Logistic Models - Calculus 2

1. The growth rate of a population, $\frac{dP}{dt}$, is plotted against the population size, $P$. The resulting graph for a logistic model is a parabola. If the parabola has its vertex at the point $(500, 20)$, which differential equation describes this model?
   Answer: $\frac{dP}{dt} = 0.08P(1-\frac{P}{1000})$ When you encounter a logistic model question involving a parabolic graph of growth rate versus population, you need to understand how the standard logistic equation creates this relationship and what the vertex tells you about the model's parameters. The logistic differential equation $\frac{dP}{dt} = rP(1-\frac{P}{K})$ can be rewritten as $\frac{dP}{dt} = rP - \frac{r}{K}P^2$. This is a quadratic function in P, which explains why plotting $\frac{dP}{dt}$ versus P gives you a parabola opening downward. The vertex at (500, 20) provides two crucial pieces of information. The x-coordinate (500) tells you where maximum growth occurs, which happens at $P = \frac{K}{2}$ for any logistic model. Therefore, $\frac{K}{2} = 500$, so $K = 1000$. The y-coordinate (20) gives you the maximum growth rate: $\frac{rK}{4} = 20$. Substituting K = 1000: $\frac{r(1000)}{4} = 20$, which gives you $r = 0.08$. This confirms that the correct equation is $\frac{dP}{dt} = 0.08P(1-\frac{P}{1000})$, which is option C. Option A has the wrong carrying capacity (K = 500 instead of 1000). Option B has the wrong growth rate (r = 0.04 instead of 0.08). Option D combines both errors, using both the wrong carrying capacity and wrong growth rate. Remember: in logistic models, maximum growth always occurs at half the carrying capacity, and you can find both parameters directly from the vertex coordinates of the parabolic growth rate graph.
2. A population $P(t)$ is modeled by the logistic differential equation $\frac{dP}{dt} = 0.05P(1 - \frac{P}{1000})$. At what population size is the population growing most rapidly, and what is this maximum rate?
   Answer: The population grows most rapidly at a size of 500, with a rate of 12.5 individuals per unit time. The logistic equation is $\frac{dP}{dt} = kP(1 - \frac{P}{L})$, where $L$ is the carrying capacity. Here, $L=1000$. The population growth rate is maximized when the population is at half the carrying capacity, so $P = L/2 = 1000/2 = 500$. To find the maximum rate, substitute $P=500$ into the differential equation: $\frac{dP}{dt} = 0.05(500)(1 - \frac{500}{1000}) = 25(1 - 0.5) = 25(0.5) = 12.5$.
3. The standard logistic model for a population $P$ is $\frac{dP}{dt} = kP(1-\frac{P}{L})$. Suppose a constant number of individuals, $H$, are harvested per unit time, leading to the model $\frac{dP}{dt} = kP(1-\frac{P}{L}) - H$. For a fish population with $k=0.8$, $L=1000$, and $H=150$, what are the equilibrium populations (where $\frac{dP}{dt}=0$)?
   Answer: $P=250$ and $P=750$ When you encounter a differential equation with harvesting, you're looking for equilibrium points where the population change equals zero. This means setting $\frac{dP}{dt} = 0$ and solving for the population values where growth balances harvesting. Setting the equation to zero: $0 = kP(1-\frac{P}{L}) - H$. Substituting the given values $k=0.8$, $L=1000$, and $H=150$: $0 = 0.8P(1-\frac{P}{1000}) - 150$ Expanding: $0 = 0.8P - 0.0008P^2 - 150$ Rearranging: $0.0008P^2 - 0.8P + 150 = 0$ Multiplying by 1250 to eliminate decimals: $P^2 - 1000P + 187,500 = 0$ Using the quadratic formula: $P = \frac{1000 \pm \sqrt{1,000,000 - 750,000}}{2} = \frac{1000 \pm \sqrt{250,000}}{2} = \frac{1000 \pm 500}{2}$ This gives $P = 750$ and $P = 250$. Choice A (300 and 700) represents calculation errors in the quadratic formula. Choice B (0 and 1000) gives the equilibrium points for the standard logistic model without harvesting - a common trap for students who forget about the harvesting term. Choice C (500 only) might result from incorrectly assuming the vertex of the parabola is the only equilibrium. Remember that harvesting modifications to population models always create quadratic equations with two equilibrium points. Always check your arithmetic carefully when applying the quadratic formula, and don't forget to include all terms when setting up your equilibrium equation.
