times one minus P over 750, for t from 0 to 120 months and P from 0 to 900, with the solution through P equal 200 at t equal 0 rising in an S shape toward the dashed line at 750, and passing 375 near t equal 25.
Check the field above the capacity
Why: Segments there must point down, so no solution can cross 750.
\[ 0.04(900)\left(1 - \frac{900}{750}\right) = 36(-0.2) = -7.2 < 0 \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 403 — Checkpoint 4.14a and b
The rabbits give you all three numbers directly: 200 to start, 4 percent per month, and a capacity of 750. Writing the model is substitution; the interesting part is sketching without solving.
Two slopes set the scale of the picture. At the start the rate is about 5.9 rabbits a month; at half the capacity it reaches its maximum of 7.5. So the curve starts steep, gets a little steeper until it crosses 375, and then flattens toward 750.
Look at the field in the figure and trace the solution with your finger along the segments. The check confirms the top of the field: at 900 rabbits the rate is minus 7.2, so segments above the capacity point down, and your sketch must never go above 750. The units matter too: time is in months, so the curve takes years to level off.
Estimation
\[ \frac{dP}{dt} = 0.04P\left(1 - \frac{P}{750}\right), \quad P(0) = 200 \]
Predict first
Following the rabbits' solution curve on the previous slide, roughly when does the population pass 375, half the capacity?
Correct: About 25 months
Why: The curve crosses the dotted line at 375 a little after month 24. The formula later in this lesson gives 25.3 months exactly. A field is good for shapes and rough times; exact times need the solution formula.
Go back to the rabbits' curve and find where it crosses the dotted line at 375. Make your estimate before revealing.
The crossing is a little after month 24, and Part 4 will derive the exact value, 25.3 months, from the formula. So the field was good to within a month or so, which is typical: a well-drawn field gets shapes, limits and rough times right.
What a field cannot give you is precision. If the question asks for the population after exactly 12 months, or the time to reach exactly 600 rabbits, you need the formula. That is the motivation for Part 3: the field tells you what the answer looks like, and the solution tells you what it is.
Worked example
Find the equilibria and their stability.
\[ P' = CP - P^2 = P(C - P) \]
Equilibria for C equal to 3
Why: Set the product to zero.
\[ P(3 - P) = 0 \;\Longrightarrow\; P = 0, \; P = 3 \]
Stability for C equal to 3
Why: Differentiate the right side and evaluate.
\[ g'(P) = 3 - 2P: \quad g'(0) = 3 > 0, \quad g'(3) = -3 < 0 \]
Equilibria for C equal to minus 3
Why: Factor out minus P.
\[ -3P - P^2 = -P(3 + P) = 0 \;\Longrightarrow\; P = 0, \; P = -3 \]
Stability for C equal to minus 3
Why: Same derivative rule.
\[ g'(P) = -3 - 2P: \quad g'(0) = -3 < 0, \quad g'(-3) = 3 > 0 \]
Figure (svg): Three downward parabolas of P prime against P: for C equal 3, crossing zero at 0 and 3; for C equal 0, touching zero only at 0; for C equal minus 3, crossing at minus 3 and 0. Open dots mark unstable equilibria, filled dots stable ones, and a half-filled marker the touching point.
Check against the logistic form
Why: With C equal to 3 this is the logistic equation with r and K both 3, whose zero is unstable and whose capacity is stable, as found.
\[ 3P\left(1 - \frac{P}{3}\right) = 3P - P^2 \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 405 — Exercises 168 to 170
The book writes the equation as P prime equals C P minus P squared. Factor it as P times C minus P and the equilibria jump out: zero and C.
For C equal to 3, the slope of the rate is 3 minus 2 P, which is positive at zero and negative at 3. So zero is unstable and 3 is stable: the ordinary logistic picture. For C equal to minus 3, the same calculation reverses both signs: zero becomes stable and minus 3 unstable. The parabolas in the figure show why: flipping the sign of C slides the parabola across the axis.
The case C equal to zero, Exercise 169, is the purple curve. It only touches the axis at zero. Just above zero the rate is negative, pushing populations down toward zero; just below, it is also negative, pushing them away. An equilibrium like that is called semistable.
Error analysis
Annotate
On: \( g(0) = r\cdot 0\left(1 - \frac{0}{K}\right) = 0 \;\Longrightarrow\; P = 0 \text{ is stable} \)
Read each annotation and find the exact point where the reasoning breaks. The first line is correct; the conclusion does not follow from it.
A rate of zero is what makes a point an equilibrium in the first place. Every equilibrium has that property, stable or not, so it cannot decide stability. Stability is a question about the neighbours: what happens to a population that starts slightly off the equilibrium?
For zero, a population slightly above it has a positive rate, so it grows away. That is instability. The quickest correct argument is the one on the earlier slide: the derivative of the rate at zero is r, which is positive. Whenever you classify an equilibrium, say which of the two tests you used, the slope or the arrows.
Section
Part 3
Concept
Because the right side depends on P alone, the equation separates. First put the bracket over a common denominator:
\[ \frac{dP}{dt} = rP\left(\frac{K - P}{K}\right) = \frac{rP(K - P)}{K} \]
Multiply by dt, divide by P times K minus P, and multiply by K, so that every P is on the left and every t on the right:
\[ \frac{K}{P(K - P)}\,dP = r\,dt \]
Dividing by P times K minus P is only allowed when neither factor is zero, which is why the two equilibria were found first: they are lost at this step.
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 399 — Step 2
Separation works whenever the rate is a function of P times a function of t. Here the function of t is just the constant 1, so the equation is separable, as every autonomous equation is.
The algebra is set up to make the next step easy. Writing the bracket over the common denominator K produces P times K minus P, which is exactly the product that partial fractions will split. Multiplying both sides by K puts a clean K in the numerator.
The warning on the slide matters. Dividing by P times K minus P assumes neither factor is zero. The two constant solutions, P equal to 0 and P equal to K, make one of them zero, so the method cannot produce them, and you must remember them separately. This is a feature of every separable equation, and a common source of lost answers.
Worked example
Split the left-hand integrand so that each piece integrates to a logarithm.
\[ \frac{K}{P(K - P)} = \frac{A}{P} + \frac{B}{K - P} \]
Clear the denominators
Why: Multiply both sides by P times K minus P.
\[ K = A(K - P) + BP \]
Put P equal to 0
Why: The B term vanishes.
\[ K = AK \;\Longrightarrow\; A = 1 \]
Put P equal to K
Why: The A term vanishes.
\[ K = BK \;\Longrightarrow\; B = 1 \]
Write the decomposition
Why: Both numerators are one.
\[ \frac{K}{P(K - P)} = \frac{1}{P} + \frac{1}{K - P} \]
Check by recombining
Why: Put the two fractions back over one denominator; the P terms cancel.
\[ \frac{1}{P} + \frac{1}{K - P} = \frac{(K - P) + P}{P(K - P)} = \frac{K}{P(K - P)} \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 400 — the decomposition left to the reader
This is the same technique as Section 3.4: two distinct linear factors in the denominator, so two constant numerators. The book leaves the verification to you; here it is.
After clearing the denominators you have an identity, true for every P. That means you may choose any convenient values of P, and the best choices are the ones that make a term vanish: P equal to 0 kills the B term and gives A equal to 1, and P equal to K kills the A term and gives B equal to 1.
