Free math worksheets

- Substitute the stated population into the rate equation. - Check which factor becomes zero. - Recall what a zero rate means for an equilibrium initial value.

1. Substitute \(N=5000\): \(N'=r\cdot5000\cdot\left(1-\frac{5000}{5000}\right)=0\). 2. Thus, \(N=5000\) is an equilibrium. A solution with \(N(0)=5000\) remains \(N(t)=5000\) for all time.

The instantaneous rate is \(0\). A solution starting at \(N=5000\) remains at the equilibrium \(N(t)=5000\).

- Equilibrium levels make the derivative equal to zero. - Set each population-dependent factor equal to zero.

1. Equilibria occur where \(P'=0\). 2. The factors are zero at \(P=0\) and \(P=300\). 3. The positive equilibrium in the logistic factor is the carrying capacity, so \(K=300\).

Equilibria: \(P=0\) and \(P=300\) Carrying capacity: \(300\)

- Use the exponential-growth formula with a fixed percent rate. - Identify resources or conditions that cannot grow without limit. - Consider what happens when the habitat is nearly at capacity.

1. The exponential model is \(P(t)=120(1.12)^t\). Thus \(P(15)=120\cdot(1.12)^{15}\approx656.83\), or about \(657\) deer. 2. Exponential growth assumes unlimited resources. In reality, food, habitat, disease, predators, and competition limit the population. 3. In a logistic model, the annual increase becomes smaller as the population approaches carrying capacity and tends toward \(0\).

a) About \(657\) deer b) The model ignores limiting factors such as food, habitat, disease, predators, or competition. c) The annual increase decreases toward \(0\) as the population approaches \(G\).

- Substitute each time into the exponential model. - Compare the prediction with the lake's total area. - Consider what happens when little open water remains.

1. \(f(10)=5\cdot(1.5)^{10}\approx288.33\,\text{m}^2\). 2. \(f(20)=5\cdot(1.5)^{20}\approx16{,}626.28\,\text{m}^2\). 3. Since the lake has only \(10{,}000\,\text{m}^2\) of surface area, Model 1 is not physically reasonable at \(t=20\). 4. Logistic models account for carrying capacities and slowing growth caused by limited space, light, nutrients, and other resources.

a) After \(10\) weeks: about \(288.33\,\text{m}^2\); after \(20\) weeks: about \(16{,}626.28\,\text{m}^2\) b) Model 1 is not reasonable at \(t=20\) because it predicts an area larger than the lake. c) Logistic growth accounts for finite resources and a carrying capacity.

- Look for the horizontal level the curve approaches as time increases. - The fastest growth occurs where the graph is visually steepest. - After estimating the saturation level, find \(80\%\) of it and locate that height on the graph.

1. The curve levels off near \(f=10\), so the saturation level is about \(10\) thousand views, or \(10{,}000\) views. 2. The curve is steepest at its change in concavity, which occurs near \(t=4\) hours. 3. Eighty percent of the saturation level is \(8\) thousand views. The graph reaches \(f=8\) near \(t=8\) hours.

a) About \(10{,}000\) views b) About \(4\) hours; this is where the graph is steepest and changes concavity. c) About \(8\) hours; this is where the graph reaches \(8\) thousand views.

- Set \(M(5000-M)=0\) to identify both equilibrium membership levels. - Use \(M(0)=800\) to determine the initial signs of \(M\) and \(5000-M\). - The product \(M(5000-M)\) is largest when the two factors are equal.

1. The zeros of the rate are \(M=0\) and \(M=5000\); the positive carrying capacity is \(5000\). 2. Since \(0<800<5000\), both factors in the rate are positive, so the quantity is initially increasing. 3. The product \(M(5000-M)\) is largest halfway between the equilibria, at \(M=2500\).

a) Carrying capacity: \(5000\,\text{members}\); equilibria: \(M=0\) and \(M=5000\) b) Initially increasing c) Fastest at \(M=2500\,\text{members}\)

- Substitute \(P=2000\) and \(P=8000\) separately into the same logistic rate expression. - Compare each current amount with its remaining capacity \(10{,}000-P\). - Use the symmetry of complementary levels around \(K/2\) to explain the relationship between the two rates.

1. At \(P=2000\), \(P'=0.2\cdot2000\cdot\left(1-\frac{2000}{10{,}000}\right)=320\). 2. At \(P=8000\), \(P'=0.2\cdot8000\cdot\left(1-\frac{8000}{10{,}000}\right)=320\). 3. The rates are equal because \(2000\) and \(8000\) are equally far from the carrying-capacity midpoint, so the logistic rate function has the same value at both levels.

The instantaneous growth rate is \(320\) at both \(P=2000\) and \(P=8000\). The rates are equal because these population levels are symmetric about the midpoint \(P=5000\), so the products \(P(10{,}000-P)\) are equal.

- Evaluate the logistic expression at \(t=8\) using the unrounded exponential value. - Divide the resulting registration count by \(3000\) to obtain the percentage of capacity. - Subtract the count from \(3000\) to find the remaining capacity.

1. \(A(8)=\frac{3000}{1+5e^{-1.6}}\approx1492.92\). 2. Using the unrounded value, the percentage of capacity reached is \(\frac{A(8)}{3000}\cdot100\%\approx49.76\%\). 3. The remaining capacity is \(3000-A(8)\approx1507.08\).

\(1492.92\) registrations, \(49.76\%\) of carrying capacity, and \(1507.08\) registrations of remaining capacity

- Set the rate equal to zero to find the equilibrium values. - Identify the positive equilibrium. - Distinguish the coefficient controlling speed from the value controlling the upper limit.

1. Set the rate equal to zero: \(0.002P(900-P)=0\). The equilibria are \(P=0\) and \(P=900\). 2. The positive equilibrium \(900\) is the carrying capacity. The coefficient \(0.002\) is a proportionality constant that controls the time scale, not the limiting value.