4. Euler's method with a step size of $\Delta t = 1$ is used to approximate the solution to the logistic differential equation $\frac{dP}{dt} = 0.5P(1-\frac{P}{100})$. If the initial population is $P_0 = 20$, what is the approximate population $P_2$ after two steps?
   Answer: 38.08 When you encounter Euler's method problems, you're using a numerical technique to approximate solutions to differential equations by taking small linear steps. For the logistic equation $\frac{dP}{dt} = 0.5P(1-\frac{P}{100})$, you'll apply the formula $P_{n+1} = P_n + \Delta t \cdot f(P_n)$ where $f(P) = 0.5P(1-\frac{P}{100})$. Starting with $P_0 = 20$ and $\Delta t = 1$: Step 1: Calculate $f(20) = 0.5(20)(1-\frac{20}{100}) = 0.5(20)(0.8) = 8$ So $P_1 = 20 + 1(8) = 28$ Step 2: Calculate $f(28) = 0.5(28)(1-\frac{28}{100}) = 0.5(28)(0.72) = 10.08$ So $P_2 = 28 + 1(10.08) = 38.08$ Answer A (36.00) likely comes from computational errors in the second step or rounding mistakes. Answer C (28.00) represents stopping after just one step instead of completing both required iterations. Answer D (39.20) suggests an arithmetic error, possibly miscalculating the growth rate in the second step or confusing decimal operations. The correct answer is B (38.08). Study tip: In Euler's method problems, organize your work step-by-step and double-check each calculation. The logistic model's growth rate decreases as population approaches carrying capacity, so verify that your growth increments are getting smaller with each step. Always count your iterations carefully to ensure you've completed the requested number of steps.
5. The time $T$ required for a logistic population to grow from $10\%$ of its carrying capacity to $90\%$ of its carrying capacity is given by $T = \frac{1}{k} \ln(81)$. If the carrying capacity is $L=500$ and the growth constant is $k=0.1$, what is the maximum growth rate of this population?
   Answer: 12.5 When you encounter logistic growth problems, remember that the maximum growth rate occurs at the inflection point, which is exactly halfway to the carrying capacity—at $P = L/2$. The logistic growth equation is $\frac{dP}{dt} = kP(1 - \frac{P}{L})$, where the growth rate depends on both the current population $P$ and how close it is to the carrying capacity $L$. To find the maximum growth rate, you need to determine when $\frac{dP}{dt}$ is largest. Taking the derivative and setting it equal to zero shows that the maximum occurs when $P = L/2$. With $L = 500$, this happens when $P = 250$. Substituting into the growth equation: $\frac{dP}{dt} = 0.1 \times 250 \times (1 - \frac{250}{500}) = 0.1 \times 250 \times 0.5 = 12.5$. Choice A (25) represents what you'd get if you forgot the $(1 - P/L)$ factor and just calculated $kP = 0.1 \times 250$. Choice C (50) is the population at maximum growth rate, not the growth rate itself—a common unit confusion. Choice D ($10\ln(81)$) incorrectly tries to use the given time formula, but that describes the time interval, not the instantaneous growth rate. The correct answer is B (12.5). Study tip: For logistic growth, always remember that maximum growth rate occurs at $P = L/2$, and the formula becomes $\frac{dP}{dt} = \frac{kL}{4}$. This gives you a quick shortcut: $\frac{0.1 \times 500}{4} = 12.5$.
6. Consider two distinct populations, A and B, both governed by logistic growth. Population A is modeled by $\frac{dP_A}{dt} = 0.1P_A(1-\frac{P_A}{100})$ and Population B by $\frac{dP_B}{dt} = 0.2P_B(1-\frac{P_B}{100})$. If both populations start with the same initial value $P_0=10$, which statement accurately compares their behavior?