The check is the one you should always do with partial fractions: add the pieces back together. The P and minus P cancel in the numerator, leaving K. It takes ten seconds, and it catches most sign errors before they spread through the rest of the solution.
Worked example
Integrate the separated equation and undo the logarithms.
\[ \int\left(\frac{1}{P} + \frac{1}{K - P}\right)dP = \int r\,dt \]
Integrate term by term
Why: The second logarithm carries a minus sign: the derivative of K minus P is minus one.
\[ \ln|P| - \ln|K - P| = rt + C \]
Combine the logarithms
Why: A difference of logarithms is the logarithm of a quotient.
\[ \ln\left|\frac{P}{K - P}\right| = rt + C \]
Exponentiate
Why: e to a sum is a product of exponentials.
\[ \left|\frac{P}{K - P}\right| = e^{C}e^{rt} \]
Drop the absolute value
Why: Absorb the sign into a new constant, which may now be negative.
\[ \frac{P}{K - P} = C_1 e^{rt} \qquad (4.9) \]
Check the logarithms by differentiating
Why: Differentiate the left side of the second line with respect to P; the chain rule restores the integrand.
\[ \frac{d}{dP}\left[\ln|P| - \ln|K - P|\right] = \frac1P - \frac{-1}{K - P} = \frac1P + \frac{1}{K - P} \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 400 — equation 4.9
Each piece of the decomposition integrates to a logarithm, but the second one needs care, and that is the step to slow down on. The inner function is K minus P, whose derivative with respect to P is minus 1, so its logarithm comes with a minus sign.
From there it is logarithm rules. A difference of logarithms is the logarithm of a quotient; exponentiating turns the constant C into a multiplying factor e to the C. When the absolute value is removed, the factor may be positive or negative, which is why a new name, C sub 1, is introduced.
Equation 4.9 is worth remembering on its own. It says that P over K minus P, the population divided by the room left, grows exponentially. You will use it three more times in this lesson: to find the constant, to find times, and to fit data.
Trap
A common line in the middle of the solution:
\[ \int\frac{dP}{K - P} = \ln|K - P| \]
Wrong. Differentiate it back and the sign is off.
Substitute u equal to K minus P. Then du is minus dP, and the minus sign comes out in front. Lose it and the two logarithms add instead of subtract, and the quotient in equation 4.9 becomes a product.
\[ \int\frac{dP}{K - P} = -\ln|K - P| + C \]
This is the single most common error in the whole derivation, and it is easy to see why: the integral of one over x is the logarithm of x, and K minus P looks like it should follow the same pattern.
Differentiate the wrong answer and the problem shows. The derivative of the logarithm of K minus P, by the chain rule, is one over K minus P times minus 1. So the wrong line is off by exactly a sign. With the sign lost, the two logarithms add, you get the logarithm of P times K minus P, and everything after that is wrong.
The safest habit is a substitution in your head: u is K minus P, du is minus dP. Or simply differentiate your antiderivative back before moving on.
Worked example
Turn equation 4.9 into a formula for P, and fix the constant with the initial population.
Multiply by K minus P
Why: Clear the fraction.
\[ P = C_1 e^{rt}(K - P) = C_1Ke^{rt} - C_1Pe^{rt} \]
Collect the P terms and factor
Why: Move the P term to the left.
\[ P\left(1 + C_1e^{rt}\right) = C_1Ke^{rt} \]
Divide
Why: This is equation 4.10.
\[ P(t) = \frac{C_1Ke^{rt}}{1 + C_1e^{rt}} \]
Find the constant from 4.9 at t equal to 0
Why: Easier than using 4.10: put P equal to the initial population.
\[ \frac{P_0}{K - P_0} = C_1e^{0} = C_1 \]
Substitute and multiply top and bottom by K minus the initial population
Why: The nested fractions clear.
\[ P(t) = \frac{P_0Ke^{rt}}{(K - P_0) + P_0e^{rt}} \qquad (4.11) \]
Check the initial value
Why: At t equal to 0 every exponential is one.
\[ P(0) = \frac{P_0K}{(K - P_0) + P_0} = \frac{P_0K}{K} = P_0 \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, pp. 400-401 — equations 4.10 and 4.11, Theorem 4.2
The goal now is algebra: get P alone. The equation from 4.9 has P on both sides once you clear the fraction, so collect the P terms on one side and factor P out. That is the same move as solving any linear equation for a variable that appears twice.
Equation 4.10 still has the unknown constant. The book's advice is good: find it from equation 4.9, not 4.10, because 4.9 at t equal to 0 gives the constant directly as the starting population over the room left.
Substituting and clearing the nested fractions gives Theorem 4.2. The check at t equal to 0 is quick and worth doing: every exponential becomes 1, the denominator becomes K, and the formula returns the starting population, as it must.
Notation
Annotate
On: \( P(t) = \frac{P_0Ke^{rt}}{(K - P_0) + P_0e^{rt}} \)
\[ P(t) = \frac{K}{1 + \frac{K - P_0}{P_0}e^{-rt}} \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 401 — Theorem 4.2
The formula has three inputs and one variable, and every behaviour you have met so far can be read off it. Step through the annotations with that in mind.
While t is small and the starting population is small compared with K, the denominator is close to K, the Ks cancel, and the formula is the exponential model. As t grows, the term with e to the r t takes over the denominator, cancels the same term in the numerator, and leaves K.
The second form below, obtained by dividing top and bottom by the starting population times e to the r t, is often easier to use. It shows the limit even more clearly: e to the minus r t dies away, leaving K over 1. It is also the form used to verify the theorem two slides from now.
Fill the middle
\[ P(t) = \frac{P_0Ke^{rt}}{\;\boxed{\phantom{K - P_0}}\; + P_0e^{rt}} \]
Fill in the blanks
The missing term in the denominator is K − P₀, the room left to grow at the start.
Why: The constant C₁ is P₀/(K − P₀). Multiplying the top and bottom of equation 4.10 by K − P₀ leaves exactly K − P₀ in the denominator, next to P₀e^(rt). Check: at t = 0 the denominator becomes K, and P(0) = P₀.
Type the missing term before you look back. This is the piece of Theorem 4.2 people most often forget or garble.
Where does it come from? The constant from equation 4.9 is the starting population over K minus the starting population. Clearing that fraction by multiplying the top and bottom by K minus the starting population leaves that exact expression in the denominator.
A self-check that works even if you have forgotten the formula: at t equal to 0 the denominator must equal K, so that the formula returns the starting population. The starting population is already there in the second term, so the missing term must be K minus the starting population.
Worked example
Confirm that the formula really solves equation 4.8, using its second form.
\[ P = \frac{K}{1 + Ae^{-rt}}, \qquad A = \frac{K - P_0}{P_0} \]
Differentiate
Why: Chain rule on K times the reciprocal; the inner derivative is minus r A e to the minus rt.
\[ P' = \frac{rKAe^{-rt}}{(1 + Ae^{-rt})^2} \]
Compute the brake
Why: Subtract P over K from one.
\[ 1 - \frac{P}{K} = 1 - \frac{1}{1 + Ae^{-rt}} = \frac{Ae^{-rt}}{1 + Ae^{-rt}} \]
Multiply by rP
Why: Multiply the two fractions.
\[ rP\left(1 - \frac PK\right) = r\cdot\frac{K}{1 + Ae^{-rt}}\cdot\frac{Ae^{-rt}}{1 + Ae^{-rt}} = \frac{rKAe^{-rt}}{(1 + Ae^{-rt})^2} \]
Check: the two sides match
Why: Line one equals line three, so the equation holds for every t, and the long-run limit is K.
\[ P' = rP\left(1 - \frac PK\right), \qquad \lim_{t\to\infty}P = \frac{K}{1 + 0} = K \]
Deriving a formula and checking it are different skills, and this check is independent of the derivation: it does not use partial fractions or logarithms at all, only differentiation.
