Problem 43 - Calculus and its applications

Question

The tortoise population, \(P(t),\) in a square mile of the Mojave Desert after \(t\) years can be approximated by the logistic equation $$ P(t)=\frac{3000}{20+130 e^{-0.214 t}} $$ a) Find the tortoise population after \(0 \mathrm{yr}, 5 \mathrm{yr}, 15 \mathrm{yr},\) and 25 yr. b) Find the rate of change in the population, \(P^{\prime}(t)\). c) Find the rate of change in the population after \(0 \mathrm{yr}\) \(5 \mathrm{yr}, 15, \mathrm{yr},\) and \(25 \mathrm{yr}\) d) What is the limiting value (see Exercise 42 ) for the population of tortoises in a square mile of the Mojave Desert?

Step-by-Step Solution

Answer

a) Populations: 20, 46, 99, 134. b) Find derivative. c) Rate of change: -3.71, -6.74, -7.23, -3.00. d) Limiting value: 150 tortoises.

Step 1: Calculate Population at t=0, 5, 15, 25

To find the tortoise population at different times, substitute the values of \( t \) into the logistic equation \( P(t) = \frac{3000}{20 + 130 e^{-0.214 t}} \):1. For \( t = 0 \): \[ P(0)=\frac{3000}{20+130e^{-0.214 \times 0}} = \frac{3000}{20+130 \times 1} = \frac{3000}{150} = 20 \]2. For \( t = 5 \): \[ P(5) = \frac{3000}{20 + 130 e^{-0.214 \times 5}} \approx \frac{3000}{20 + 130 \times 0.348} \approx \frac{3000}{65.24} \approx 45.98 \]3. For \( t = 15 \): \[ P(15) = \frac{3000}{20 + 130 e^{-0.214 \times 15}} \approx \frac{3000}{20 + 130 \times 0.080} \approx \frac{3000}{30.40} \approx 98.68 \]4. For \( t = 25 \): \[ P(25) = \frac{3000}{20 + 130 e^{-0.214 \times 25}} \approx \frac{3000}{20 + 130 \times 0.018} \approx \frac{3000}{22.34} \approx 134.27 \]The population values at \( t = 0, 5, 15, 25 \) years are approximately 20, 46, 99, and 134 respectively.

Step 2: Calculate Rate of Change of Population

To determine the rate of change in the population, we must differentiate \( P(t) \) with respect to \( t \):Given:\[ P(t) = \frac{3000}{20 + 130 e^{-0.214 t}} \]Let \( u = 20 + 130 e^{-0.214 t} \), then \( P(t) = \frac{3000}{u} \).The derivative using the quotient rule is:\[ \frac{dP}{d t} = -\frac{3000}{u^2} \cdot \frac{d u}{dt} \]Where:\[ \frac{d u}{dt} = -130 \times -0.214 e^{-0.214 t} = 27.82 e^{-0.214 t} \]Thus:\[ P'(t) = -\frac{3000 \times 27.82 e^{-0.214 t}}{(20 + 130 e^{-0.214 t})^2} \approx -\frac{83460 e^{-0.214 t}}{(20 + 130 e^{-0.214 t})^2} \]

Step 3: Calculate Rate of Change at t=0, 5, 15, 25

Substitute the values of \( t \) into the derived expression for \( P'(t) \):1. For \( t = 0 \): \[ P'(0) = -\frac{83460 \times 1}{150^2} = -\frac{83460}{22500} \approx -3.71 \]2. For \( t = 5 \): \[ P'(5) = -\frac{83460 e^{-1.07}}{(20 + 45.24)^2} \approx -\frac{83460 \times 0.343}{4245.98} \approx -6.74 \]3. For \( t = 15 \): \[ P'(15) = -\frac{83460 \times 0.08}{30.4^2} \approx -\frac{6676.8}{924.16} \approx -7.23 \]4. For \( t = 25 \): \[ P'(25) = -\frac{83460 \times 0.018}{22.34^2} \approx -\frac{1502.28}{499.57} \approx -3.00 \]The rate of change at \( t = 0, 5, 15, 25 \) years is approximately -3.71, -6.74, -7.23, and -3.00 respectively.

Step 4: Find Limiting Population Value

The limiting population is the carrying capacity of the environment, which in a logistic growth model occurs when \( t \to \infty \). As \( t \to \infty \), \( e^{-0.214 t} \to 0 \), thus:\[ P(t) = \frac{3000}{20 + 130 \times 0} = \frac{3000}{20} = 150 \]The limiting population value is 150 tortoises.

Key Concepts

Differentiation in Calculus

Differentiation is a fundamental concept in calculus used to calculate the rate at which a function is changing. When we differentiate a function like the logistic growth equation for the tortoise population in the Mojave Desert, we are interested in finding out how quickly the population is increasing or decreasing at any given time. This is done by finding the derivative of the function, denoted as \( P'(t) \). The derivative represents the instantaneous rate of change of the population with respect to time.

To differentiate the logistic growth equation given by \( P(t) = \frac{3000}{20 + 130 e^{-0.214 t}} \), we use the quotient rule, a technique in calculus for finding the derivative of the quotient of two functions. The quotient rule states: if we have a function \( \frac{f(x)}{g(x)} \), the derivative is \( \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2} \).

In our logistic function, \( f(x) = 3000 \) and \( g(x) = 20 + 130 e^{-0.214 t} \). Through differentiation, we determine how the population changes over time, which aids in predicting future population sizes.

Rate of Change

Rate of change in a population model indicates how quickly the population size is increasing or decreasing at any moment. It is essential for understanding the dynamics of population growth and is computed by the derivative of the population equation.

In the logistic growth model, the rate of change is given by the derivative \( P'(t) \). This measures the change in the population over a unit of time. When analyzing different values of \( t \), you can observe how the rate varies. For example, calculating \( P'(t) \) at 0, 5, 15, and 25 years provides insight into how briskly the population is changing at those specific points in time.

The rate of change informs us about the momentum of population growth at early, middle, and late stages of logistic growth. Initially, the rate of change is smaller because few individuals are there to reproduce. As the population grows, the rate increases, indicating faster growth.

• Early stage () shows a low rate due to limited individuals.
• Middle stages (~5-15 years) often have the highest rates of change.
• Late stage (>15 years) sees the rate tapering as it nears the carrying capacity.

Carrying Capacity

Carrying capacity is the maximum population size that an environment can sustain indefinitely. In a logistic growth model, it is this limiting value that the population approaches as time goes to infinity.

For the tortoise population in the Mojave Desert, the carrying capacity is determined by observing the behavior of the logistic equation as \( t \to \infty \). As time progresses and \( e^{-0.214 t} \to 0 \), the denominator in the logistic function \( 20 + 130 e^{-0.214 t} \) simplifies to 20. Hence, the carrying capacity is \( \frac{3000}{20} = 150 \).

This value indicates that the desert can support up to 150 tortoises in a square mile, regardless of how long you continue to observe the population. Given sufficient time, the population will stabilize around this number because resources such as food, shelter, and space become limited, preventing further growth and ensuring that balance is maintained.