   Answer: Population B reaches its inflection point at an earlier time than Population A. When you encounter logistic growth problems, focus on identifying the key parameters and understanding how they affect population dynamics over time. Both populations follow the standard logistic model $\frac{dP}{dt} = rP(1-\frac{P}{K})$ where $r$ is the intrinsic growth rate and $K$ is the carrying capacity. For both populations, $K = 100$, but Population A has $r = 0.1$ while Population B has $r = 0.2$. The inflection point occurs when the population reaches half its carrying capacity ($P = K/2 = 50$). At this point, growth rate is maximized. Since Population B has twice the intrinsic growth rate of Population A, it will reach any given population level faster, including the inflection point at $P = 50$. This makes answer B correct. Let's examine why the other options are wrong. Answer A is backwards: the maximum growth rate occurs at the inflection point and equals $\frac{rK}{4}$. For Population A: $\frac{0.1 \times 100}{4} = 2.5$. For Population B: $\frac{0.2 \times 100}{4} = 5.0$. Population B has the higher maximum growth rate. Answer C is incorrect because the rate of approach to carrying capacity depends on the intrinsic growth rate $r$ – Population B approaches faster. Answer D is wrong because both populations have identical carrying capacities of 100. Remember: in logistic growth comparisons, a higher intrinsic growth rate means faster progression through all phases of growth, including reaching the inflection point earlier, while the carrying capacity depends only on the environmental limits, not the growth rate.
7. The population of a certain bacteria follows the logistic model $\frac{dP}{dt} = 0.04P(1 - \frac{P}{5000})$. The term $\frac{1}{P}\frac{dP}{dt}$ represents the relative (or per capita) growth rate. How does this relative growth rate change as the population $P$ increases from a small value towards the carrying capacity?
   Answer: It is a linearly decreasing function of $P$. The relative growth rate is given by $\frac{1}{P}\frac{dP}{dt}$. From the logistic equation, we can divide by $P$ to find this rate: $\frac{1}{P}\frac{dP}{dt} = 0.04(1 - \frac{P}{5000}) = 0.04 - \frac{0.04}{5000}P$. This expression is a linear function of $P$ with a negative slope ($-\frac{0.04}{5000}$). Therefore, as the population $P$ increases, the relative growth rate decreases linearly.
8. In a logistic growth model, the relative growth rate is defined as $\frac{1}{P}\frac{dP}{dt}$. For the equation $\frac{dP}{dt} = 0.02P\left(1 - \frac{P}{800}\right)$, at what population level is the relative growth rate equal to 75% of its maximum value?
   Answer: $P = 200$ The relative growth rate is $\frac{1}{P}\frac{dP}{dt} = \frac{1}{P} \cdot 0.02P\left(1 - \frac{P}{800}\right) = 0.02\left(1 - \frac{P}{800}\right)$. The maximum relative growth rate occurs when $P = 0$, giving a maximum of 0.02. We want 75% of this maximum: $0.02\left(1 - \frac{P}{800}\right) = 0.75 \times 0.02 = 0.015$. Solving: $1 - \frac{P}{800} = 0.75$, so $\frac{P}{800} = 0.25$, which gives $P = 200$. Choice A represents 50% of maximum relative growth rate. Choice B results from incorrect percentage calculation. Choice D corresponds to 62.5% of maximum relative growth rate.
9. A population follows the modified logistic model $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) - H$, where $H$ represents a constant harvesting rate. If $r = 0.3$, $K = 1000$, and $H = 50$, what are the equilibrium population levels?