Start from the second form, K over one plus A e to the minus r t. Its derivative comes from the chain rule on a reciprocal. Separately, compute one minus P over K; the ones cancel in a way that leaves a single clean fraction. Multiply by r P, and the result is identical to the derivative.
Since the two sides match for every t, the formula solves the equation, and since it returns the starting population at t equal to 0, it solves the initial-value problem. The limit at the end is the long-run behaviour, now proved rather than read off a picture.
Ranking
Put in order
Order the steps of solving dP/dt = rP(1 − P/K) with P(0) = P₀.
Why: The equilibria come first because dividing by P(K − P) loses them. The constant is easiest to find at equation 4.9, before solving for P, which is the order the book uses.
Order the steps before revealing. The middle of the sequence is forced by the algebra; the interesting decisions are at the two ends.
At the start, the equilibria come first, because the separation step divides by P times K minus P and would silently lose them. At the end, finding the constant from equation 4.9 before solving for P is a choice of convenience, and it is the one the book recommends because it is shorter.
If you can reproduce these six steps from memory, you can solve any logistic problem, including ones with numbers as awkward as the deer's. The next slide does exactly that.
Worked example
900,000 deer in 2004. Unchecked, the herd doubles every 3 years; the land supports 27 deer per square mile over 39,732 square miles.
Growth rate from the doubling time
Why: Doubling in 3 years means e to the 3r is 2. This is 0.23105; the book rounds it up to 0.2311, and the deck keeps the book's value so the numbers match.
\[ e^{3r} = 2 \;\Longrightarrow\; r = \frac{\ln 2}{3} \approx 0.23105 \]
Carrying capacity from the density
Why: Deer per square mile times square miles.
\[ K = 27 \times 39{,}732 = 1{,}072{,}764 \]
Write the initial-value problem
Why: Part (a).
\[ \frac{dP}{dt} = 0.2311P\left(1 - \frac{P}{1{,}072{,}764}\right), \quad P(0) = 900{,}000 \]
Find the constant from equation 4.9
Why: Put t equal to 0.
\[ C_2 = \frac{900{,}000}{1{,}072{,}764 - 900{,}000} = \frac{900{,}000}{172{,}764} = \frac{25{,}000}{4799} \approx 5.209 \]
Write the solution, equation 4.10
Why: Part (b).
\[ P(t) = \frac{1{,}072{,}764\,C_2e^{0.2311t}}{1 + C_2e^{0.2311t}} \]
Divide top and bottom by the constant
Why: One over it is 4799 over 25,000, exactly 0.19196.
\[ P(t) = \frac{1{,}072{,}764\,e^{0.2311t}}{0.19196 + e^{0.2311t}} \]
Check the initial value
Why: At t equal to 0 the exponential is 1, and 1.19196 times 900,000 is exactly the capacity.
\[ P(0) = \frac{1{,}072{,}764}{1.19196} = 900{,}000 \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, pp. 395-397 — Example 4.14a and b
This example starts from words, so the first job is extracting the three parameters. The doubling time gives r: if the herd doubles in 3 years without a brake, e to the 3 r is 2. That comes to 0.23105. The book writes 0.2311, a rounding slip in the fourth place; the deck keeps the book's value so that every number matches the textbook.
The capacity is a density times an area, 27 deer per square mile over 39,732 square miles, which is 1,072,764 deer. The starting population is the 2004 estimate. With those, part (a) is just equation 4.8 with numbers.
For part (b), equation 4.9 at t equal to 0 gives the constant, 900,000 over 172,764, which simplifies to 25,000 over 4799. The neat simplification at the end, 0.19196, is exact, and so is the check: 1.19196 times 900,000 is precisely the capacity.
Worked example
Predict the population in 2007 and compare it with the unchecked doubling.
\[ P(t) = \frac{1{,}072{,}764\,e^{0.2311t}}{0.19196 + e^{0.2311t}} \]
Evaluate the exponential
Why: Three years at rate 0.2311.
\[ e^{0.2311(3)} = e^{0.6933} \approx 2.0003 \]
Substitute
Why: Put it into the numerator and the denominator.
\[ P(3) \approx \frac{1{,}072{,}764(2.0003)}{0.19196 + 2.0003} \]
Divide
Why: Round to whole deer.
\[ P(3) \approx 978{,}830 \]
Figure (svg): The deer population in thousands for t from 0 to 20 years, rising from 900 and flattening under a dashed line at the carrying capacity 1072.8, with a dot at t equal 3 marking 978.8 thousand.
Compare with the doubling
Why: Unchecked growth would give 1.8 million; the logistic herd adds only about 79 thousand deer.
\[ 900{,}000 \times 2 = 1{,}800{,}000 \quad\text{vs}\quad 978{,}830 \]
Check against the rate
Why: Three years of growth averages about 26 thousand a year, less than the 33,496 a year the herd starts with, as it must be for a slowing herd.
\[ \frac{978{,}830 - 900{,}000}{3} \approx 26{,}277 < 33{,}496 = P'(0) \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 398 — Example 4.14c and Figure 4.21
The book asks this because the doubling time was quoted as 3 years, and it wants you to see how far the logistic prediction falls from a doubling. Evaluate the formula carefully, one move at a time.
The exponent is 0.6933, and e to that power is almost exactly 2: over three years, the unbraked exponential really would double. But the logistic formula divides by 0.19196 plus that 2, and the result is about 978,830 deer, a gain of under 79 thousand.
Look at the figure: the herd starts well up the flat part of the curve. The check links this back to Part 1. The herd's rate in 2004 was about 33,496 deer a year, and the average over the next three years is about 26,277, smaller, because the rate keeps falling as the herd grows. A prediction and a rate that disagree would signal an arithmetic error.
Worked example
Suppose the herd reached 1,200,000 deer. Same equation, new initial condition.
Find the constant from equation 4.9
Why: Now K minus the initial population is negative, so the constant is negative.
\[ C_2 = \frac{1{,}200{,}000}{1{,}072{,}764 - 1{,}200{,}000} = \frac{1{,}200{,}000}{-127{,}236} \approx -9.4313 \]
Substitute into equation 4.10
Why: The general solution is unchanged.
\[ P(t) = \frac{1{,}072{,}764\,C_2e^{0.2311t}}{1 + C_2e^{0.2311t}} \]
Divide top and bottom by minus the constant
Why: The numerator constant is 1,072,764 times 9.4313.
\[ P(t) \approx \frac{10{,}117{,}552\,e^{0.2311t}}{9.43129\,e^{0.2311t} - 1} \]
Figure (svg): The deer population in thousands for t from 0 to 20 years, starting at 1200 and falling, concave up, toward the dashed carrying capacity line at 1072.8 without reaching it.
