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Chapter 7: Problem 60

Let \(y(t)\) be the population of a species that is being harvested. Consider the harvesting model \(y^{\prime}(t)=0.008 y-h, y(0)=y_{0},\) where \(h>0\) is the annual harvesting rate and \(y_{0}\) is the initial population of the species. a. If \(y_{0}=2000,\) what harvesting rate should be used to maintain a constant population of \(y=2000\) for \(t \geq 0 ?\) b. If the harvesting rate is \(h=200 /\) year, what initial population ensures a constant population for \(t \geq 0 ?\)

Short Answer

Expert verified
Short Answer: For part (a), to maintain a constant population of 2000 with the given harvesting model, the harvesting rate should be 16 per year. For part (b), the initial population that ensures a constant population when the harvesting rate is 200 per year is 25,000.

Step by step solution

01

Find the harvesting rate.

Since we know that \(y^{\prime}(t)=0.008y-h\) and \(y^{\prime}(t)=0\), we can set up the following equation to solve for the harvesting rate (\(h\)), with \(y=2000\) and \(y_{0}=2000\): \[0 = 0.008(2000) - h\]
02

Solve for the harvesting rate (\(h\)).

Solving the equation for \(h\), we obtain: \[h = 0.008(2000)\] \[h = 16 \text{ per year}\] Thus, the harvesting rate should be 16 per year to maintain a constant population of 2000 for \(t \geq 0\). For part b: Step 1: We want to find what initial population (\(y_{0}\)) would ensure a constant population when \(h=200\) per year.
03

Integrate the differential equation.

We are given the differential equation \(y^{\prime}(t)=0.008y-h\) with \(h = 200\). We can rewrite the equation as: \[y^{\prime}(t) - 0.008y = -200\] We now integrate both sides of this equation: \[\int(y^{\prime}(t) - 0.008y)dt = -200\int dt\]
04

Find a general solution.

Integrating both sides, we get: \[y(t) = Ce^{0.008t} - \frac{200}{0.008}\] Now, we have to determine the value of \(C\). To do this, we will use the information that the population is constant for \(t \geq 0\):
05

Determine \(C\).

Since the population is constant, we know that \(y^{\prime}(t)=0\). Thus: \[0 = Ce^{0.008t} - \frac{200}{0.008}\] We can set \(y(0)=y_{0}\) and solve for \(C\), where \(y_{0}\) is the initial population we want to find: \[0 = Ce^{0.008(0)} - \frac{200}{0.008}\] \[0 = C - \frac{200}{0.008}\] Thus, \(C = \frac{200}{0.008}\).
06

Find the initial population \(y_{0}\).

Now that we have found \(C\), we can plug it back into the general solution: \[y(t) = e^{0.008t}\cdot\frac{200}{0.008} - \frac{200}{0.008}\] Setting \(t = 0\), we get the initial population: \[y_{0} = e^{0.008(0)}\cdot\frac{200}{0.008} - \frac{200}{0.008}\] \[y_{0} = \frac{200}{0.008}\] So, the initial population that ensures a constant population for \(t \geq 0\) when the harvesting rate is 200 per year is \(y_{0} = \frac{200}{0.008} = 25000\).

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Key Concepts

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

Differential Equation
A differential equation is a mathematical equation that relates a function with its derivatives. In the context of a harvesting model, we use differential equations to describe the change in population over time as it is influenced by both growth and harvesting. For the model given, the differential equation is:\[ y'(t) = 0.008y - h \]- **\(y'(t)\):** Represents the rate of change of the population over time.- **\(0.008y\):** Reflects natural growth, where 0.008 is the growth rate of the population.- **\(-h\):** Denotes the harvesting effect, taking individuals from the population at a constant rate \(h\).
The primary goal of solving such differential equations in a harvesting context is to find conditions allowing for a constant population, meaning the birth and harvesting rates balance each other out.
Initial Population
Initial population refers to the starting number of individuals in a species at the beginning of the observation period (\(t=0\)). In the given exercise, this initial population is denoted as \(y_0\). Knowing the initial population is crucial because it impacts decisions about sustainable harvesting rates or any required adjustments to keep the population steady over time.- To maintain a constant population of 2000 with a harvesting rate of \(16\) per year, the initial population must also begin precisely at 2000.- When determining the necessary initial population for different harvesting rates, it鈥檚 important to calculate this number to ensure the desired balance between the number of harvested individuals and population growth.
Overall, identifying the right initial population helps in setting correct harvest strategies and avoiding depletion or overpopulation of the species.
Harvesting Rate
Harvesting rate implies the number of individuals removed from the population per unit time, often set on an annual basis (as indicated by \(h\)). In maintaining a constant population over time, the harvesting rate should precisely counterbalance the natural growth of the population:- **Balancing Growth:** To maintain 2000 individuals, calculate \(h\) using the equation \(y'(t)=0.008y-h\). Setting \(y'(t) = 0\): \[0 = 0.008 \times 2000 - h \] Solving this gives \(h = 16\), thus the population stays constant at 2000 if 16 individuals are harvested per year.- **Adjusting for Larger \(h\):** If \(h=200\), calculate the necessary \(y_0\) for balance: \[ y_0 = \frac{h}{0.008} = 25000 \] Starting with 25000 ensures stability with such a high harvesting rate.The correct harvesting rate is key to ensuring long-term viability of the species while also meeting human or economic needs efficiently.