   Answer: $P = 200$ (unstable) and $P = 800$ (stable) At equilibrium, $\frac{dP}{dt} = 0$, so $0.3P\left(1 - \frac{P}{1000}\right) - 50 = 0$. Expanding: $0.3P - 0.0003P^2 = 50$, which gives $0.0003P^2 - 0.3P + 50 = 0$. Multiplying by 10000/3: $P^2 - 1000P + 166667 = 0$. Using the quadratic formula: $P = \frac{1000 \pm \sqrt{1000000 - 666668}}{2} = \frac{1000 \pm \sqrt{333332}}{2} \approx \frac{1000 \pm 577}{2}$. This gives $P \approx 200$ and $P \approx 800$. For stability, check $f'(P) = 0.3 - 0.0006P$. At $P = 200$: $f'(200) = 0.18 > 0$ (unstable). At $P = 800$: $f'(800) = -0.18 < 0$ (stable).
10. A population $P(t)$ follows a logistic model $\frac{dP}{dt} = kP(1-\frac{P}{1000})$. The carrying capacity is 1000. At $t=0$, the population is 100. At $t=5$, the population is 250. Which expression represents the growth constant $k$?
    Answer: $\frac{1}{5}\ln(3)$ When you encounter a logistic growth problem, you're working with a differential equation that models population growth with environmental limits. The key is solving for the unknown parameter using the given data points. The logistic equation $\frac{dP}{dt} = kP(1-\frac{P}{1000})$ has the general solution $P(t) = \frac{1000}{1 + Ae^{-kt}}$, where $A$ is determined by initial conditions. With $P(0) = 100$, we get $100 = \frac{1000}{1 + A}$, so $A = 9$. This gives us $P(t) = \frac{1000}{1 + 9e^{-kt}}$. Using the second condition $P(5) = 250$: $250 = \frac{1000}{1 + 9e^{-5k}}$. Solving: $1 + 9e^{-5k} = 4$, so $9e^{-5k} = 3$, which means $e^{-5k} = \frac{1}{3}$. Taking the natural logarithm: $-5k = \ln(\frac{1}{3}) = -\ln(3)$. Therefore, $k = \frac{1}{5}\ln(3)$. Choice A gives $5\ln(3)$, which would result from incorrectly multiplying by 5 instead of dividing. Choice B represents $\frac{1}{5}\ln(9/7)$, which might come from algebraic errors when manipulating the fraction $\frac{1000}{250}$. Choice C gives $\frac{1}{5}\ln(7/3)$, possibly from confusing the relationship between the exponential terms. The correct answer is D: $\frac{1}{5}\ln(3)$. Remember: with logistic growth problems, always set up the general solution first, then use both given conditions systematically. Don't rush the algebra—small errors in fraction manipulation are common traps in these problems.
11. A logistic differential equation can be written as $\frac{dP}{dt} = rP(1-\frac{P}{L})$ or as $\frac{dP}{dt} = aP - bP^2$. If the two equations describe the same population dynamics, what is the carrying capacity $L$ in terms of the constants $a$ and $b$?
    Answer: $L = a/b$ When you encounter logistic differential equations in different forms, the key is recognizing they describe the same underlying population dynamics with a carrying capacity - the maximum sustainable population. To find the relationship between the two forms, expand the first equation: $\frac{dP}{dt} = rP(1-\frac{P}{L}) = rP - \frac{r}{L}P^2$. Now compare this with the second form $\frac{dP}{dt} = aP - bP^2$. Since these equations describe identical population dynamics, their coefficients must match: the coefficient of $P$ gives us $r = a$, and the coefficient of $P^2$ gives us $\frac{r}{L} = b$. Substituting $r = a$ into the second equation: $\frac{a}{L} = b$. Solving for $L$: $L = \frac{a}{b}$. You can verify this makes biological sense: when $P = L = \frac{a}{b}$, substituting into the second equation gives $\frac{dP}{dt} = a \cdot \frac{a}{b} - b \cdot (\frac{a}{b})^2 = \frac{a^2}{b} - \frac{a^2}{b} = 0$, confirming zero growth at carrying capacity. Looking at the wrong answers: (A) $L = ab$ has incorrect units and would make the carrying capacity unrealistically large. (B) $L = \frac{b}{a}$ inverts the correct relationship - this would actually represent the reciprocal of carrying capacity. (C) $L = a$ ignores the $b$ parameter entirely, which doesn't account for the population's self-limiting behavior. Remember: when converting between equivalent differential equation forms, match coefficients term by term. The logistic model's carrying capacity always appears in the denominator of the limiting term.