Check the start and the long run
Why: At t equal to 0 the formula returns 1.2 million, and as t grows it tends to the capacity from above.
\[ P(0) = \frac{10{,}117{,}552}{8.43129} \approx 1{,}200{,}000, \qquad P(t) \to \frac{10{,}117{,}552}{9.43129} \approx 1{,}072{,}764 \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, pp. 398-399 — Example 4.14d and Figure 4.22
Nothing in the method changes when the starting population is above K, except one sign. K minus the starting population is now negative, so the constant is negative: about minus 9.43.
The book then multiplies top and bottom by a negative constant to make the formula look tidy, giving a numerator constant of about 10,117,552 and a denominator of 9.43129 times the exponential, minus 1. The book prints the numerator as 10,117,551, cutting off the decimal instead of rounding; the difference is one deer in ten million and changes nothing visible.
The check does two things. At t equal to 0 the formula returns 1.2 million, and as t grows the minus 1 becomes negligible, leaving the numerator constant over 9.43129, which is the capacity. The figure shows the decline that the phase line predicted: steep at first, then gently settling onto K.
Worked example
Solve the rabbits' initial-value problem and predict the population after 12 months.
\[ r = 0.04, \quad K = 750, \quad P_0 = 200 \]
Substitute into Theorem 4.2
Why: K minus the initial population is 550.
\[ P(t) = \frac{200(750)e^{0.04t}}{550 + 200e^{0.04t}} \]
Simplify
Why: Divide the top and bottom by 50.
\[ P(t) = \frac{3000e^{0.04t}}{11 + 4e^{0.04t}} \]
Put t equal to 12
Why: The exponent is 0.48.
\[ P(12) = \frac{3000e^{0.48}}{11 + 4e^{0.48}} \approx \frac{4848.2}{17.464} \approx 277.6 \]
Figure (svg): Two curves from 200 rabbits over 60 months: the exponential 200 e to the 0.04 t climbing past 2000, and the logistic solution rising to about 600 and bending toward the dashed line at 750. Dots mark month 12: 323 exponential, 278 logistic.
Check against the exponential model
Why: With no brake the rabbits would number about 323; the logistic answer, about 278, must be smaller, and it is.
\[ 200e^{0.48} \approx 323.2 > 277.6 \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 403 — Checkpoint 4.14c and d
This is Theorem 4.2 applied directly, with small numbers. K minus the starting population is 550, and dividing the top and bottom by 50 gives the clean form with 3000 on top and 11 plus 4 times the exponential below.
The time unit is the thing to watch. The growth rate is per month, so one year is t equal to 12, not t equal to 1. The exponent is then 0.48, and the formula gives about 277.6, which you would report as about 278 rabbits.
The figure compares the two models from the same start. After one year the exponential would give 323 rabbits, so the brake has already cost 45. That is the check: the logistic value must be below the exponential one, because the bracket is always less than one while the population is below capacity.
Section
Part 4
Concept
You do not need the formula to find where the solution bends. Differentiate the equation itself with respect to t, using the chain rule:
\[ P'' = \frac{d}{dt}\left[rP - \frac{r}{K}P^2\right] = \left(r - \frac{2r}{K}P\right)P' \]
\[ P'' = r\left(1 - \frac{2P}{K}\right)\cdot rP\left(1 - \frac{P}{K}\right) \]
| population | sign of P′′ | shape |
|---|---|---|
| 0 < P < K/2 | positive | concave up: speeding up |
| P = K/2 | zero | inflection: fastest growth |
| K/2 < P < K | negative | concave down: slowing |
| P > K | positive | concave up: falling, ever more gently |
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 402 — the point of inflection
Here is a shortcut the book does not take. Instead of differentiating the solution formula twice, differentiate the equation. The rate is a function of P, and P is a function of t, so the chain rule gives the second derivative as the derivative of the rate with respect to P, times the first derivative.
Both factors are already familiar. The derivative of the rate is r times one minus 2 P over K, which changes sign at half the capacity. The first derivative is the rate itself, which changes sign at the capacity.
The table lists the four possibilities. Below half the capacity, both are positive and the curve bends up. Between half and full, the first factor turns negative and the curve bends down. Above K, both are negative, their product is positive, and the falling curve bends up. So the inflection happens exactly when the population crosses half the capacity, with no formula needed.
Picture it
Figure (svg): Left: the S-shaped solution with K equal 10, r equal 0.5 and P0 equal 0.5, drawn orange while concave up and blue while concave down, with a dot where it crosses P equal 5 at t equal 5.89. Right: its rate dP/dt against t, a bell-shaped hump peaking at 1.25 at the same time 5.89.
The solution is steepest at the moment it crosses half the capacity, and that is exactly where it changes from concave up to concave down.
The two panels show the same solution. On the left, the population, coloured orange while it bends upward and blue once it bends downward. On the right, its rate, plotted against time.
The dot on the left is where the colour changes, at the moment the population crosses 5, half the capacity. The dot on the right is the top of the hump. They sit at the same time, 5.89, and that is not a coincidence: the steepest point of a curve is where its slope is largest, and the slope is largest where the second derivative is zero.
The hump is symmetric. The population takes as long to climb from near zero to half the capacity as it then takes to approach the capacity from half. Epidemics and product adoptions often show this same bell of new cases per week.
Worked example
Find when the solution of Theorem 4.2 reaches its inflection point, for a population that starts below half its capacity.
Set the population equal to half the capacity
Why: The inflection is at K over 2.
\[ \frac{P_0Ke^{rt}}{(K - P_0) + P_0e^{rt}} = \frac{K}{2} \]
Cross-multiply and cancel K
Why: Both sides are positive.
\[ 2P_0e^{rt} = (K - P_0) + P_0e^{rt} \]
Isolate the exponential
Why: Subtract the initial population times the exponential.
\[ P_0e^{rt} = K - P_0 \;\Longrightarrow\; e^{rt} = \frac{K - P_0}{P_0} \]
Take logarithms
Why: Divide by r.
\[ t^{*} = \frac{1}{r}\ln\frac{K - P_0}{P_0} \]
Check with the rabbits
Why: The field reading was about 25 months.
\[ t^{*} = \frac{1}{0.04}\ln\frac{550}{200} = 25\ln 2.75 \approx 25.29 \text{ months} \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, pp. 402-403 — solving for the inflection time
Knowing the inflection happens at half the capacity, the time is found by setting the formula equal to K over 2 and solving. This is shorter than the book's route, which sets the second derivative of the formula to zero, and it arrives at the same answer.
After cross-multiplying, the starting population times the exponential appears on both sides and combines, leaving e to the r t equal to the room left divided by the starting population. A logarithm finishes it.
The check closes the loop with Part 2. From the rabbits' direction field you estimated about 25 months; the formula gives 25 times the logarithm of 2.75, which is 25.29 months. A field reading confirmed by a formula is about as secure as an answer gets.
Worked example
One bacterium, growth rate 20 percent per hour, carrying capacity one million cells. How long until there are 500,000 cells?
\[ r = 0.2, \quad K = 10^6, \quad P_0 = 1 \]
Notice the target
Why: 500,000 is half the capacity, so the answer is the inflection time.
\[ 500{,}000 = \frac{K}{2} \]
Use the inflection formula
Why: One over 0.2 is 5.
\[ t^{*} = \frac{1}{0.2}\ln\frac{10^6 - 1}{1} = 5\ln 999{,}999 \]
Evaluate
Why: The logarithm of a million is about 13.8155.
\[ t^{*} \approx 5(13.8155) \approx 69.08 \text{ hours} \]
Check by substituting back into Theorem 4.2
Why: At this time e to the rt is 999,999, so the numerator is exactly half the denominator times the capacity.
\[ P = \frac{1 \cdot 10^6 \cdot 999{,}999}{999{,}999 + 999{,}999} = 500{,}000 \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 405 — Exercise 175
The key step is noticing what the target is. 500,000 cells is exactly half of a million, so the question is really asking for the inflection time, and the formula from the previous slide applies at once.