The carrying capacity is \(900\), the positive value that makes the rate zero. The number \(0.002\) is a proportionality constant that controls the time scale.

- Substitute \(t=0\) into the stated logistic form. - Use the initial value to solve for \(A\). - Leave the positive rate constant \(r\) unspecified.

1. Use the initial value: \(3000=\frac{12{,}000}{1+A}\). 2. Therefore, \(1+A=4\), so \(A=3\). 3. The family is \(P(t)=\frac{12{,}000}{1+3e^{-rt}}\), where \(r>0\).

\(P(t)=\frac{12{,}000}{1+3e^{-rt}}\), where \(r>0\)

- Determine the sign of each factor in the rate equation. - Use the sign of the product to determine the direction of change. - Compare the current quantity with the carrying capacity.

1. At \(Q=12{,}000\), the factor \(Q\) is positive and the factor \(10{,}000-Q\) is negative. 2. Therefore, \(Q'<0\), so the model predicts that the quantity will decrease toward the carrying capacity \(10{,}000\).

\(Q'<0\). The model predicts a decrease toward the carrying capacity \(10{,}000\).

- Distinguish the initial direction of change from long-term behavior. - Identify what happens to the remaining-capacity factor as the quantity grows. - Use the positive equilibrium to describe the limiting behavior.

1. A positive initial derivative describes only the solution's local direction of change. 2. In a logistic model, the remaining-capacity factor decreases as the quantity approaches the carrying capacity. 3. The derivative approaches \(0\), so the solution levels off at the carrying capacity rather than growing without bound.

A positive initial derivative describes only local behavior. As the quantity approaches the carrying capacity, the remaining-capacity factor approaches \(0\), the growth rate decreases, and the solution levels off.

- In logistic growth, identify the population level where the growth rate is greatest. - Use half the carrying capacity for the inflection-point population. - Substitute that population and the measured growth rate into the logistic equation.

1. For a nonconstant logistic solution, the change from accelerating to slowing growth occurs at the inflection point, where \(P=\frac{800}{2}=400\) animals. 2. Substitute \(P=400\) and \(P'=48\): \(48=r(400)\left(1-\frac{400}{800}\right)=200r\). 3. Therefore, \(r=0.24\,\text{year}^{-1}\).

The population is \(400\) animals, and \(r=0.24\,\text{year}^{-1}\).

- Set the entire logistic rate equal to zero to find equilibria. - Determine the sign of each factor on the two population intervals. - A large population does not guarantee positive growth when the carrying-capacity factor is negative. - Connect the stable positive equilibrium with the environmental interpretation.

1. Equilibria occur when \(0.6P\left(1-\frac{P}{600}\right)=0\), so \(P=0\) and \(P=600\). 2. For \(0<P<600\), both factors are positive, so \(P'>0\) and the population increases. For \(P>600\), the factor \(1-\frac{P}{600}\) is negative, so \(P'<0\) and the population decreases. 3. Theo’s statement is false. Starting above \(600\) produces a negative growth rate, so the model moves the population downward toward \(600\). 4. The carrying capacity \(600\) is the long-term population level supported by the environment in this model.

a) \(P=0\) and \(P=600\) b) Increasing for \(0<P<600\); decreasing for \(P>600\) c) Above \(600\), the derivative is negative, so the population decreases toward \(600\). d) The modeled environmental limit is \(600\) animals.

- Substitute each listed population directly into the logistic rate equation. - The carrying capacity is the positive population where the growth rate becomes zero. - Compare populations that are equally far below and above half the carrying capacity.

1. At \(P=125\), \(P'=0.4\cdot125\left(1-\frac{125}{500}\right)=50\cdot\frac{3}{4}=37.5\), so A is \(37.5\). 2. At \(P=250\), \(P'=0.4\cdot250\left(1-\frac{250}{500}\right)=100\cdot\frac{1}{2}=50\), so B is \(50\). 3. At \(P=375\), \(P'=0.4\cdot375\left(1-\frac{375}{500}\right)=150\cdot\frac{1}{4}=37.5\), so C is \(37.5\). 4. The carrying capacity is \(500\) animals. The largest listed growth rate occurs at \(P=250\), half the carrying capacity. Populations \(125\) below and above \(250\) have equal growth rates.

a) A: \(37.5\), B: \(50\), C: \(37.5\) animals per year b) \(500\) animals c) The greatest listed growth rate occurs at \(P=250\). The rates are symmetric about half the carrying capacity.

- Divide the logistic equation by the current population to identify the relative rate. - Compare the stated fraction of \(r\) with the carrying-capacity factor. - Use the absolute growth measurement after finding \(K\). - Interpret the positive equilibrium in the fish-population context.

1. The relative growth rate is \(\frac{P'}{P}=r\left(1-\frac{P}{K}\right)\). At \(P=150\), \(r\left(1-\frac{150}{K}\right)=\frac{r}{2}\), so \(K=300\) fish. 2. Then \(45=150r\left(1-\frac{150}{300}\right)=75r\), so \(r=0.6\,\text{year}^{-1}\). 3. The carrying capacity \(K=300\) means the model predicts a long-term environmental limit of \(300\) fish.

\(K=300\) fish and \(r=0.6\,\text{year}^{-1}\). The carrying capacity is the model’s long-term environmental limit.

- Read the starting y-value and the horizontal level approached by the curve. - Locate where the graph changes from concave up to concave down. - Distinguish a positive growth rate from an increasing growth rate.

1. The graph begins at about \(100\) fish and levels off near \(500\) fish, so the initial population is about \(100\) and the carrying capacity is about \(500\). 2. Logistic growth is fastest at the inflection point, where the population is half the carrying capacity. The graph changes concavity near \(P=250\), at about \(t=17.3\) years. 3. Ana’s statement is false. The population increases throughout, but its growth rate increases only before the inflection point. After that point, the graph is concave down, so the growth rate decreases while remaining positive.

a) Initial population: about \(100\) fish; carrying capacity: about \(500\) fish b) About \(17.3\) years, at a population of about \(250\) fish c) The growth rate increases before the inflection point and decreases afterward, even though the population continues to rise.