12. A rumor spreads through a school of 1200 students according to the logistic model $\frac{dR}{dt} = 0.002R(1200 - R)$, where $R(t)$ is the number of students who have heard the rumor at time $t$ days. If 20 students initially know the rumor, approximately how long will it take for 90% of the school to hear it?
    Answer: 3.4 days The logistic solution is $R(t) = \frac{1200}{1 + Ae^{-kt}}$ where $k = 0.002 \cdot 1200 = 2.4$. From $R(0) = 20$: $20 = \frac{1200}{1+A}$, so $A = 59$. We want $R(t) = 0.9 \times 1200 = 1080$. Setting up: $1080 = \frac{1200}{1 + 59e^{-2.4t}}$. Solving: $1 + 59e^{-2.4t} = \frac{1200}{1080} = \frac{10}{9}$, so $59e^{-2.4t} = \frac{1}{9}$, giving $e^{-2.4t} = \frac{1}{531}$. Therefore $t = \frac{\ln(531)}{2.4} \approx \frac{6.27}{2.4} \approx 3.4$ days. Choice A uses 95% instead of 90%. Choice B results from computational error in the exponential. Choice D confuses the calculation with 95% coverage.
13. The number of students at a university who have heard a rumor is modeled by $N(t) = \frac{5000}{1+499e^{-0.8t}}$. How many students had heard the rumor initially, and how many had heard it when the rumor was spreading the fastest?
    Answer: Initially 10 students; 2500 students when spreading fastest. This question involves a logistic growth model, which commonly appears when studying population dynamics or diffusion processes. When you see this type of function, you need to find two key points: the initial value and the inflection point where growth is fastest. To find the initial number of students, substitute $t = 0$ into $N(t) = \frac{5000}{1+499e^{-0.8t}}$: $N(0) = \frac{5000}{1+499e^{0}} = \frac{5000}{1+499(1)} = \frac{5000}{500} = 10$ For logistic models, the rumor spreads fastest at the inflection point, which occurs when $N(t)$ equals half the carrying capacity (the horizontal asymptote). As $t \to \infty$, $e^{-0.8t} \to 0$, so $N(t) \to \frac{5000}{1+0} = 5000$. Therefore, the rumor spreads fastest when $N(t) = \frac{5000}{2} = 2500$ students have heard it. Looking at the wrong answers: Choice A correctly identifies 10 initial students but incorrectly claims maximum spread occurs at 5000 students—this is actually the final equilibrium, not the fastest growth point. Choice B gets the inflection point right (2500) but wrongly calculates the initial value as 1 student. Choice C incorrectly uses 499 (the coefficient from the denominator) as the initial value and mistakes the carrying capacity for the inflection point. Study tip: For logistic models $\frac{L}{1+ae^{-kt}}$, remember that initial value comes from substituting $t=0$, the carrying capacity is $L$, and fastest growth occurs at $L/2$.
14. The rate of change of a population $y$ is modeled by the differential equation $\frac{dy}{dt} = 3y - 0.01y^2$. What is the carrying capacity of the population, and at what population size is the growth rate maximized?
    Answer: The carrying capacity is 300, and the growth rate is maximized when the population is 150. To find the carrying capacity, we need to write the equation in the standard form $\frac{dy}{dt} = ky(1 - \frac{y}{L})$ or $\frac{dy}{dt} = ky(L-y)$. Factoring the given equation: $\frac{dy}{dt} = y(3 - 0.01y) = 0.01y(300 - y)$. Comparing this to the form $\frac{dy}{dt} = k'y(L-y)$, we can identify the carrying capacity $L = 300$. The growth rate is maximized at half the carrying capacity, which is $y = L/2 = 300/2 = 150$.
15. The logistic equation $\frac{dP}{dt} = kP(1-\frac{P}{L})$ is a nonlinear differential equation. However, it can be transformed into a linear differential equation using the substitution $y = 1/P$. What is the resulting linear differential equation for $y$?