One over the growth rate is 5 hours. The ratio inside the logarithm is the room left over the starting population, 999,999 over 1. Its logarithm is about 13.8, so the answer is about 69.08 hours, just under three days.
The check substitutes back into Theorem 4.2. At that time the exponential equals 999,999, so the denominator is twice 999,999 and the numerator is a million times 999,999, and the ratio is exactly half a million. Notice how most of the 69 hours is spent while the colony is tiny: the doubling time is only about 3.5 hours, but it takes nearly twenty doublings to get from one cell to half a million.
Trap
Asked for the inflection of the herd that starts at 1.2 million:
\[ t^{*} = \frac{1}{r}\ln\frac{K - P_0}{P_0} \]
Wrong to expect an answer: the formula breaks.
With the initial population above K, the fraction is negative and its logarithm does not exist. Above the capacity the concavity table says concave up for all time: the curve falls steeply, then gently, and never bends the other way.
\[ \frac{K - P_0}{P_0} = \frac{-127{,}236}{1{,}200{,}000} < 0 \]
The inflection formula looks as if it always gives a time. It does not: its logarithm needs a positive argument, and that only happens when the population starts below the capacity.
For the herd of 1.2 million, the room left is negative, the ratio is negative, and the formula has no answer. That is not a failure of the method; it is the correct answer. The concavity table from earlier says that above the capacity the curve is concave up for all time, so there is no point where it bends the other way.
The figure for Example 4.14(d) shows it: a decline that is steep at first and flattens, with no change of bending. When a formula refuses to give a value, ask what the refusal means before forcing a number out of it.
Picture it
Figure (svg): Two curves through 900 thousand deer at t equal 0, for t from minus 20 to 10 years. The exponential curve 900 e to the 0.2311 t shoots off the top of the graph after t equal 3. The logistic curve is an S that follows the exponential closely before about t equal minus 12 and flattens under the dashed capacity line; a dot marks its inflection at t equal minus 7.14, half the capacity.
Run the deer model backward from 2004. About twenty years earlier the herd was small and growing at nearly the full exponential rate; its fastest growth came about seven years before 2004, near 1997, at half the capacity.
\[ t^{*} = \frac{1}{0.2311}\ln\frac{172{,}764}{900{,}000} \approx -7.14 \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 402 — Figure 4.23
This is Figure 4.23. Both curves pass through 900 thousand deer in 2004, and the horizontal axis runs twenty years into the past and ten into the future.
On the left the two curves nearly agree. Twenty years before 2004 the logistic herd is only about 52 thousand, under a twentieth of the capacity, and its relative growth rate is about 22 percent, close to the full 23. That is the book's point: early growth looks exponential.
The dot marks the inflection. Using the formula with a starting population above half the capacity gives a negative time, about minus 7.14: the herd grew fastest around 1997. By 2004 the logistic herd was nearly full, while the exponential curve, sharing the same start and rate, is about to leave the top of the graph.
Comparison
Comparison matrix
| property | exponential dP/dt = rP | logistic dP/dt = rP(1 − P/K) |
|---|---|---|
| equilibria | P = 0 only | P = 0 and P = K |
| relative growth rate | constant r | r(1 − P/K), falling in P |
| long-run behaviour | grows without bound | approaches K |
| inflection point | none | at P = K/2, if P₀ < K/2 |
| solution | P₀e^(rt) | P₀Ke^(rt)/((K − P₀) + P₀e^(rt)) |
Fill in every blank before revealing. Each row compares one property, and each blank is something you have derived rather than something to memorise.
Two rows are worth checking against pictures you have seen. The relative growth rate row is the flat line versus the falling line from Part 1. The inflection row carries a condition: the logistic curve shows its inflection for positive time only if it starts below half the capacity, as the next slide explores.
The last row, the solution, is where students most often go wrong under time pressure. If you can write Theorem 4.2 from memory and check it at t equal to 0, you have it.
Counterexample
Discussion prompt
Someone claims: every logistic solution that starts between 0 and K has an S-shape for t at least 0. Find a starting population that breaks the claim, and say why.
Think about what the inflection time formula says when the starting population is between half the capacity and the capacity. Try to find a counterexample before revealing.
With a start at eight tenths of the capacity, the ratio in the logarithm is one quarter, and its logarithm is negative. The inflection time is before the clock started. On the graph, the part of the S that bends upward has already happened; from t equal to 0 onward you see only the part that bends down.
The Kentucky deer are exactly this case: 900 thousand is 84 percent of the capacity. That is why Figure 4.21 of the book shows a curve that only flattens. A logistic population always has an inflection somewhere on its full history, if it started below capacity, but you only see it if you catch the population early enough.
Tweak it
Parameter explorer
The curve is the solution of Theorem 4.2. Change one parameter at a time. Which one moves the ceiling, which one changes only how quickly the S rises, and what happens when the start is above K?
\[ r = {r}, \quad K = {K}, \quad P_0 = {P0} \]
Move one slider at a time and describe what changes before moving the next. The goal is to connect each parameter to one feature of the curve.
The capacity K sets where the curve levels off, and nothing else sets it. The growth rate r changes how fast the rise happens: double it and the whole S is squeezed into half the time, but it rises to the same height. The starting population decides which part of the S you see.
Now push the starting population above K. The curve falls to K instead of rising. Put it exactly at K and the curve is flat. Put it between half the capacity and the capacity and there is no visible S, only a bend downward. These are the same three cases as the direction field, now in one picture you control.
Intuition
The same fact, K over 2, shows up in two places, and it helps to see them as one.
\[ \underbrace{g'(P) = 0 \text{ at } P = \tfrac K2}_{\text{the rate is largest}} \quad\Longleftrightarrow\quad \underbrace{P'' = g'(P)\,P' = 0}_{\text{the curve inflects}} \]
Read against P, the rate parabola peaks at half the capacity. Read against t, that peak is the moment the curve is steepest, which is the inflection. Below half, each extra organism adds more births than it costs; above half, crowding costs more than it adds.
You have now met half the capacity in two different guises, and it is worth seeing why they are the same fact.
Read against P, the rate is a parabola, and its peak is where its derivative with respect to P vanishes. Read against t, the second derivative of the population is that same derivative times the rate. So the second derivative vanishes exactly when the rate is at its peak. The steepest point of the curve and the peak of the rate are one event, seen from two axes.
There is also a biological reading. Below half the capacity, one more animal adds more births than it costs in crowding; above half, it costs more than it adds. That balance point is why fisheries and forests aim to keep stocks near half their natural capacity, which Part 5 makes precise.
Section
Part 5
Concept
A logistic model needs r, K and the initial population. Take logarithms in equation 4.9 and the S becomes a line:
\[ \ln\frac{P}{K - P} = \ln C_1 + rt \]
Figure (svg): The whooping crane counts from 1940 to 2000 transformed to ln of P over 10000 minus P and plotted against years since 1940: seven dots rising from minus 6.12 to minus 3.64, roughly along a straight line through the first and last dots with slope 0.0413.
