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Seventeen adults came ashore from the British ship HMS Bounty in 1790 to settle on the uninhabited South Pacific island Pitcairn. The population, \(P(t),\) of the island \(t\) years after 1790 can be approximated by the logistic equation $$ P(t)=\frac{3400}{17+183 e^{-0.982 t}} $$ a) Find the population of the island after \(10 \mathrm{yr}, 50 \mathrm{yr}\) and \(75 \mathrm{yr}\) b) Find the rate of change in the population, \(P^{\prime}(t)\). c) Find the rate of change in the population after \(10 \mathrm{yr}, 50 \mathrm{yr},\) and \(75 \mathrm{yr}\) d) What is the limiting value for the population of Pitcairn? (The limiting value is the number to which the population gets closer and closer but never reaches.)

Short Answer

Expert verified
After 10 years, the population is approximately 183. The rate of change approaches 0 over time, and the limiting population is 200.

Step by step solution

01

Substitute and Calculate for Part (a)

To find the population at a specific time, substitute the given values of \(t\) into the logistic function. 1. For \(t=10\):\[ P(10) = \frac{3400}{17 + 183 e^{-0.982 \times 10}} \]Calculate \( e^{-0.982 \times 10} \) and simplify:\[ P(10) = \frac{3400}{17 + 183 \times 0.000208} \approx 183 \]2. For \(t=50\):\[ P(50) = \frac{3400}{17 + 183 e^{-0.982 \times 50}} \]Calculate \( e^{-0.982 \times 50} \) and simplify:\[ P(50) \approx \frac{3400}{17} = 200 \]3. For \(t=75\):\[ P(75) = \frac{3400}{17 + 183 e^{-0.982 \times 75}} \]Calculate \( e^{-0.982 \times 75} \) and simplify:\[ P(75) \approx \frac{3400}{17} = 200 \]
02

Differentiate the Function for Part (b)

To find the rate of change of the population, differentiate the logistic equation with respect to \(t\).Given:\[ P(t) = \frac{3400}{17 + 183 e^{-0.982t}} \]Apply the quotient rule: \[ P'(t) = \frac{d}{dt} \left( \frac{3400}{17 + 183 e^{-0.982t}} \right) \]Use:\[ P'(t) = \frac{-3400 \times (-0.982) \times 183 e^{-0.982t}}{(17 + 183 e^{-0.982t})^2} \]Simplify:\[ P'(t) = \frac{3400 \times 180.036 \times e^{-0.982t}}{(17 + 183 e^{-0.982t})^2} \]
03

Calculate the Rate of Change for Part (c)

Substitute \(t=10\), \(t=50\), and \(t=75\) into \(P'(t)\) from Step 2.1. For \(t=10\):\[ P'(10) = \frac{3400 \times 180.036 \times 0.000208}{(17 + 183 \times 0.000208)^2} \approx 0.676 \]2. For \(t=50\):Calculate \( e^{-0.982 \times 50} \) and substitute:\[ P'(50) \approx 0 \] (As \( e^{-0.982 \times 50} \) approaches zero)3. For \(t=75\):Likewise,\[ P'(75) \approx 0 \] (As \( e^{-0.982 \times 75} \) approaches zero)
04

Determine the Limiting Value for Part (d)

Determine what \(P(t)\) approaches as \(t\) approaches infinity.As \(t\) approaches infinity, \(e^{-0.982t}\) approaches zero.So, \[ P(t) \approx \frac{3400}{17} = 200 \]Thus, the limiting value of the population is 200.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Population Growth
Understanding population growth through the lens of a logistic equation provides a dynamic view of how populations evolve over time. When seventeen adults settled on Pitcairn Island in 1790, their initial population marked the beginning of a growth process influenced by natural constraints. Unlike linear growth, where populations expand at a constant rate, logistic growth considers the environmental limits, leading to an "S" shaped curve. In this case, the formula: \[ P(t) = \frac{3400}{17 + 183 e^{-0.982t}} \]illustrates how the population starts slowly at first, accelerates, and then levels off as it nears its carrying capacity. This pattern emerges because as resources such as space and food become scarcer, the growth rate slows down, preventing infinite growth in a finite world. After substituting specific values of \(t\) like 10, 50, and 75 years, we calculate the population at those times, showing how the number of inhabitants expands initially but then stabilizes, reflecting real-world limits.
Differential Calculus
Using differential calculus, we dive deeper into the rate at which Pitcairn Island's population changes. Differential calculus allows us to calculate the derivative of the population function, denoted as \(P'(t)\). This derivative provides insights into the instantaneous rate of change, effectively showing how quickly or slowly the population is growing at any given time.Calculating the derivative involves the application of the quotient rule, since our original formula is a fraction. The rate of change derived is:\[ P'(t) = \frac{3400 \times 180.036 \times e^{-0.982t}}{(17 + 183 e^{-0.982t})^2} \]With this formula, we substitute values such as \(t = 10, 50,\) and \(75\) years, offering a closer look at how growth transitions over time. Initially, the rate is measurable, but as time passes, the effect diminishes as environmental factors and earlier growth slow the expansion, highlighting the pivotal role of calculus in modeling natural phenomena.
Limiting Value
The concept of a limiting value is crucial in understanding logistic growth, as it denotes the maximum population size that Pitcairn Island can sustain indefinitely. As time \(t\) tends towards infinity, the factor \(e^{-0.982t}\) approaches zero, simplifying our logistic equation. This end behavior points the population function towards a constant value, showing mathematical stabilization:\[ \lim_ {t \to \infty} P(t) = \frac{3400}{17} = 200 \]This value, known as the carrying capacity, is where the population levels out permanently because resources can only support so many individuals. It reflects a balance between the number of people the environment can support and the natural constraints that limit further growth. The idea of a limiting value is essential for predicting long-term population dynamics, showcasing how mathematical models apply directly to ecological and environmental studies.

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Most popular questions from this chapter

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