    Answer: $\frac{dy}{dt} + ky = \frac{k}{L}$ When you encounter a nonlinear differential equation like the logistic equation, substitution methods can often transform it into a more manageable linear form. The key is carefully applying the chain rule when making the substitution. Starting with $\frac{dP}{dt} = kP(1-\frac{P}{L})$ and the substitution $y = \frac{1}{P}$, we need to find $\frac{dy}{dt}$. Since $y = P^{-1}$, using the chain rule: $\frac{dy}{dt} = -P^{-2}\frac{dP}{dt} = -\frac{1}{P^2}\frac{dP}{dt}$ From our substitution, $P = \frac{1}{y}$, so $P^2 = \frac{1}{y^2}$. Therefore: $\frac{dy}{dt} = -y^2\frac{dP}{dt}$ Substituting the original logistic equation: $\frac{dy}{dt} = -y^2 \cdot kP(1-\frac{P}{L}) = -y^2 \cdot k \cdot \frac{1}{y}(1-\frac{1}{yL}) = -ky(1-\frac{1}{yL})$ Expanding: $\frac{dy}{dt} = -ky + \frac{k}{L}$ Rearranging: $\frac{dy}{dt} + ky = \frac{k}{L}$ This confirms answer C is correct. Answer A has the wrong sign on the $ky$ term and wrong constant term. Answer B has both terms with incorrect signs compared to our result. Answer D has the correct constant term but the wrong sign on the $ky$ term. Remember: when using substitutions in differential equations, always apply the chain rule carefully and track your signs. The goal is transforming a nonlinear equation into the standard linear form $\frac{dy}{dt} + p(t)y = q(t)$.
16. For the logistic differential equation $\frac{dP}{dt} = kP(L-P)$, the second derivative is $\frac{d^2P}{dt^2} = k(L-2P)\frac{dP}{dt}$. If a population is at $P = L/4$, which statement accurately describes the population's growth at that moment?
    Answer: The growth rate is positive and increasing. When analyzing logistic differential equations, you need to understand both the growth rate (first derivative) and how that rate is changing (second derivative). The logistic model describes population growth that starts slowly, accelerates, then slows as it approaches a carrying capacity. At $P = L/4$, let's examine what's happening. First, substitute into the original equation: $\frac{dP}{dt} = kP(L-P) = k \cdot \frac{L}{4} \cdot (L - \frac{L}{4}) = k \cdot \frac{L}{4} \cdot \frac{3L}{4} = \frac{3kL^2}{16}$. Since $k > 0$ and $L > 0$, this is positive, so the population is growing. Now check the second derivative: $\frac{d^2P}{dt^2} = k(L-2P)\frac{dP}{dt} = k(L - 2 \cdot \frac{L}{4})\frac{dP}{dt} = k(L - \frac{L}{2})\frac{dP}{dt} = k \cdot \frac{L}{2} \cdot \frac{dP}{dt}$. Since both terms are positive, the second derivative is positive, meaning the growth rate itself is increasing. Choice A is wrong because while the growth rate is positive, it's increasing, not decreasing. Choice C is incorrect because maximum growth occurs at $P = L/2$ (where the second derivative equals zero), not at $P = L/4$. Choice D is wrong because the growth rate is positive, not negative. The answer is B: the growth rate is positive and increasing. Study tip: In logistic growth, the inflection point at $P = L/2$ divides two phases: accelerating growth below $L/2$ and decelerating growth above $L/2$. Always check both derivatives to understand the complete behavior.
17. A population of bacteria follows the logistic model $\frac{dP}{dt} = 0.3P\left(1 - \frac{P}{800}\right)$, where $P(t)$ is the population at time $t$ hours. If the initial population is 50 bacteria, what is the population when the growth rate is maximum?