- Recall where the logistic inflection point lies relative to \(K\). - Use the maximum-growth formula \(rK/4\). - Include population and time units in the parameters.

1. In a logistic model, the inflection point occurs at half the carrying capacity, not at the carrying capacity itself. Thus \(K/2=250\), so \(K=500\) fish. 2. The growth rate is maximal at the inflection point and equals \(rK/4\). 3. Therefore, \(50=r(500)/4=125r\), so \(r=0.40\) per year.

Morgan confused the inflection level with the carrying capacity. The model has \(K=500\) fish and \(r=0.40\) per year.

- Determine the sign of each factor in the differential equation at \(P=200\). - For logistic growth, compare the two factors \(P\) and \(1-P/K\) as the population moves from \(0\) to \(K\).

1. At \(P=200\), both \(P\) and \(1-\frac{P}{800}\) are positive, so \(P'>0\). The population is increasing. 2. Logistic growth is fastest at half the carrying capacity. 3. Half of \(800\) is \(400\), so the absolute growth rate is largest at \(P=400\).

a) Increasing b) \(P=400\)

- Compare the long-run shapes rather than just the common starting value. - A logistic curve has a finite upper limiting level in this setting. - Read that limiting level from the y-axis.

1. Curve \(q\) bends over and approaches a horizontal level, while curve \(p\) keeps steepening and leaves the top of the viewing window. 2. Therefore, \(q\) is the logistic curve. 3. The horizontal level approached by \(q\) is about \(500\), so the carrying capacity is approximately \(500\).

Curve \(q\) is logistic. Its carrying capacity is about \(500\). A supporting feature is that \(q\) levels off toward a horizontal limiting value, unlike \(p\).

- Evaluate the model at \(t=0\) and as \(t\to\infty\). - Consider how the denominator changes as \(t\) increases. - Isolate the exponential expression before taking a logarithm. - The growth rate is the first derivative of the height function.

1. \(h(0)=\frac{3}{6}=0.5\,\text{m}\). As \(t\to\infty\), \(e^{-0.2t}\to 0\), so \(h(t)\to 3\,\text{m}\). 2. The function \(e^{-0.2t}\) strictly decreases. Therefore, the positive denominator \(1+5e^{-0.2t}\) strictly decreases, while the numerator remains positive and constant. Thus, the quotient strictly increases. 3. Set \(h(t)=1.5\). Then \(1+5e^{-0.2t}=2\), so \(e^{0.2t}=5\). Therefore, \(t=5\ln 5\approx 8.05\) weeks. 4. Differentiate: \(h'(t)=\frac{3e^{-0.2t}}{(1+5e^{-0.2t})^2}\). Thus, \(h'(10)\approx 0.144\,\text{m/week}\).

a) Initial height: \(0.5\,\text{m}\); theoretical maximum: \(3\,\text{m}\) b) The positive denominator decreases as \(t\) increases, so the quotient increases. c) \(t=5\ln 5\approx 8.05\,\text{weeks}\) d) \(h'(10)\approx 0.144\,\text{m/week}\)

- At the stated occupancy, the two rate factors are the current index and the unused portion of the \(4000\)-unit capacity. - The product of those factors is largest when they are equal. - Compare the given occupancy with that balanced level after finding it.

1. Substitute \(P=1000\): \(P'=0.0005\cdot1000\cdot(4000-1000)=1500\). 2. The fastest-growth level is halfway between the equilibria: \(P=\frac{4000}{2}=2000\). 3. The maximum rate is \(0.0005\cdot2000\cdot(4000-2000)=2000\). 4. Since \(1000<2000\), the given occupancy level is below the fastest-growth level.

a) \(1500\,\text{index units per year}\) b) Maximum rate: \(2000\,\text{index units per year}\) at \(P=2000\) c) Below the fastest-growth level

- For the current enrollment rate, pair the \(6000\) active passes with the remaining capacity in the \(9000\)-pass system. - Maximize the logistic product by balancing active and remaining pass counts. - Determine whether the stated enrollment lies before or after that balance point.

1. Substitute \(A=6000\): \(A'=0.0001\cdot6000\cdot(9000-6000)=1800\). 2. The fastest-growth level is halfway between the equilibria: \(A=\frac{9000}{2}=4500\). 3. The maximum rate is \(0.0001\cdot4500\cdot(9000-4500)=2025\). 4. Since \(6000>4500\), the given enrollment is above the fastest-growth level.

a) \(1800\,\text{passes per week}\) b) Maximum rate: \(2025\,\text{passes per week}\) at \(A=4500\) c) Above the fastest-growth level

- Set the logistic formula equal to \(2000\) and then to \(6000\) to obtain the two corresponding times. - Use logarithms after isolating the exponential term in each equation. - Subtract the earlier time from the later time and report the positive elapsed duration.

1. Set \(P(t)=2000\): \(2000=\frac{8000}{1+7e^{-0.3t}}\). This gives \(e^{-0.3t}=\frac{3}{7}\), so \(t_1=\frac{\ln\left(\frac{7}{3}\right)}{0.3}\approx2.82\,\text{days}\). 2. Set \(P(t)=6000\): \(6000=\frac{8000}{1+7e^{-0.3t}}\). This gives \(e^{-0.3t}=\frac{1}{21}\), so \(t_2=\frac{\ln21}{0.3}\approx10.15\,\text{days}\). 3. The elapsed time is \(t_2-t_1=\frac{\ln9}{0.3}\approx7.32\,\text{days}\).