Given K, the log-odds of the data should lie on a line whose slope is r. The whooping crane counts, with K equal to 10,000, nearly do. But the transformation needs K before you start, and that is the parameter the data know least about.
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, pp. 406-407 — the whooping crane table, Exercises 203 to 207
Fitting a model means choosing its parameters so that it matches measurements. For the logistic model there are three, and they are not equally easy to measure.
Taking the logarithm of equation 4.9 turns it into a straight line in t, whose slope is the growth rate. That gives a practical test: transform the data, and if the points line up, the logistic model fits and the slope gives r. The whooping crane counts, transformed with a capacity of 10,000, fall roughly along a line with slope 0.0413, with a dip in the middle years.
The catch is in the transformation itself: it needs K before you start. The next few slides use the transformation with K given, and then ask what happens when K is really unknown.
Step zero
\[ P(0) = 2, \quad P(5) = 8, \quad K = 25 \]
Discussion prompt
Exercise 177: two monkeys are put on an island; five years later there are 8, and the capacity is estimated at 25. Before computing anything, which parameter is missing, and which equation gives it most directly?
Before any calculation, list what you know and what you need. Here you know the starting population, the capacity, and one later observation. The growth rate is missing, and the question itself asks for a time.
Equation 4.9 is the natural tool, because it has the exponential isolated: the population over the room left equals the starting value of that ratio times e to the r t. Each known data point gives one equation with one unknown.
So the plan is two passes through the same equation: first with the observation at 5 years to find r, then with the target of 16 to find t. Planning this before starting saves you from solving Theorem 4.2 for t, which is possible but far messier.
Worked example
\[ P_0 = 2, \quad K = 25, \quad P(5) = 8 \]
The constant from the initial population
Why: Equation 4.9 at t equal to 0.
\[ C_1 = \frac{2}{25 - 2} = \frac{2}{23} \]
Use the second observation
Why: Put t equal to 5 and P equal to 8.
\[ \frac{8}{17} = \frac{2}{23}e^{5r} \;\Longrightarrow\; e^{5r} = \frac{92}{17} \approx 5.412 \]
Solve for r
Why: Take logarithms and divide by 5.
\[ r = \frac15\ln\frac{92}{17} \approx 0.3377 \]
Set P equal to 16
Why: Equation 4.9 again, now with the unknown t.
\[ \frac{16}{9} = \frac{2}{23}e^{rt} \;\Longrightarrow\; e^{rt} = \frac{184}{9} \approx 20.44 \]
Solve for t
Why: Divide the logarithm by r.
\[ t = \frac{\ln(184/9)}{0.3377} \approx \frac{3.0177}{0.3377} \approx 8.94 \text{ years} \]
Figure (svg): The fitted logistic curve for the monkeys, t from 0 to 20 years: through the data points (0, 2) and (5, 8), passing 16 at t equal 8.94, and levelling toward the dashed line at 25.
Check in Theorem 4.2
Why: With e to the rt equal to 184 over 9, the formula must give 16.
\[ P = \frac{2(25)e^{rt}}{23 + 2e^{rt}} = \frac{50(184/9)}{23 + 368/9} = \frac{9200}{575} = 16 \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 405 — Exercise 177
Follow the plan from the previous slide. The constant is 2 over 23, the ratio at the start. At 5 years the ratio is 8 over 17, and dividing gives e to the 5 r equal to 92 over 17, so r is about 0.338 per year.
The second pass uses the same equation with 16 monkeys, where the ratio is 16 over 9. Dividing by the constant gives e to the r t equal to 184 over 9, about 20.4, and the logarithm divided by r gives about 8.94 years.
The check is exact rather than approximate: with e to the r t equal to 184 over 9, Theorem 4.2 gives 9200 over 575, which is exactly 16. The figure shows the whole fitted curve through both observations. Notice that 16 is past half the capacity, so the monkeys reach it after their fastest growth, at around 7 years.
Worked example
22 cranes in 1940, 256 in 2000, capacity assumed 10,000. Fit r through those two counts and predict 2010, when the real count was 437.
The constant from 1940
Why: t equal to 0 is 1940.
\[ C_1 = \frac{22}{9978} \approx 0.0022049 \]
Use the 2000 count
Why: t equal to 60.
\[ \frac{256}{9744} = C_1e^{60r} \;\Longrightarrow\; e^{60r} \approx 11.916 \]
Solve for r
Why: Divide the logarithm by 60.
\[ r = \frac{\ln 11.916}{60} \approx 0.04130 \]
Predict t equal to 70
Why: Theorem 4.2; the exponential is e to the 2.891, about 18.008.
\[ P(70) = \frac{22(10{,}000)(18.008)}{9978 + 22(18.008)} \approx 382 \]
Check against the 2010 count
Why: The model is 55 cranes short, about 13 percent: the real flock grew faster after 2000 than the fitted curve.
\[ 437 - 382 = 55, \qquad \frac{55}{437} \approx 0.13 \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, pp. 406-407 — Exercises 203 and 207
This is the same two-point method as the monkeys, with real data. The whooping crane recovery began with 22 birds in 1940 and reached 256 by 2000; the exercise tells you to assume a capacity of 10,000.
The steps are identical: the constant from 1940, then e to the 60 r from the 2000 count, then r of about 0.0413 per year. The prediction for 2010 comes from Theorem 4.2 with t equal to 70 and is about 382 cranes.
The check here is against reality, and reality disagrees: the 2010 count was 437. The model is 13 percent low. A fit through two points ignores the other five, and the population did not grow at a constant log-odds rate. Exercise 207 asks you to compare this with the threshold and Gompertz models, which is how model choice is actually done: fit several, and test each against data it was not fitted to.
Picture it
Figure (svg): Left: the seven whooping crane counts from 1940 to 2000 with three logistic fits through the first and last counts, using capacities 1000, 10000 and 100000; all three pass close to the data and predict 352, 382 and 385 cranes in 2010, where the real count 437 is shown as an open dot. Right: the same three fits run to year 150, where they reach about 954, 5194 and 9275.
Refit the cranes with capacities of 1000 and 100,000 instead. All three curves pass through the early counts and predict 352, 382 and 385 for 2010. Run them forward and they split: by year 150 they stand near 950, 5200 and 9300.
Here is the deepest point of the fitting section. Refit the same two counts with a capacity of 1000, and again with 100,000. On the left, all three curves thread through the seven counts almost equally well, and their forecasts for 2010 differ by only 33 cranes.
On the right, the same three curves run forward to year 150, and they are nothing alike. One levels off below a thousand; one is past five thousand and still climbing; one is past nine thousand.
The reason is the brake. When the population is a few hundred, the bracket is almost exactly 1 for any capacity in the thousands, so the data carry almost no information about K. Early data pin down r; only data from the flattening phase can pin down K.
Real world
Discussion prompt
A wildlife agency wants a single number: how many whooping cranes will there be in 2090? Using the three fits on the previous slide, what should an honest forecast say, and what data would sharpen it?
Write your own answer before revealing: imagine you have to sign your name under it.
A single number would hide the fact that the model's most important parameter is unknown. The responsible forecast gives the range the plausible capacities produce, and states that the capacity is the dominant uncertainty.
