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Chapter 2: Problem 21

Competition between populations: In this exercise we consider the question of competition between two populations that vie for resources but do not prey on each other. Let \(m\) be the size of the first population and \(n\) the size of the second (both measured in thousands of animals), and assume that the populations coexist eventually. Here is an example of one common model for the interaction: Per capita growth rate for \(m=5(1-m-n)\), Per capita growth rate for \(n=6(1-0.7 m-1.2 n)\). a. An isocline is formed by the points at which the per capita growth rate for \(m\) is zero. These are the solutions of the equation \(5(1-m-n)=0\). Find a formula for \(n\) in terms of \(m\) that describes this isocline. b. The points at which the per capita growth rate for \(n\) is zero form another isocline. Find a formula for \(n\) in terms of \(m\) that describes this isocline. c. At an equilibrium point the per capita growth rates for \(m\) and for \(n\) are both zero. If the populations reach such a point, they will remain there indefinitely. Use your answers to parts \(a\) and \(b\) to find the equilibrium point.

Short Answer

Expert verified
The equilibrium point is \((m, n) = (0.4, 0.6)\).

Step by step solution

01

Solve for the Isocline of Population m

To find the isocline for population \(m\), we need to solve the equation where its per capita growth rate is zero: \(5(1-m-n)=0\). Divide both sides by 5 to simplify: \(1-m-n=0\). Then solve for \(n\) in terms of \(m\): \(n = 1 - m\). This is the isocline for \(m\).
02

Solve for the Isocline of Population n

For the isocline of population \(n\), use the equation where its per capita growth rate is zero: \(6(1-0.7m-1.2n) = 0\). Dividing by 6 gives \(1-0.7m-1.2n = 0\). Solve for \(n\) in terms of \(m\) by first moving the terms around: \(1 = 0.7m + 1.2n\). This simplifies to \(n = \frac{1-0.7m}{1.2}\). This is the isocline for \(n\).
03

Find the Equilibrium Point from Both Isoclines

An equilibrium point is found where both isoclines intersect; thus where both conditions are true. Set the isoclines equal to find intersection: \(1 - m = \frac{1-0.7m}{1.2}\). Multiply everything by 1.2 to clear the fraction: \(1.2(1 - m) = 1 - 0.7m\). This simplifies to \(1.2 - 1.2m = 1 - 0.7m\). Solve for \(m\) by combining like terms: \(1.2m - 0.7m = 1.2 - 1\) which gives \(0.5m = 0.2\). Divide by 0.5 to find \(m = 0.4\). Now substitute \(m = 0.4\) back into \(n = 1 - m\): \(n = 1 - 0.4 = 0.6\). The equilibrium point is \((m, n) = (0.4, 0.6)\).

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Key Concepts

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

Isocline
In population dynamics, an **isocline** is a crucial concept that helps us understand how two populations interact without preying on each other as they compete for resources. An isocline is essentially a set of points where a specific variable, like the per capita growth rate of a population, equals zero. For population models, isoclines are often visualized on a graph to show how populations will behave at different sizes over time.

For example, in the problem given, we have two populations: population "m" and population "n". Each has its own isocline determined by setting the per capita growth rate to zero.
  • For population "m", the per capita growth rate equation is given as: \(5(1 - m - n) = 0\). Solving this, the isocline for "m" is described by the formula \(n = 1 - m\).
  • For population "n", the growth rate is expressed by: \(6(1 - 0.7m - 1.2n) = 0\). Solving, the isocline for "n" becomes \(n = \frac{1 - 0.7m}{1.2}\).
These isoclines represent the combinations of "m" and "n" that result in no growth for each population. Points on an isocline indicate population levels where that particular population neither grows nor shrinks. The intersection of the isoclines subsequently points us toward the equilibrium point.
Equilibrium Point
An **equilibrium point** is a fundamental concept in understanding the balance within ecosystems, particularly in population dynamics. It denotes a condition where both populations reach a stable size, ceasing to change over time when not disrupted by external factors. This is achieved when the per capita growth rates for all interacting populations are zero simultaneously. At this point, neither population will grow or shrink, regardless of its starting size.

In the provided example, the equilibrium point arises where the isoclines for both populations "m" and "n" intersect. Mathematically, this intersection occurs when both conditions of zero growth rates for each population are satisfied.
  • For the given populations, setting the isoclines equal, as derived from the isoclines: \(1 - m = \frac{1 - 0.7m}{1.2}\).
  • By solving, we find the crossing point at \(m = 0.4\) and substituting this into population "m"'s isocline, yields \(n = 0.6\).
Thus, the equilibrium point is \((m, n) = (0.4, 0.6)\). This balance point suggests that if both populations reach these sizes, they remain constant unless external impacts disrupt the system's dynamics. Understanding equilibrium points helps ecologists establish stability within ecological management and control population sizes effectively.
Per Capita Growth Rate
The **per capita growth rate** represents the average rate at which individuals within a population reproduce over a given time, factoring in births and deaths relative to the existing population size. It is a crucial measurement in population dynamics as it determines whether a population is growing, declining, or remaining stable.