\(7.32\,\text{days}\)

- Use \(P(0)=750\) in the logistic form to determine the shape constant \(A\). - At half the carrying capacity, the denominator of the logistic fraction has a simple required value. - Use the condition at \(t=5\) to solve for \(r\), then substitute both constants back into the model.

1. Use \(P(0)=750\): \(750=\frac{6000}{1+A}\), so \(A=7\). 2. At half the carrying capacity, \(3000=\frac{6000}{1+7e^{-5r}}\), so \(7e^{-5r}=1\). 3. Thus \(r=\frac{\ln7}{5}\approx0.3892\,\text{week}^{-1}\). 4. The model is \(P(t)=\frac{6000}{1+7e^{-\left(\frac{\ln7}{5}\right)t}}\).

\(A=7\), \(r=\frac{\ln7}{5}\,\text{week}^{-1}\), and \(P(t)=\frac{6000}{1+7e^{-\left(\frac{\ln7}{5}\right)t}}\)

- Substitute \(N=1500\) into the differential equation to find the instantaneous rate. - The logistic growth rate as a function of \(N\) is maximized at half of \(6000\). - Compare \(1500\) with that midpoint to decide whether the rate is still increasing as membership grows.

1. Substitute \(N=1500\): \(N'=0.4\cdot1500\cdot\left(1-\frac{1500}{6000}\right)=450\). 2. The logistic growth rate is greatest at half the carrying capacity, \(N=3000\). 3. Since \(1500<3000\), the growth rate initially increases as \(N\) rises from \(1500\).

a) \(450\,\text{members per time unit}\) b) The growth rate is initially increasing because \(1500\) is below the midpoint \(3000\).

- Insert \(P=300\) and \(P'=84\) into the logistic rate equation. - Isolate the factor \(1-300/K\) before solving for \(K\). - Check that the resulting capacity is greater than \(300\) and reproduces the stated rate.

1. Substitute the observation: \(84=0.4\cdot300\cdot\left(1-\frac{300}{K}\right)\). 2. Divide by \(120\): \(0.7=1-\frac{300}{K}\). 3. Hence, \(\frac{300}{K}=0.3\), so \(K=1000\). 4. Substitution gives \(0.4\cdot300\cdot(1-0.3)=84\), confirming the result.

\(K=1000\)

- Use the fact that logistic growth is fastest at half the carrying capacity. - Substitute \(P=\frac{K}{2}\) into each rate equation. - Compare the resulting maximum rates by forming a ratio.

1. For \(P'=rP\left(1-\frac{P}{K}\right)\), the maximum growth rate occurs at \(P=\frac{K}{2}\). 2. Model A's maximum rate is \(0.3\cdot\frac{K}{2}\cdot\frac{1}{2}=\frac{0.3K}{4}\). 3. Model B's maximum rate is \(0.5\cdot\frac{K}{2}\cdot\frac{1}{2}=\frac{0.5K}{4}\). 4. Their ratio is \(\frac{0.5}{0.3}=\frac{5}{3}\), so Model B's maximum growth rate is \(\frac{5}{3}\) times Model A's.

Model A's maximum growth rate is \(\frac{0.3K}{4}\), and Model B's is \(\frac{0.5K}{4}\). Model B's maximum is \(\frac{5}{3}\) times Model A's.

- Substitute the observed population and rate into the model to find \(k\). - Logistic growth is fastest at half the carrying capacity. - Evaluate the rate at that midpoint.

1. Substitute the observation: \(600=k\cdot1000\cdot(4000-1000)\), so \(k=0.0002\,(\text{participant}\cdot\text{week})^{-1}\). 2. The fastest-growth level is half the carrying capacity: \(P=2000\). 3. The maximum rate is \(0.0002\cdot2000\cdot(4000-2000)=800\,\text{participants per week}\).

\(k=0.0002\,(\text{participant}\cdot\text{week})^{-1}\), and the maximum predicted rate is \(800\,\text{participants per week}\) at \(P=2000\).

- Differentiate the relationship between active users and students who are not yet active. - Rewrite each relative rate using the complementary number of students. - The two listed user counts exchange the roles of active and remaining users. - Interpret carrying capacity using the stated maximum market size.

1. Since \(R=1000-P\), \(R'=-P'=-0.5P\left(1-\frac{P}{1000}\right)\). 2. Using \(P=1000-R\), this becomes \(R'=-0.5R\left(1-\frac{R}{1000}\right)\). 3. The relative rates are \(\frac{P'}{P}=0.5\frac{R}{1000}\) and \(\left|\frac{R'}{R}\right|=0.5\frac{P}{1000}\). 4. At \(P=250\), the rates are \(\frac{3}{8}\,\text{month}^{-1}\) and \(\frac{1}{8}\,\text{month}^{-1}\), respectively. At \(P=750\), they are \(\frac{1}{8}\,\text{month}^{-1}\) and \(\frac{3}{8}\,\text{month}^{-1}\), respectively. 5. The carrying capacity \(1000\) is the model’s upper limit for the number of active users, corresponding to the full potential student market.

a) \(R'=-0.5R\left(1-\frac{R}{1000}\right)\) b) At \(P=250\): \(P'/P=\frac{3}{8}\,\text{month}^{-1}\), \(|R'/R|=\frac{1}{8}\,\text{month}^{-1}\). At \(P=750\): \(P'/P=\frac{1}{8}\,\text{month}^{-1}\), \(|R'/R|=\frac{3}{8}\,\text{month}^{-1}\). c) The carrying capacity is the maximum modeled market of \(1000\) active users.

- Treat the measured population increase as a quadratic condition on \(P\). - Compare each solution with half the carrying capacity. - Concavity indicates whether the growth rate itself is increasing or decreasing. - Think about the symmetry of the logistic rate as a function of population.