The remedy is to measure the capacity some other way, not from the population curve. That is exactly what Kentucky did for the deer in Example 4.14: sustainable density times area. For cranes, that would mean surveying wetland habitat. Or, if you can wait, collect data until the growth visibly slows, which is when the curve starts to reveal its ceiling.
Concept
Take the trout of Exercise 179: logistic growth with r equal to 0.4 and K equal to 10,000, and 400 fish caught every year.
\[ P' = 0.4P\left(1 - \frac{P}{10{,}000}\right) - 400 \]
Figure (svg): The trout growth rate against the population, for P from 0 to 10500: the dashed parabola 0.4 P times one minus P over 10000, peaking at 1000, and the same parabola lowered by 400. The lowered one crosses zero at 1127, marked with an open dot, and at 8873, marked with a filled dot.
The catch lowers the whole rate curve. The capacity drops, and a second, unstable equilibrium appears: a threshold below which the fishing outpaces the breeding.
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 405 — Exercise 179
Exercises 179 to 188 extend the model by taking animals out. With a fixed catch of 400 trout a year, the rate is the logistic parabola minus 400: the same shape, lowered.
Look at where the lowered curve crosses zero. The right crossing, near 8873, is the new stable population: fishing has reduced the pond's long-run stock from 10,000. The left crossing, near 1127, is new, and it is unstable.
That second equilibrium is the important one. Below it, the fish breed more slowly than they are caught, and the population collapses to zero. Harvesting has created an extinction threshold that the plain logistic model never had, which is why fisheries managers watch stock levels, not just catch sizes.
Worked example
\[ g(P) = 0.4P - 0.00004P^2 - 400 \]
Set the rate to zero and clear decimals
Why: Multiply by minus 25,000.
\[ P^2 - 10{,}000P + 10{,}000{,}000 = 0 \]
Quadratic formula
Why: Half of 10,000 is 5000.
\[ P = 5000 \pm \sqrt{5000^2 - 10^7} = 5000 \pm \sqrt{1.5 \times 10^7} \]
Evaluate
Why: The square root is about 3873.
\[ P_1 \approx 1127, \qquad P_2 \approx 8873 \]
Classify by the slope of g
Why: Differentiate once.
\[ g'(P) = 0.4 - 0.00008P \]
Check the signs
Why: Positive at the lower root, negative at the upper, symmetric about 5000.
\[ g'(1127) \approx 0.310 > 0 \;(\text{unstable}), \quad g'(8873) \approx -0.310 < 0 \;(\text{stable}) \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 405 — Exercises 179 and 180
Setting the rate to zero gives a quadratic. Multiplying by minus 25,000 clears the decimals and makes the leading coefficient 1, which keeps the arithmetic clean.
The quadratic formula gives 5000 plus or minus the square root of 15 million, about 3873. Notice the symmetry about 5000: the equilibria sit equally far either side of half the capacity, because the parabola is symmetric about its peak.
Stability comes from the slope of the rate, exactly as for the plain logistic equation. The derivative is 0.4 minus 0.00008 P, positive at the lower root and negative at the upper one. So 1127 is unstable and 8873 is stable, matching the arrows you would draw from the figure on the previous slide.
Concept
The catch f simply lowers the parabola. Equilibria exist only while the lowered parabola still reaches zero, that is, while f is at most the parabola's peak, which Exercise 200 found:
\[ f_{\max} = \frac{rK}{4} = \frac{0.4(10{,}000)}{4} = 1000 \text{ fish per year} \]
Catch 600 a year and the equilibria move to about 1838 and 8162; catch more than 1000 and every population declines to zero. Fisheries call this the maximum sustainable yield, and it is taken at half the capacity.
\[ f = 600: \quad P = 5000 \pm \sqrt{10^7} \approx 1838, \; 8162 \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 405 — Exercises 181 and 182
Raise the catch and the parabola sinks further. The two equilibria slide toward each other: at a catch of 600 they are near 1838 and 8162. When the catch equals the peak of the parabola, they meet at 5000, half the capacity.
The peak is the number from Exercise 200, r times K over 4, which for the trout is 1000 fish a year. Beyond that, the lowered parabola is below zero everywhere, so the rate is negative for every population and the pond is fished out, however many fish it starts with.
This is the mathematical core of the maximum sustainable yield used in fisheries management. In practice nobody fishes at exactly that value, because a small error in r or K, or one bad year, tips the stock over the edge. The mathematics explains the caution.
Prediction
\[ P' = 0.4P\left(1 - \frac{P}{10{,}000}\right) - 0.1P \]
Predict first
Exercise 184: instead of a fixed catch, a tenth of the fish are caught each year. Where does the trout population settle?
Correct: 7500
Why: Collect the linear terms: 0.4P minus 0.1P is 0.3P, and the equation becomes 0.3P(1 − P/7500). It is still logistic, with a smaller rate 0.3 and a smaller capacity 7500, and 7500 is stable. Unlike a fixed catch, proportional fishing never creates an extinction threshold.
\[ 0.3P - 0.00004P^2 = 0.3P\left(1 - \frac{P}{7500}\right) \]
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 405 — Exercise 184
Commit to a prediction, and think about what kind of term is being subtracted this time.
The catch is now a tenth of the stock, a term proportional to P, not a constant. Collect the terms that are proportional to P: 0.4 P minus 0.1 P is 0.3 P. The equation becomes a logistic equation again, with rate 0.3 and a capacity you find by factoring: 7500.
So the pond settles at 7500 fish, and there is no extinction threshold, because zero is still the only other equilibrium and it is still unstable. Exercise 185 raises the fishing fraction to 0.4, which pushes the capacity down to zero; beyond that the population dies out whatever the start. Proportional fishing degrades gracefully; fixed catches can collapse suddenly.
Concept
Some species need a minimum number to survive. The book's student project adds a threshold T to the model, here for Montana elk with capacity 25,000, threshold 5000 and growth rate 0.18:
\[ \frac{dP}{dt} = -rP\left(1 - \frac PK\right)\left(1 - \frac PT\right) \qquad (4.12) \]
Figure (svg): Top: the rate curve for the elk threshold model, negative between 0 and 5000, positive between 5000 and 25,000 with a hump near 17,600, and negative beyond 25,000. Bottom: the phase line, with filled dots at 0 and 25,000, an open dot at 5000, arrows pointing left below 5000, right between 5000 and 25,000, and left above 25,000.
Now there are three equilibria, 0, T and K, and the threshold between them is unstable: above it the herd grows to K, below it the herd dies out.
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 404 — Student Project and equation 4.12
The book's student project adds a second bracket, one minus P over T, for a minimum viable population. The elk numbers are from the project: capacity 25,000, threshold 5000, growth rate 18 percent a year. The minus sign in front makes the signs come out right, as the figure confirms.
Read the rate curve: negative below 5000, positive between 5000 and 25,000, negative again above. So there are three equilibria. Zero and the capacity are stable, and the threshold between them is unstable.
The biological meaning is stark. A herd just above 5000 slowly recovers toward 25,000. A herd just below slides to extinction, even though nothing is hunting it: there are simply too few animals to find mates and sustain themselves. This is the reasoning behind endangered species lists, and why conservation often aims to get a population past a threshold rather than just to protect it.
