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Chapter 8: Problem 14

Make a sketch of the population function (as a function of time) that results from the following growth rate functions. Assume the population at time \(t=0\) begins at some positive value.

Short Answer

Expert verified
Based on these general observations, the population function's sketch will start at a point on the positive y-axis corresponding to \(P_0\) and will either have an upward slope if the growth rate function is positive or a downward slope if the growth rate function is negative. Without specific information about the growth rate function, we cannot draw the exact population function graph.

Step by step solution

01

Identify the growth rate functions

The exercise does not specify the growth rate functions, as it asks to sketch the "following growth rate functions." Since the problem does not provide any specific growth rates, let's consider a general case. Let \(r(t)\) be the growth rate function with respect to time \(t\) (we don't have a specific function since it's not provided in the exercise).
02

Integrate the growth rate function

To find the population function, we need to integrate the growth rate function with respect to \(t\). Let \(P(t)\) be the population function. Then, \(P(t) = \int r(t) dt + C\), where \(C\) is the constant of integration.
03

Consider the initial population

The exercise states that the population at time \(t=0\) begins at some positive value. Let this initial population be \(P_0\). We can now use this information to find the constant of integration \(C\). At \(t=0\), \(P(0) = P_0\). So, \(P_0 = \int r(0) dt + C\). To find \(C\), we need to evaluate the integral at \(t=0\). However, since no detailed information about the growth rate function \(r(t)\) is provided, we can not evaluate this integral. Therefore, we'll represent the population function as: \(P(t) = \int r(t) dt + P_0\).
04

Sketch the population function

With the given information, it's impossible to sketch the exact population function since we don't have a specific growth rate function \(r(t)\). However, we can conclude the general form of the function: If the growth rate function is positive (i.e., the population is increasing with time), the integral would be positive, and hence, the population function would have an upward slope. If the growth rate function is negative (i.e., the population is decreasing with time), the integral would be negative, and hence, the population function would have a downward slope. In both cases, the initial population will be at a positive value of \(P_0\).

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Most popular questions from this chapter

One possible model that describes the free fall of an object in a gravitational field subject to air resistance uses the equation \(v^{\prime}(t)=g-b v,\) where \(v(t)\) is the velocity of the object for \(t \geq 0\), \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(b>0\) is a constant that involves the mass of the object and the air resistance. a. Verify by substitution that a solution of the equation, subject to the initial condition \(v(0)=0,\) is \(v(t)=\frac{g}{b}\left(1-e^{-b t}\right)\). b. Graph the solution with \(b=0.1 s^{-1}\). c. Using the graph in part (c), estimate the terminal velocity \(\lim _{t \rightarrow \infty} v(t)\).

An object in free fall may be modeled by assuming that the only forces at work are the gravitational force and air resistance. By Newton's Second Law of Motion (mass \(\times\) acceleration \(=\) the sum of the external forces), the velocity of the object satisfies the differential equation $$\underbrace {m}_{\text {mass}}\quad \cdot \underbrace{v^{\prime}(t)}_{\text {acceleration }}=\underbrace {m g+f(v)}_{\text {external forces}}$$ where \(f\) is a function that models the air resistance (assuming the positive direction is downward). One common assumption (often used for motion in air) is that \(f(v)=-k v^{2},\) where \(k>0\) is a drag coefficient. a. Show that the equation can be written in the form \(v^{\prime}(t)=g-a v^{2},\) where \(a=k / m\) b. For what (positive) value of \(v\) is \(v^{\prime}(t)=0 ?\) (This equilibrium solution is called the terminal velocity.) c. Find the solution of this separable equation assuming \(v(0)=0\) and \(0 < v^{2} < g / a\) d. Graph the solution found in part (c) with \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) \(m=1,\) and \(k=0.1,\) and verify that the terminal velocity agrees with the value found in part (b).

Solve the differential equation for Newton's Law of Cooling to find the temperature in the following cases. Then answer any additional questions. A glass of milk is moved from a refrigerator with a temperature of \(5^{\circ} \mathrm{C}\) to a room with a temperature of \(20^{\circ} \mathrm{C}\). One minute later the milk has warmed to a temperature of \(7^{\circ} \mathrm{C}\). After how many minutes does the milk have a temperature that is \(90 \%\) of the ambient temperature?

The equation \(y^{\prime}(t)+a y=b y^{p},\) where \(a, b,\) and \(p\) are real numbers, is called a Bernoulli equation. Unless \(p=1,\) the equation is nonlinear and would appear to be difficult to solve-except for a small miracle. By making the change of variables \(v(t)=(y(t))^{1-p},\) the equation can be made linear. Carry out the following steps. a. Letting \(v=y^{1-p},\) show that \(y^{\prime}(t)=\frac{y(t)^{p}}{1-p} v^{\prime}(t)\). b. Substitute this expression for \(y^{\prime}(t)\) into the differential equation and simplify to obtain the new (linear) equation \(v^{\prime}(t)+a(1-p) v=b(1-p),\) which can be solved using the methods of this section. The solution \(y\) of the original equation can then be found from \(v\).

Solve the differential equation for Newton's Law of Cooling to find the temperature in the following cases. Then answer any additional questions. A pot of boiling soup \(\left(100^{\circ} \mathrm{C}\right)\) is put in a cellar with a temperature of \(10^{\circ} \mathrm{C}\). After 30 minutes, the soup has cooled to \(80^{\circ} \mathrm{C}\). When will the temperature of the soup reach \(30^{\circ} \mathrm{C} ?\)
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