1. Solve \(0.3P\left(1-\frac{P}{1000}\right)=72\). This simplifies to \(P^2-1000P+240{,}000=0\). 2. Factoring gives \((P-400)(P-600)=0\), so \(P=400\) or \(P=600\) birds. 3. Logistic solutions are concave up below \(K/2=500\) and concave down above \(500\). Therefore, growth is accelerating at \(P=400\) and slowing at \(P=600\). 4. The logistic growth-rate expression is a downward-opening quadratic in \(P\) that is symmetric about \(P=K/2\). Populations equally far below and above \(500\) therefore have the same absolute growth rate.

a) \(P=400\) birds or \(P=600\) birds b) Growth is accelerating at \(400\) birds and slowing at \(600\) birds. c) The two populations are equally far from half the carrying capacity, where the logistic growth-rate quadratic is symmetric.

- Substitute each population fraction directly into the logistic rate. - Differentiate the rate equation with respect to time. - Factor out the first derivative before comparing the two cases.

1. At either \(P=\frac{K}{3}\) or \(P=\frac{2K}{3}\), \(P'=rK\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)=\frac{2rK}{9}\). 2. Differentiating the logistic equation gives \(P''=rP'\left(1-\frac{2P}{K}\right)\). 3. At \(P=\frac{K}{3}\), \(\frac{P''}{P'}=\frac{r}{3}\). At \(P=\frac{2K}{3}\), \(\frac{P''}{P'}=-\frac{r}{3}\).

At both population levels, \(P'=\frac{2rK}{9}\). The ratios are \(\frac{P''}{P'}=\frac{r}{3}\) at \(P=\frac{K}{3}\) and \(\frac{P''}{P'}=-\frac{r}{3}\) at \(P=\frac{2K}{3}\).

- View the logistic growth rate as a quadratic function of population. - Identify the axis of symmetry from the paired population levels. - Use the maximum value of the rate quadratic to find \(r\). - Interpret the positive equilibrium using the wildlife context.

1. The logistic growth-rate quadratic is symmetric about \(P=\frac{K}{2}\). 2. Since equal-rate populations are symmetric about \(300\), \(\frac{K}{2}=300\), so \(K=600\) animals. 3. The maximum rate is \(\frac{rK}{4}\), so \(90=\frac{600r}{4}=150r\). Therefore, \(r=0.6\,\text{year}^{-1}\). 4. The carrying capacity means the model predicts a long-term environmental limit of \(600\) animals.

\(K=600\) animals and \(r=0.6\,\text{year}^{-1}\). The carrying capacity is the model’s long-term environmental limit.

- Evaluate both differential equations at the common initial value. - Differentiate each rate equation to obtain a second derivative. - Interpret equal first derivatives and unequal second derivatives geometrically.

1. The logistic rate is \(P'(0)=0.5\cdot100\cdot0.9=45\), while \(E'(0)=0.45\cdot100=45\). 2. For the logistic model, \(P''=0.5P'\left(1-\frac{2P}{1000}\right)\), so \(P''(0)=0.5\cdot45\cdot0.8=18\). 3. For the exponential model, \(E''=0.45E'=0.45^2E\), so \(E''(0)=20.25\). 4. The models have the same initial value and tangent slope, but the exponential model has greater initial upward curvature.

\(P'(0)=E'(0)=45\), \(P''(0)=18\), and \(E''(0)=20.25\). The models share an initial tangent, but the exponential model bends upward more strongly.

- Differentiate the logistic rate with respect to time. - Factor out \(P'\) before using the measured ratio. - Substitute the population into the original equation after finding \(r\).

1. Differentiating the logistic equation gives \(\frac{P''}{P'}=r\left(1-\frac{2P}{1000}\right)\) when \(P'\ne0\). 2. At \(P=250\), \(0.10=r(1-0.5)=\frac{r}{2}\), so \(r=0.20\,\text{year}^{-1}\). 3. Then \(P'=0.20\cdot250\cdot0.75=37.5\) animals per year.

\(r=0.20\,\text{year}^{-1}\), and \(P'=37.5\) animals per year.

- Relate the non-equilibrium inflection level of a logistic model to its carrying capacity. - Use \(\frac{P'}{P}\) to express the relative growth rate. - Consider where the logistic growth-rate expression is largest between the two equilibria. - Interpret the positive equilibrium in the population context.

1. For logistic growth, the non-equilibrium inflection level is \(P=\frac{K}{2}\). Since that level is \(400\), \(K=800\) birds. 2. The relative growth rate is \(\frac{P'}{P}=r\left(1-\frac{P}{K}\right)\). At \(P=100\), \(0.24=r\left(1-\frac{100}{800}\right)=\frac{7}{8}r\), so \(r=\frac{48}{175}\,\text{year}^{-1}\). 3. On \(0\le P\le K\), the positive growth rate \(rP\left(1-\frac{P}{K}\right)\) is largest at \(P=\frac{K}{2}\). Thus the maximum is \(\frac{rK}{4}=\frac{48}{175}\cdot200=\frac{384}{7}\) birds per year. 4. The carrying capacity \(K=800\) birds is the population level toward which the model tends when the initial population is positive and below \(K\).

\(K=800\) birds, \(r=\frac{48}{175}\,\text{year}^{-1}\approx0.2743\,\text{year}^{-1}\), and the maximum positive growth rate on \(0\le P\le K\) is \(\frac{384}{7}\approx54.86\) birds per year. The carrying capacity is \(800\) birds.

- View relative growth rate as a linear function of population. - Use the two measurements to find the slope of that line. - Evaluate the logistic growth rate at half the carrying capacity. - Interpret the positive equilibrium in the deer-population context.

