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Applications of Differential Equations

Population Dynamics

(continued from last page...)

I am not going to tell you what the [[1,1,2]] trick looks like. How many times have you used it now?

Finding the Value of the Growth Constant, k

We now address the value of k. It is a measure of the rate at which the population grows. To determine k we can use any one of the data points we haven't used yet (we can't use the first one since we already used it to find the initial condition.) Remember, each data point represents a tp-pair that the solution of the problem should theoretically pass through. This means that substituting any one of them into our general solution, (the solution we just found,) should theoretically satisfy it, and produce an equation whose only remaining variable is the k we seek.

So which data point should we use? Perhaps the last one would best capture the spread of the data. Using it, we need to plug in 2010 into malthus as the t-value, and set the result equal to 310383948., (i.e. the p-value). This will give us an equation in which only k remains. The equation would look something like this in Mathematica form:

(malthus /. t -> 2010) == 310383948.

This says to take malthus, substitute 2010 in place of t and form an equation by setting the resulting quantity equal to 310383948.. However, don't do it yet—hold on!

How do we solve the resulting equation? You may recall that Mathematica's standard equation solving command is Solve. (We learned how to use it in the introductory notebooks.) Let's go and use it to solve for k...


Compass If you're lost, impatient, want an overview of this laboratory assignment, or maybe even all three, you can click on the compass button on the left to go to the table of contents for this laboratory assignment.
 
 

ODE Laboratories: A Sabbatical Project by Christopher A. Barker

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