Another change of variable problem

In summary, the population has a periodic growth rate of A[1 + sin(t/(2*pi))] and there is no threshold. The initial population is Yo = Y knot = K/2.
  • #1
spoon
12
0
I'm also attempting another problem...

A population has a periodic growth rate r(t) = A[1 + sin(t/(2*pi))], but otherwise follows the logistic population model with carrying capacity K. There is no threshold and the initial population is Yo = Y knot = K/2.

a. Modify the basic logistic equation for this population.
b. Use a change of variable z(t) = y/K to find a initial value equation in z(t). Then Solve for z(t).

So far I have:
Using the logistic growth equation: dy/dt = r(1-y/k)y
Leading to:
integral of [dy/((1-y/k)y)] = rt +c
substituting for r:
integral of [dy/((1-y/k)y)] = t*A[1 + sin(t/(2*pi))] + c

I can't really follow how to use the change of variables well so part b is the most confusing part.
 
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  • #2
spoon said:
So far I have:
Using the logistic growth equation: dy/dt = r(1-y/k)y
Leading to:
integral of [dy/((1-y/k)y)] = rt +c
substituting for r:
integral of [dy/((1-y/k)y)] = t*A[1 + sin(t/(2*pi))] + c

ah,
as a friend of mine likes to say, "Classic mistake."
if r = r(t) you can't integrate it as if it were a constant then
plug in the value. you need:

[tex] \int \ldots dy = C + \int r(t) dt [/tex]

I can't really follow how to use the change of variables well so part b is the most confusing part.

There shouldn't be anything tricky to a change of variables.
(except, of course, deciding what varibles to use.)
It's a plug-and-chug operation.

if y = k z then
y' = ?

now, just replace every occurence of y with (kz) and every
occurence of y' with ?.
 
  • #3
Thanks for the point about r(t), so I fixed that part, but I'm still confused as how to go about the rest...
If I substitute y with kz and y' with ?, then how do I integrate the following:
?/[(1-z)kz]
Because I could integrate it if the "?" were a "dz" right?
 
  • #4
I guess I was unclear. it's your job to find out what
to put in in place of the ?.

What I thought i had hinted at
was if y = kz then y' = kz', because
k is a constant.
 
  • #5
Sorry, looking back that was pretty clear...
So then it would be the integral of:
dz/(1-z)z = ln(1-z) + ln(z)
Then set this equal to:
C + Integral( r(t))
and just solve for z?
 
Last edited:

1. What is a change of variable problem?

A change of variable problem is a mathematical problem in which the independent variable of a function is replaced with a new variable, typically to simplify the problem or make it easier to solve.

2. Why do we use change of variable in problem solving?

Change of variable allows us to transform a problem into a new form that may be easier to solve or understand. It can also help us to find new solutions or insights to a problem.

3. What are some common examples of change of variable problems?

Some common examples of change of variable problems include integration problems in calculus, transformation of equations in algebra, and substitution in geometry.

4. How do you choose the right variable for a change of variable problem?

Choosing the right variable for a change of variable problem depends on the specific problem at hand. Some strategies for choosing a variable include looking for patterns, simplifying expressions, or using known identities or formulas.

5. Can change of variable be used in other fields besides mathematics?

Yes, change of variable can be used in various fields such as physics, engineering, and economics. It is a useful tool for problem solving and can be applied in many different contexts.

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