Matching
Match the pairs
Why: The phase line of equation 4.12 has arrows pointing away from T = 5000 on both sides and toward 0 and K = 25,000. The project's 18,000 elk lie between T and K, so they rise to the capacity; 3000 elk are below the threshold and die out.
Use the phase line under the threshold figure. Place each starting population on it, follow the arrow, and see where it stops.
The 3000 elk are below the threshold, where the arrow points left, so they decline to zero. The 5000 elk sit exactly on the unstable equilibrium: in the model they stay there, but any disturbance at all sends them one way or the other. The project's 18,000 elk are between the threshold and the capacity, so they rise to 25,000. The 30,000 elk fall back to 25,000.
Notice how much you could answer without solving the equation, which the book says is very difficult to do explicitly. That is the power of the phase line: long-run behaviour from signs alone.
Section
Part 6
Pattern
Figure (svg): A flow diagram for a logistic problem: read r, K and P0 from the words; set the rate to zero to get the equilibria 0 and K; then either branch to the phase line for qualitative questions, or to Theorem 4.2 for numbers; from Theorem 4.2, solve P of t equals a target for a time, or use K over 2 for the fastest growth.
Every problem in this section follows the same path, and the diagram lays it out. Start by extracting the three parameters from the words, converting percentages to decimals and doubling times to rates.
Then decide what kind of question you have. Where does it end up, does it grow or shrink, is an equilibrium stable: those need only the equilibria and the phase line, and no formula at all. How many at a given time, or when does it reach a target: those need Theorem 4.2, with the constant from equation 4.9.
Two shortcuts are worth keeping. A target of half the capacity is the inflection time, with its own formula. And the fastest growth rate is always r times K over 4, at half the capacity.
Check
Check your understanding
For dP/dt = 0.5P(1 − P/40), which statement is correct?
Answer: B
Why: The rate is positive between 0 and 40 and negative above 40, so populations move away from 0 and toward 40 from both sides. Equivalently, the slope of the rate is 0.5 at zero and minus 0.5 at forty.
Answer before revealing, and decide which test you are using: the arrows on the phase line or the slope of the rate.
The equilibria are 0 and 40, and the rate is positive between them. So arrows point away from 0 and toward 40, from both sides. Zero is unstable and 40 is stable, as for every logistic equation with positive r.
If you picked D, notice the trap: 20 is special, but it is where the rate is largest, not where it is zero. Equilibria come from setting the rate to zero, never from maximising it.
Check
Check your understanding
A logistic population has r = 0.3 per year and K = 800. At what population is it growing fastest, and how fast?
Answer: B
Why: The rate is a downward parabola in P with roots 0 and 800, so it peaks halfway, at 400. There the rate is rK over 4, which is 0.3 times 800 over 4, or 60 per year.
Answer before revealing. Two numbers are asked for, a population and a rate, and the distractors are built from mixing them up.
The rate is a downward parabola with roots at 0 and 800, so it peaks midway at 400. The peak rate is r times K over 4, which is 60 per year. Equivalently, at 400 the rate is 0.3 times 400 times the bracket one half.
Choice C is the most tempting wrong answer: it forgets the brake, computing 0.3 times 400 as if the growth were exponential. At half the capacity the brake is half on, which halves the rate.
Check
Check your understanding
Which function solves dP/dt = 0.1P(1 − P/500) with P(0) = 100?
Answer: A
Why: Theorem 4.2 gives 100 times 500 times the exponential, over 400 plus 100 times the exponential. Dividing the top and bottom by 100 leaves 500 times the exponential over 4 plus the exponential. At t equal to 0 that is 500 over 5, which is 100.
Answer before revealing, then check your choice at t equal to 0, which eliminates two distractors immediately.
Theorem 4.2 with a start of 100 and a capacity of 500 has 400 in the denominator next to 100 times the exponential. Dividing everything by 100 gives choice A. At t equal to 0 it gives 500 over 5, which is 100, as it should.
Choice D is a subtle one. It is the second form of the solution with the sign of the exponent flipped. It does start at 100, but it then shrinks toward zero instead of growing toward 500. The long-run limit is the second check that catches it: a correct logistic solution starting below K must approach K.
Explain it to yourself
Discussion prompt
In two or three sentences: why does a logistic population grow fastest at half its carrying capacity? Give one reason in terms of the rate as a function of P, and one in terms of the shape of the curve P(t).
Write your explanation in your own words before revealing. This is one of the homework transfer prompts, and it is worth getting both halves right.
The first reason is about the rate as a function of P. It is a product of two factors pulling in opposite directions: the population itself, which grows, and the room left, which shrinks. A product like that is largest when they balance, and for a downward parabola that is exactly halfway between the roots.
The second reason is about the curve against time. Its steepest point is where the slope stops increasing and starts decreasing, which is where the second derivative changes sign. The chain-rule formula for the second derivative shows that happens when P crosses half the capacity. Same point, two arguments.
Exit ticket
\[ K = 200, \quad r = 0.02, \quad P_0 = 50 \]
Discussion prompt
Exercise 173: a park's deer have carrying capacity 200, growth rate 2 percent per year and 50 deer at the start. Write the population at any time, and say when it grows fastest.
OpenStax Calculus Volume 2, §4.4 The Logistic Equation §4.4, p. 405 — Exercise 173
Try this on your own, then compare with the revealed answer. It is Exercise 173 from the book, and it uses the whole lesson in a few lines.
Theorem 4.2 with a start of 50 and a capacity of 200 has 150 in the denominator; dividing top and bottom by 50 gives 200 times the exponential over 3 plus the exponential. The fastest growth is at 100 deer, half the capacity, and the inflection formula gives 50 times the logarithm of 3, about 54.9 years.
A growth rate of 2 percent is slow, which is why the time is long. If you got a different form of the formula, check it at t equal to 0 and at large t: if it gives 50 and approaches 200, it is equivalent.
Recap
| idea | the mathematics | what it tells you |
|---|---|---|
| logistic equation | dP/dt = rP(1 − P/K) | exponential when small, braked near K |
| equilibria | P = 0 (unstable), P = K (stable) | every positive start ends at K |
| solution (Theorem 4.2) | P = P₀Ke^(rt) / ((K − P₀) + P₀e^(rt)) | numbers at any time; C₁ = P₀/(K − P₀) |
| inflection | P = K/2 at t = (1/r) ln((K − P₀)/P₀) | fastest growth, rate rK/4 |
| fitting | ln(P/(K − P)) = ln C₁ + rt | early data fix r, barely K |
\[ \frac{dP}{dt} = rP\left(1 - \frac PK\right) \quad\Longrightarrow\quad P(t) = \frac{K}{1 + \frac{K - P_0}{P_0}e^{-rt}} \]
Next, Section 4.5 meets first-order linear equations, which need a new method: the integrating factor.
Stewart, Calculus: Early Transcendentals 8e, §9.4 Models for Population Growth §9.4, pp. 610-619 — the same material in Stewart
The table collects the five ideas of the lesson. The equation adds one brake to exponential growth; its two equilibria and their stability tell you the long-run fate of every positive population; Theorem 4.2 gives exact numbers; the inflection at half the capacity is where growth is fastest; and fitting data pins down the growth rate far better than the capacity.
The displayed line is worth memorising in its second form, K over one plus a constant times e to the minus r t, because the limit K is visible in it at a glance.
Stewart covers the same material in Section 9.4. The next section, 4.5, turns to first-order linear equations, which are not separable in general and need a new tool.
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