1. The relative rate is \(r\left(1-\frac{P}{K}\right)=r-\frac{r}{K}P\), a linear function of \(P\). 2. The relative rate decreases by \(0.06\) per year as the population increases by \(200\) deer, so \(\frac{r}{K}=0.0003\) per deer-year. 3. Using \(P=200\), \(0.18=r-0.0003\cdot200=r-0.06\), so \(r=0.24\) per year. 4. Then \(K=\frac{0.24}{0.0003}=800\) deer. 5. The absolute growth rate is largest at \(P=K/2=400\), where \(P'=\frac{rK}{4}=48\) deer per year. The carrying capacity \(K=800\) is the population level the environment is modeled to support in the long run.

\(r=0.24\) per year, \(K=800\) deer, and the maximum absolute growth rate is \(48\) deer per year. The carrying capacity is the model's long-term environmental limit.

- Replace population by its odds relative to the remaining capacity. - Compute the multiplicative change in odds over each population interval. - Equal odds factors correspond to equal elapsed times.

1. In a logistic model, the odds \(\frac{P}{1000-P}\) grow exponentially at rate \(r\). 2. From \(P=100\) to \(P=200\), the odds change from \(\frac19\) to \(\frac14\), a factor of \(\frac94\). 3. From \(P=800\) to \(P=900\), the odds change from \(4\) to \(9\), also a factor of \(\frac94\). 4. Therefore, both elapsed times satisfy \(e^{r\Delta t}=\frac94\), so \(\Delta t=\frac{1}{r}\ln\frac94\).

Both intervals last \(\frac{1}{r}\ln\frac94\) time units.

- Use the absolute-rate measurement to determine \(r\). - Convert each relative rate into a fish-population level. - Compare logistic odds at those two population levels.

1. At \(P=100\), \(16=100r\left(1-\frac{100}{500}\right)=80r\), so \(r=0.20\) per year. 2. The relative rate is \(0.20\left(1-\frac{P}{500}\right)\). 3. A relative rate of \(0.16\) per year corresponds to \(P=100\), and a relative rate of \(0.04\) per year corresponds to \(P=400\). 4. The logistic odds change from \(\frac{100}{400}=\frac14\) to \(\frac{400}{100}=4\), a factor of \(16\). 5. Thus \(e^{0.20\Delta t}=16\), so \(\Delta t=\frac{\ln16}{0.20}=20\ln2\) years.

\(20\ln2\approx13.86\) years.

- Use the rate at half the carrying capacity to find \(r\). - Set the logistic rate equal to \(28\) turtles per year. - Expect two population levels symmetric about half the carrying capacity.

1. At \(P=350\), \(35=350r\left(1-\frac12\right)=175r\), so \(r=0.20\) per year. 2. Set \(0.20P\left(1-\frac{P}{700}\right)=28\). 3. Simplifying gives \(P^2-700P+98{,}000=0\). 4. The quadratic formula gives \(P=\frac{700\pm\sqrt{98{,}000}}{2}=350\pm70\sqrt5\).

\(P=350-70\sqrt5\approx193.48\) turtles and \(P=350+70\sqrt5\approx506.52\) turtles.

- Divide the logistic equation by the population to use the relative rate. - Differentiate the logistic rate and factor out \(P'\). - Compare population odds at the current and inflection levels.

1. At \(P=300\), \(0.12=r\left(1-\frac13\right)=\frac23r\), so \(r=0.18\) per year. 2. The absolute rate is \(P'=0.18\cdot300\cdot\frac23=36\) birds per year. 3. Logistic acceleration is \(P''=rP'\left(1-\frac{2P}{900}\right)\), so \(P''=0.18\cdot36\cdot\frac13=2.16\) birds per year squared. 4. The odds increase from \(\frac{300}{600}=\frac12\) to \(\frac{450}{450}=1\), a factor of \(2\). 5. Thus \(e^{0.18\Delta t}=2\), so \(\Delta t=\frac{50\ln2}{9}\) years.

\(r=0.18\) per year, \(P''=2.16\) birds per year squared, and \(\Delta t=\frac{50\ln2}{9}\approx3.85\) years.

- Convert each stated fraction of capacity into population odds. - Use exponential change of odds to find the full elapsed time. - The multiplier over half the time is the square root of the full multiplier.

1. Logistic odds \(\frac{P}{K-P}\) grow by the factor \(e^{rt}\). 2. The odds are \(\frac14\) at \(t=0\) and \(4\) at \(t=T\), so \(e^{0.30T}=16\). 3. Therefore, \(T=\frac{\ln16}{0.30}=\frac{40\ln2}{3}\) years. 4. At time \(T/2\), the odds multiplier is \(e^{0.30T/2}=\sqrt{16}=4\). 5. Starting from odds \(\frac14\), the midpoint odds are \(1\), which means \(P(T/2)=K/2\).

\(T=\frac{40\ln2}{3}\approx9.24\) years, and \(P(T/2)=K/2\).

- Use the relative rate at one-quarter of carrying capacity to find \(r\). - Complementary population levels have the same absolute logistic growth rate. - Use the sign of \(1-2P/K\) to interpret acceleration.

1. At \(P=200\), \(0.15=r\left(1-\frac14\right)=\frac34r\), so \(r=0.20\) per year. 2. At \(P=200\), \(P'=0.20\cdot200\cdot0.75=30\) bison per year, and \(P''=0.20\cdot30\cdot0.5=3\) bison per year squared. Growth is speeding up. 3. At \(P=600\), \(P'=0.20\cdot600\cdot0.25=30\) bison per year, and \(P''=0.20\cdot30\cdot(-0.5)=-3\) bison per year squared. Growth is slowing down.

\(r=0.20\) per year. At \(P=200\): \(P'=30\) bison per year and \(P''=3\) bison per year squared, so growth is accelerating. At \(P=600\): \(P'=30\) bison per year and \(P''=-3\) bison per year squared, so growth is decelerating.

- Identify the population level of maximum logistic growth. - Compare population odds at the initial and maximum-growth times. - Apply the same odds multiplier over the next equal time interval.