For example, in the interactions between two populations depicted in the scenario, the per capita growth rates are specified by distinct equations.
  • For population "m", the growth rate given is: \(5(1 - m - n)\). Here, the multiplier 5 implies a specific growth influence on the population, moderated by its current size and that of the competing population "n".
  • For population "n", the growth rate is: \(6(1 - 0.7m - 1.2n)\). Similarly, the factor 6 represents growth potential, adjusted by interaction terms designed to reflect the competition dynamics.
The per capita growth rates capture how both populations influence each other. Notably, as resources are shared and competition dynamics play out, these rates inform how likely one population is to outcompete the other. When these growth rates hit zero, populations reach equilibrium if ample resources exist for sustainable survival of both. Otherwise, adjustments or declines occur based on competitive interactions between them.

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Most popular questions from this chapter

The profit \(P\), in thousands of dollars, that a manufacturer makes is a function of the number \(N\) of items produced in a year, and the formula is $$ P=-0.2 N^{2}+3.6 N-9 . $$ a. Express using functional notation the profit at a production level of 5 items per year, and then calculate that value. b. Determine the two break-even points for this manufacturer-that is, the two production levels at which the profit is zero. c. Determine the maximum profit if the manufacturer can produce at most 20 items in a year.

The economic order quantity model tells a company how many items at a time to order so that inventory costs will be minimized. The number \(Q=Q(N, c, h)\) of items that should be included in a single order depends on the demand \(N\) per year for the product, the fixed cost \(c\) in dollars associated with placing a single order (not the price of the item), and the carrying cost \(h\) in dollars. (This is the cost of keeping an unsold item in stock.) The relationship is given by $$ Q=\sqrt{\frac{2 N c}{h}} $$ a. Assume that the demand for a certain item is 400 units per year and that the carrying cost is \(\$ 24\) per unit per year. That is, \(N=400\) and \(h=24\). i. Find a formula for \(Q\) as a function of the fixed ordering \(\operatorname{cost} c\), and plot its graph. For this particular item, we do not expect the fixed ordering costs ever to exceed \(\$ 25\). ii. Use the graph to find the number of items to order at a time if the fixed ordering cost is \$6 per order. iii. How should increasing fixed ordering cost affect the number of items you order at a time? b. Assume that the demand for a certain item is 400 units per year and that the fixed ordering cost is \(\$ 14\) per order. i. Find a formula for \(Q\) as a function of the carrying cost \(h\) and make its graph. We do not expect the carrying cost for this particular item ever to exceed \(\$ 25\). ii. Use the graph to find the optimal order size if the carrying cost is \(\$ 15\) per unit per year. iii. How should an increase in carrying cost affect the optimal order size? iv. What is the average rate of change per dollar in optimal order size if the carrying cost increases from \(\$ 15\) to \(\$ 18 ?\) v. Is this graph concave up or concave down? Explain what that tells you about how optimal order size depends on carrying costs.

The cost of making a can is determined by how much aluminum \(A\), in square inches, is needed to make it. This in turn depends on the radius \(r\) and the height \(h\) of the can, both measured in inches. You will need some basic facts about cans. See Figure \(2.110\). The surface of a can may be modeled as consisting of three parts: two circles of radius \(r\) and the surface of a cylinder of radius \(r\) and height \(h\). The area of these circles is \(\pi r^{2}\) each, and the area of the surface of the cylinder is \(2 \pi r h\). The volume of the can is the volume of a cylinder of radius \(r\) and height \(h\), which is \(\pi r^{2} h\). In what follows, we assume that the can must hold 15 cubic inches, and we will look at various cans holding the same volume. a. Explain why the height of any can that holds a volume of 15 cubic inches is given by $$ h=\frac{15}{\pi r^{2}} \text {. } $$ b. Make a graph of the height \(h\) as a function of \(r\), and explain what the graph is showing. c. Is there a value of \(r\) that gives the least height \(h\) ? Explain. d. If \(A\) is the amount of aluminum needed to make the can, explain why $$ A=2 \pi r^{2}+2 \pi r h $$ e. Using the formula for \(h\) from part a, explain why we may also write \(A\) as $$ A=2 \pi r^{2}+\frac{30}{r} $$

In a chemical reaction, the reaction rate \(R\) is a function of the concentration \(x\) of the product of the reaction. For a certain secondorder reaction between two substances, we have the formula $$ R=0.01 x^{2}-x+22 . $$ Here \(x\) is measured in moles per cubic meter and \(R\) is measured in moles per cubic meter per second. a. Make a graph of \(R\) versus \(x\). Include concentrations up to 100 moles per cubic meter. b. Use functional notation to express the reaction rate when the concentration is 15 moles per cubic meter, and then calculate that value. c. The reaction is said to be in equilibrium when the reaction rate is 0 . At what two concentrations is the reaction in equilibrium?

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