# Exponential growth problem

• August 6th 2008, 04:58 PM
c47v3770
Exponential growth problem
Guys,

I can't seem to solve this problem. Maybe you guys can help me out?
Meanwhile, I will be looking for my old math books :)

"Consider a population that begins growing exponentially at a rate of 15% per year and then follows a logistic growth pattern. If the carrying capacity is 100 million, find the actual growth rate when the population is 10 million, 50 million and 90 million."

Thanks a bunch..
• August 6th 2008, 06:36 PM
TKHunny
It's a little tough to know what to say, since you posted in this generic category. Had you posted in Algebra, there would be a good answer. Had you posted in Differential Equations, there would be simpler answer. The nature of the question rather leans toward the differential equation formulation:

$\frac{dP}{dt}\;=\;r*P\left(1-\frac{P}{K}\right)$

The Initial Growth Rate is r = 0.15
The Carying Capacity is K = 100 (in millions)

You have only to substitute your given population values and calculate the actual growth at that level. Calculating the Growth Rate is a matter of dividing by the population again. At P = 80, I get 3%.
• August 6th 2008, 07:51 PM
c47v3770
Quote:

Originally Posted by TKHunny
It's a little tough to know what to say, since you posted in this generic category. Had you posted in Algebra, there would be a good answer. Had you posted in Differential Equations, there would be simpler answer. The nature of the question rather leans toward the differential equation formulation:

$\frac{dP}{dt}\;=\;r*P\left(1-\frac{P}{K}\right)$

The Initial Growth Rate is r = 0.15
The Carying Capacity is K = 100 (in millions)

You have only to substitute your given population values and calculate the actual growth at that level. Calculating the Growth Rate is a matter of dividing by the population again. At P = 80, I get 3%.

thank you.
• August 8th 2008, 05:37 PM
c47v3770
So P= population right?

So when I substitute to find the actual growth for 10 million, I get:

0.15 x 10.000.000(1-10.000.000/100.000.000)

Am I correct?

and then you said that "Calculating the Growth Rate is a matter of dividing by the population again."

So I divide the result from the equation above by the population (10.000.000) ???

• August 8th 2008, 06:46 PM
TKHunny
You have it. The notation is a little awkward because the definition is not quite in the terms you want. Rather then multiply by the population, just to divide by it again, you can just leave it out if all you want is the rate.
• August 8th 2008, 06:55 PM
c47v3770
Quote:

Originally Posted by TKHunny
You have it. The notation is a little awkward because the definition is not quite in the terms you want. Rather then multiply by the population, just to divide by it again, you can just leave it out if all you want is the rate.

"Rather then multiply by the population, just to divide by it again, you can just leave it out if all you want is the rate."

I don't really understand what you said. Did you mean to say that I don't have to multiply and divide if all I want is the rate?

Thank you!
• August 8th 2008, 07:29 PM
TKHunny
The Definitions

$\frac{dP}{dt}$ is the Growth over some unit of time.

$\frac{\frac{dP}{dt}}{P}$ is the Growth Rate over some unit of time.

The Algebra

$\frac{dP}{dt}\;=\;rP\left(1-\frac{P}{K}\right)$ is the Growth over some unit of time.

$\frac{\frac{dP}{dt}}{P}\;=\;\frac{rP(1-\frac{P}{K})}{P}\;=\;r\left(1-\frac{P}{K}\right)$ is the Growth Rate over some unit of time.

In the last statement, we see that it is not necessary to calculate the Growth, if all you need is the Growth Rate. The algebra has provided us with a shorter route to calculate the Growth Rate.
• August 9th 2008, 08:35 AM
c47v3770
Quote:

Originally Posted by TKHunny
The Definitions

$\frac{dP}{dt}$ is the Growth over some unit of time.

$\frac{\frac{dP}{dt}}{P}$ is the Growth Rate over some unit of time.

The Algebra

$\frac{dP}{dt}\;=\;rP\left(1-\frac{P}{K}\right)$ is the Growth over some unit of time.

$\frac{\frac{dP}{dt}}{P}\;=\;\frac{rP(1-\frac{P}{K})}{P}\;=\;r\left(1-\frac{P}{K}\right)$ is the Growth Rate over some unit of time.

In the last statement, we see that it is not necessary to calculate the Growth, if all you need is the Growth Rate. The algebra has provided us with a shorter route to calculate the Growth Rate.

Awesome!
So I just use the formula:

$\frac{\frac{dP}{dt}}{P}\;=\;\frac{rP(1-\frac{P}{K})}{P}\;=\;r\left(1-\frac{P}{K}\right)$

and substitute?

Sorry if I am asking too many questions..
• August 9th 2008, 11:38 AM
TKHunny
Go forward! No need to step gingerly. You have thought it through. You have done the algebra. Abandon anything except confidence.
• August 9th 2008, 01:29 PM
c47v3770
Thanks so much!

Quick question: do I also have to substitute for

dP
____
dt
______ =
P

You have been of great help :)
• August 9th 2008, 06:03 PM
TKHunny
Please don't. That's what the problem wants. Once you have it, leave it alone. (Happy)

I warned you up front that the notation was a little awkward. You are probaly happy with something like x = 2. Yea! I'm done! I found the value of x.

When we seed (dP/dt)/P and we get (dP/dt)/P = 2.4%, it's not quite so obvious that we are done.