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Most popular questions from this chapter

Apply Simpson's Rule to the following integrals. It is easiest to obtain the Simpson's Rule approximations from the Trapezoid Rule approximations, as in Example \(7 .\) Make \(a\) table similar to Table 7.8 showing the approximations and errors for \(n=4,8,16,\) and \(32 .\) The exact values of the integrals are given for computing the error. \(\int_{0}^{\pi} e^{-t} \sin t d t=\frac{1}{2}\left(e^{-\pi}+1\right)\)

Powers of sine and cosine It can be shown that \(\int_{0}^{\pi / 2} \sin ^{n} x d x=\int_{0}^{\pi / 2} \cos ^{n} x d x=\) \(\left\\{\begin{array}{ll}\frac{1 \cdot 3 \cdot 5 \cdots(n-1)}{2 \cdot 4 \cdot 6 \cdots n} \cdot \frac{\pi}{2} & \text { if } n \geq 2 \text { is an even integer } \\ \frac{2 \cdot 4 \cdot 6 \cdots(n-1)}{3 \cdot 5 \cdot 7 \cdots n} & \text { if } n \geq 3 \text { is an odd integer. }\end{array}\right.\)

Use the indicated methods to solve the following problems with nonuniform grids. A hot-air balloon is launched from an elevation of 5400 ft above sea level. As it rises, its vertical velocity is computed using a device (called a variometer) that measures the change in atmospheric pressure. The vertical velocities at selected times are shown in the table (with units of \(\mathrm{ft} / \mathrm{min}\) ). $$\begin{array}{|l|c|c|c|c|c|c|c|} \hline t \text { (min) } & 0 & 1 & 1.5 & 3 & 3.5 & 4 & 5 \\ \hline \begin{array}{l} \text { Velocity } \\ \text { (ft/min) } \end{array} & 0 & 100 & 120 & 150 & 110 & 90 & 80 \\ \hline \end{array}$$ a. Use the Trapezoid Rule to estimate the elevation of the balloon after five minutes. Remember that the balloon starts at an elevation of \(5400 \mathrm{ft}\) b. Use a right Riemann sum to estimate the elevation of the balloon after five minutes. c. A polynomial that fits the data reasonably well is $$g(t)=3.49 t^{3}-43.21 t^{2}+142.43 t-1.75$$ Estimate the elevation of the balloon after five minutes using this polynomial.

An open cylindrical tank initially filled with water drains through a hole in the bottom of the tank according to Torricelli's Law (see figure). If \(h(t)\) is the depth of water in the tank for \(t \geq 0,\) then Torricelli's Law implies \(h^{\prime}(t)=2 k \sqrt{h}\), where \(k\) is a constant that includes the acceleration due to gravity, the radius of the tank, and the radius of the drain. Assume that the initial depth of the water is \(h(0)=H\). a. Find the general solution of the equation. b. Find the solution in the case that \(k=0.1\) and \(H=0.5 \mathrm{m}\). c. In general, how long does it take the tank to drain in terms of \(k\) and \(H ?\)

Another Simpson's Rule formula is \(S(2 n)=\frac{2 M(n)+T(n)}{3},\) for \(n \geq 1 .\) Use this rule to estimate \(\int_{1}^{e} 1 / x d x\) using \(n=10\) subintervals.
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