1. Maximum logistic growth occurs at half the carrying capacity, so \(P(4)=500\) birds. 2. The population odds are \(\frac{200}{800}=\frac{1}{4}\) at \(t=0\) and \(1\) at \(t=4\). 3. Thus \(e^{4r}=4\), giving \(r=\frac{\ln 2}{2}\) per year. 4. From \(t=4\) to \(t=8\), the odds are multiplied by \(4\) again, so the odds at \(t=8\) are \(4\). 5. Solving \(\frac{P(8)}{1000-P(8)}=4\) gives \(P(8)=800\) birds.

\(r=\frac{\ln 2}{2}\) per year, and \(P(8)=800\) birds.

- Use the growth-rate measurement to find the rate constant. - Convert the two population levels to logistic odds. - Relate the odds factor to elapsed time.

1. At \(P=200\), \(24=200r\left(1-\frac{1}{3}\right)=\frac{400r}{3}\), so \(r=0.18\) per year. 2. At \(P=200\), the odds are \(\frac{200}{400}=\frac{1}{2}\). At \(P=400\), the odds are \(2\). 3. The odds increase by a factor of \(4\), so \(e^{0.18\Delta t}=4\). 4. Therefore, \(\Delta t=\frac{\ln 4}{0.18}=\frac{100\ln 2}{9}\) years.

\(r=0.18\) per year, and \(\Delta t=\frac{100\ln 2}{9}\approx7.70\) years.

- Use the maximum logistic rate to find \(r\). - Convert the stated relative rate into a population level. - Use the sign of \(1-2P/K\) to interpret the acceleration.

1. The maximum rate is \(rK/4\), so \(45=900r/4\) and \(r=0.20\) per day. 2. The relative rate condition is \(0.20\left(1-\frac{P}{900}\right)=0.05\), giving \(P=675\) colonies. 3. Then \(P'=0.20\cdot675\cdot0.25=33.75\) colonies per day. 4. Also, \(P''=rP'\left(1-\frac{2P}{900}\right)=0.20\cdot33.75\cdot(-0.5)=-3.375\) colonies per day squared. The negative acceleration means the population is still growing, but its growth rate is decreasing.

\(P=675\) colonies, \(P'=33.75\) colonies per day, and \(P''=-3.375\) colonies per day squared. The growth rate is decreasing.

- Write the relative rate by dividing the logistic equation by the population. - Use the first measurement to determine the intrinsic rate. - Maximum absolute logistic growth occurs at half the carrying capacity.

1. The relative rate is \(\frac{P'}{P}=r\left(1-\frac{P}{900}\right)\). 2. At \(P=300\), \(0.12=r\left(1-\frac{1}{3}\right)=\frac{2r}{3}\), so \(r=0.18\) per year. 3. Setting \(0.06=0.18\left(1-\frac{P}{900}\right)\) gives \(P=600\) seals. 4. Maximum absolute growth occurs at \(P=450\) and equals \(\frac{rK}{4}=\frac{0.18\cdot900}{4}=40.5\) seals per year.

\(r=0.18\) per year; the relative growth rate is \(0.06\) per year at \(P=600\) seals; the maximum absolute growth rate is \(40.5\) seals per year.

- Use the relative rate at one-quarter of capacity to find \(r\). - Evaluate the logistic rate at half capacity for the maximum. - Compare population odds at the two stated levels.

1. At \(P=300\), \(0.15=r\left(1-\frac{1}{4}\right)=\frac{3}{4}r\), so \(r=0.20\) per year. 2. The maximum absolute growth rate is \(\frac{rK}{4}=\frac{0.20\cdot1200}{4}=60\) bears per year. 3. The odds are \(\frac{1}{3}\) at \(P=300\) and \(3\) at \(P=900\), a factor of \(9\). 4. Thus \(e^{0.20\Delta t}=9\), so \(\Delta t=5\ln 9=10\ln 3\) years.

\(r=0.20\) per year, the maximum growth rate is \(60\) bears per year, and \(\Delta t=10\ln 3\approx10.99\) years.

- For a logistic solution, compare the odds ratio \(P/(K-P)\) at \(P=3000\) and \(P=9000\). - The ratio changes exponentially with rate parameter \(0.25\), so the unknown starting time cancels in an elapsed-time comparison. - Use a logarithm of the ratio change to obtain the additional time.

1. A separated logistic solution satisfies \(\ln\left(\frac{P}{12{,}000-P}\right)=0.25t+C\). 2. The odds ratio changes from \(\frac{3000}{9000}=\frac{1}{3}\) to \(\frac{9000}{3000}=3\). 3. Therefore, \(0.25\Delta t=\ln3-\ln\left(\frac{1}{3}\right)=\ln9\). 4. \(\Delta t=\frac{\ln9}{0.25}\approx8.79\,\text{time units}\).

\(\Delta t\approx8.79\,\text{time units}\)

- Write one equation from the observed rate at \(P=200\) and another from the same rate at \(P=800\). - Equate the expressions and solve for the positive capacity \(K\) before finding \(k\). - Substitute the resulting constants into both original rate observations as a check.

1. Write the two rate equations: \(480=k\cdot200(K-200)\) and \(480=k\cdot800(K-800)\). 2. Since the rates are equal and \(k>0\), \(200(K-200)=800(K-800)\). Solving gives \(K=1000\). 3. Substitute \(K=1000\): \(480=k\cdot200\cdot800\), so \(k=0.003\). 4. Both substitutions give the stated rate \(480\), confirming the constants.

\(K=1000\) and \(k=0.003\)

- Use the common initial value to determine the same logistic shape constant for both plans. - Set each plan’s logistic formula equal to \(3000\) and solve the two time equations with their different rate parameters. - Subtract the smaller time from the larger to find how much sooner the faster plan reaches the target.