    Answer: 400 bacteria The growth rate $\frac{dP}{dt}$ is maximum when its derivative equals zero. Taking the derivative: $\frac{d^2P}{dt^2} = 0.3\left(1 - \frac{2P}{800}\right)\frac{dP}{dt} = 0$. Since $\frac{dP}{dt} \neq 0$ at the maximum growth point, we need $1 - \frac{2P}{800} = 0$, which gives $P = 400$. This occurs at half the carrying capacity. Choice A uses $\frac{K}{4}$ instead of $\frac{K}{2}$. Choice C uses $\frac{3K}{4}$. Choice D incorrectly uses the carrying capacity itself.
18. For the logistic model $\frac{dP}{dt} = kP(L-P)$ with $k>0$ and $L>0$, there are two equilibrium points, $P=0$ and $P=L$. Considering initial populations $P_0 > 0$, which statement correctly describes the stability of these equilibria?
    Answer: $P=0$ is an unstable equilibrium and $P=L$ is a stable equilibrium. When analyzing equilibrium stability in differential equations, you need to examine what happens to solutions when they start near each equilibrium point. For the logistic equation $\frac{dP}{dt} = kP(L-P)$, think about the sign of the derivative on either side of each equilibrium. At $P = 0$: If you start with a small positive population $P_0 > 0$, then $P > 0$ and $(L-P) > 0$ since $P < L$. This makes $\frac{dP}{dt} = kP(L-P) > 0$, so the population grows away from zero. Since solutions move away from $P = 0$, this equilibrium is unstable. At $P = L$: If you start slightly below $L$, then $P > 0$ and $(L-P) > 0$, so $\frac{dP}{dt} > 0$ and the population increases toward $L$. If you start slightly above $L$, then $P > 0$ but $(L-P) < 0$, making $\frac{dP}{dt} < 0$, so the population decreases toward $L$. Since solutions from both sides converge to $P = L$, this equilibrium is stable. Choice A incorrectly claims both equilibria are unstable. Choice B reverses the stability—it suggests populations would decay to zero, contradicting the growth away from $P = 0$. Choice C incorrectly claims both are stable, missing that $P = 0$ repels nearby solutions. Choice D correctly identifies the behavior: $P = 0$ is unstable (populations grow away) and $P = L$ is stable (the carrying capacity that populations approach). Study tip: For logistic models, remember that zero population is always unstable with positive growth rates, while the carrying capacity acts as a stable attractor.
19. To solve the logistic differential equation $\frac{dP}{dt} = kP(L-P)$ using separation of variables, one must integrate $\int \frac{dP}{P(L-P)}$. What is the correct partial fraction decomposition of the integrand $\frac{1}{P(L-P)}$?
    Answer: $\frac{1}{L}\left(\frac{1}{P} + \frac{1}{L-P}\right)$ We set up the partial fraction decomposition as $\frac{1}{P(L-P)} = \frac{A}{P} + \frac{B}{L-P}$. Multiplying by the common denominator $P(L-P)$ gives $1 = A(L-P) + BP$. To solve for $A$ and $B$, we can substitute convenient values for $P$. If $P=0$, we get $1 = A(L-0) \implies A = 1/L$. If $P=L$, we get $1 = B(L) \implies B = 1/L$. Substituting these back gives $\frac{1/L}{P} + \frac{1/L}{L-P} = \frac{1}{L}\left(\frac{1}{P} + \frac{1}{L-P}\right)$.
20. The solution to the logistic differential equation $\frac{dP}{dt} = kP(L-P)$ is of the form $P(t) = \frac{L}{1+Ae^{-Lkt}}$. For a fish population with carrying capacity $L=800$ and initial population $P(0)=100$, which equation must be solved to find the time $t$ when the population reaches 600?
    Answer: $600 = \frac{800}{1+7e^{-800kt}}$ First, we determine the constant $A$ using the initial condition $P(0)=100$ and carrying capacity $L=800$. The formula for $A$ is $A = \frac{L-P_0}{P_0}$. So, $A = \frac{800-100}{100} = \frac{700}{100} = 7$. This gives the specific solution $P(t) = \frac{800}{1+7e^{-800kt}}$. To find the time when the population reaches 600, we set $P(t) = 600$, which results in the equation $600 = \frac{800}{1+7e^{-800kt}}$.

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