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Chapter 8: Problem 12

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
Answer: The population function derived from the given linear growth rate function is \(P(t) = \frac{1}{2}k(t^2) + P_0\). Graphically, it would be a parabola opening upwards, starting at an initial population of \(P_0\) and increasing with time. The growth rate would be increasing as well since the growth rate function has a positive slope (kt).

Step by step solution

01

Identify the growth rate function

In this case, we are provided with a linear growth rate function: \(r(t) = kt\), where \(k\) is a constant.
02

Integrate the growth rate function

To find the population function, we'll need to integrate the growth rate function with respect to time. Let \(P(t)\) be the population function. Then, the relationship between the population function and the growth rate function can be described as: $$\frac{dP(t)}{dt} = r(t)$$ Now we must integrate the growth rate function: $$P(t) = \int r(t) dt = \int kt dt$$ $$P(t) = \frac{1}{2}k(t^2) + C$$ where \(C\) is the constant of integration.
03

Determine the initial population value

We are given that the population at time \(t=0\) is some positive value. Therefore, \(P(0) > 0\). Let's use \(P_0\) as the initial population value. To find the value of \(C\), we set \(t = 0\) and \(P(t) = P_0\). So, $$P(0) = P_0 = \frac{1}{2}k(0^2) + C$$ Thus, \(C = P_0\).
04

Write the final population function

After determining the constant of integration, we can now write the final population function as: $$P(t) = \frac{1}{2}k(t^2) + P_0$$
05

Sketch the population function

To sketch the population function, note the following points: 1. The initial population at time \(t = 0\) is \(P_0\). 2. The population function is a quadratic function with a positive coefficient for the \(t^2\) term. Therefore, its graph will be a parabola opening upwards. 3. As time increases, the population will grow at an increasing rate, since the growth rate function is a linear function with a positive slope (\(kt\)). With these points, you can now sketch the population function, showing a parabolic shape starting at \(P_0\) and increasing with time.

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Most popular questions from this chapter

Explain how the growth rate function determines the solution of a population model.

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. $$u^{\prime}(x)=\csc u \cos \frac{x}{2}, u(\pi)=\frac{\pi}{2}$$

A differential equation of the form \(y^{\prime}(t)=f(y)\) is said to be autonomous (the function \(f\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(f\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations. a. Find the equilibrium solutions. b. Sketch the direction field, for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(2-y)$$

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. $$y^{\prime}(x)=\frac{1+x}{2-y}, y(1)=1$$

Solve the following initial value problems. When possible, give the solution as an explicit function of \(t\) $$y^{\prime}(t)=\frac{\cos ^{2} t}{2 y}, y(0)=-2$$
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