Aimathic – Logistic models – Free Math Worksheets

Logistic models

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52847512

A fenced forest area initially contains \(120\) deer. During the first years, the population grows by \(12\%\) per year. a) Find the population after \(15\) years under an unrestricted exponential-growth model. b) After \(50\) years, the exponential model predicts more than \(34{,}000\) deer. Explain why that long-term prediction is unrealistic, and name two ecological factors that limit population growth. c) In a logistic-growth model with carrying capacity \(G\), describe what happens to the annual population increase as the population approaches \(G\).

Hints

- 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.

Solution

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 \(0\).

Answer

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 \(G\).

52847612

A new aquatic plant begins by covering \(5\,\text{m}^2\) of a lake whose total surface area is \(10{,}000\,\text{m}^2\). Researchers consider two models: Model 1: exponential growth, \(f(t)=5(1.5)^t\), where \(t\) is measured in weeks. Model 2: logistic growth that accounts for the lake's finite surface area. a) Use Model 1 to find the covered area after \(10\) weeks and after \(20\) weeks. b) Evaluate whether Model 1 is reasonable at \(t=20\), given the size of the lake. c) Explain why logistic growth is generally more appropriate than unrestricted exponential growth for biological spread.

Hints

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

Solution

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.

Answer

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.

53448312

During the first several hours after a video is uploaded, its total number of views is modeled by the logistic function \(f(t)=\frac{10}{1+9e^{-0.5t}}\). Here, \(t\) is measured in hours and \(f(t)\) is measured in thousands of views. a) Find the time \(t_m\) when the number of views is increasing most rapidly. b) Find the maximum growth rate in views per hour. c) When does the video reach \(80\%\) of its saturation level of \(10{,}000\) views? Verify your result using the graph.

Hints

- Where is a logistic curve steepest? - How is the growth rate related to the derivative? - Which value in the model gives the saturation level? - Use the target value on the vertical axis to estimate the corresponding time on the graph.

Solution

1. A logistic function grows most rapidly at its inflection point, where it reaches half its saturation value. Set \(f(t)=5\): \(\frac{10}{1+9e^{-0.5t}}=5\). 2. Solve: \(1+9e^{-0.5t}=2\), so \(e^{-0.5t}=\frac{1}{9}\). Therefore, \(t_m=2\ln(9)\approx 4.39\) hours. 3. Differentiate: \(f'(t)=\frac{45e^{-0.5t}}{(1+9e^{-0.5t})^2}\). At \(t=t_m\), \(e^{-0.5t_m}=\frac{1}{9}\), so \(f'(t_m)=\frac{5}{4}=1.25\) thousand views per hour, or \(1250\) views per hour. 4. To find when the video reaches \(80\%\) of saturation, set \(f(t)=8\): \(\frac{10}{1+9e^{-0.5t}}=8\). 5. Then \(1+9e^{-0.5t}=1.25\), so \(e^{-0.5t}=\frac{1}{36}\). Thus \(t=2\ln(36)\approx 7.17\) hours. On the graph, the curve reaches \(f(t)=8\) at about \(t=7.2\).

Answer

a) \(t_m=2\ln(9)\approx 4.39\) hours b) \(1250\) views per hour c) \(t=2\ln(36)\approx 7.17\) hours

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