1. Since \(P(0)=400\), both plans have the form \(P(t)=\frac{4000}{1+9e^{-rt}}\). 2. Set \(P=3000\): \(1+9e^{-rt}=\frac{4}{3}\), so \(e^{-rt}=\frac{1}{27}\) and \(t=\frac{\ln27}{r}\). 3. Plan A reaches the level at \(t_A=\frac{\ln27}{0.15}\approx21.97\,\text{time units}\). 4. Plan B reaches it at \(t_B=\frac{\ln27}{0.22}\approx14.98\,\text{time units}\). 5. Plan B reaches the level \(t_A-t_B\approx6.99\,\text{time units}\) sooner.

Plan A: \(21.97\,\text{time units}\) Plan B: \(14.98\,\text{time units}\) Difference: Plan B reaches the level \(6.99\,\text{time units}\) sooner.

- Differentiate the reciprocal expression carefully. - Substitute the given rate equation before simplifying. - Use the initial population to find the transformed initial value.

1. Differentiate \(Z=1200P^{-1}-1\): \(Z'=-\frac{1200P'}{P^2}\). 2. Substitute the logistic equation: \(Z'=-0.4\left(\frac{1200}{P}-1\right)=-0.4Z\). 3. Since \(Z(0)=\frac{1200}{300}-1=3\), the exponential solution is \(Z(t)=3e^{-0.4t}\). 4. Solving \(\frac{1200}{P}-1=3e^{-0.4t}\) for \(P\) gives \(P(t)=\frac{1200}{1+3e^{-0.4t}}\).

a) \(Z'=-0.4Z\) b) \(P(t)=\frac{1200}{1+3e^{-0.4t}}\)

- For logistic growth, connect the maximum absolute growth rate with the population \(K/2\). - Use the relative growth rate \(P'/P\) at \(P=300\) to obtain a second equation. - The statement that \(300\) is below the fastest-growth level selects between the two algebraic values of \(K\). - Interpret carrying capacity as an environmental limit in the model, not as a guaranteed exact count.

1. A logistic model grows fastest at \(P=\frac{K}{2}\), where \(P'=\frac{rK}{4}\). Thus \(\frac{rK}{4}=180\), so \(r=\frac{720}{K}\). 2. At \(P=300\), the relative-rate condition gives \(r\left(1-\frac{300}{K}\right)=0.45\). 3. Substituting \(r=\frac{720}{K}\) gives \(K^2-1600K+480{,}000=0\), so \(K=400\) or \(K=1200\). 4. Because \(300<\frac{K}{2}\), the carrying capacity must exceed \(600\). Therefore, \(K=1200\) fish and \(r=0.6\,\text{year}^{-1}\). 5. The fastest growth occurs at \(P=\frac{K}{2}=600\) fish. The carrying capacity means the model predicts that environmental limits stabilize the population near \(1200\) fish over time.

a) \(K=1200\) fish and \(r=0.6\,\text{year}^{-1}\) b) \(600\) fish c) The model’s long-term limiting population is \(1200\) fish.

- First determine when the quantity reaches half the carrying capacity. - Substitute times equally spaced before and after that moment. - Rewrite the two exponential factors as reciprocals before adding.

1. Setting \(P(t_0)=450\) gives \(1+8e^{-0.3t_0}=2\), so \(8e^{-0.3t_0}=1\) and \(t_0=\frac{\ln 8}{0.3}=\frac{10}{3}\ln 8\). 2. Since \(8e^{-0.3t_0}=1\), \(P(t_0+s)=\frac{900}{1+e^{-0.3s}}\) and \(P(t_0-s)=\frac{900}{1+e^{0.3s}}\). 3. Let \(u=e^{0.3s}\). Then the sum is \(\frac{900u}{u+1}+\frac{900}{u+1}=900\).

a) \(t_0=\frac{10}{3}\ln 8\) b) \(P(t_0-s)+P(t_0+s)=900\) for every admissible \(s\).

- Express the growth rate using the fraction of carrying capacity. - Solve the resulting quadratic and choose the value below one-half. - Compare population odds at the two levels to obtain the elapsed time.

1. The maximum rate is \(\frac{rK}{4}\). Let \(u=P/K\). Half the maximum gives \(u(1-u)=\frac{1}{8}\). 2. Solving gives \(u=\frac{1}{2}\pm\frac{\sqrt{2}}{4}\). The lower branch uses \(u_1=\frac{1}{2}-\frac{\sqrt{2}}{4}\). 3. For a logistic solution, the odds \(\frac{P}{K-P}\) grow by the factor \(e^{r\Delta t}\). At \(P_1\), the odds are \(3-2\sqrt{2}\); at \(K/2\), they are \(1\). 4. Thus \(e^{r\Delta t}=\frac{1}{3-2\sqrt{2}}=(1+\sqrt{2})^2\), so \(\Delta t=\frac{2\ln(1+\sqrt{2})}{r}\).

a) \(P_1=K\left(\frac{1}{2}-\frac{\sqrt{2}}{4}\right)\) b) \(\Delta t=\frac{2\ln(1+\sqrt{2})}{r}\)

- Substitute the stated fraction of carrying capacity into the logistic equation. - Differentiate the relative growth rate with respect to time. - Track units when using the product and quotient of the two parameters.

1. At \(P=\frac{K}{3}\), \(40=r\left(\frac{K}{3}\right)\left(\frac23\right)=\frac{2rK}{9}\), so \(rK=180\) fish per year. 2. The relative growth rate is \(g=r\left(1-\frac{P}{K}\right)\), so \(g'=-\frac{r}{K}P'\). 3. Using \(g'=-0.02\) per year squared and \(P'=40\) fish per year gives \(\frac{r}{K}=0.0005\) per fish-year. 4. Multiplying \(rK=180\) by \(\frac{r}{K}=0.0005\) gives \(r^2=0.09\), so the positive rate constant is \(r=0.30\) per year. 5. Then \(K=180/0.30=600\) fish.

\(r=0.30\) per year and \(K=600\) fish.