# Logistic Growth

• Apr 10th 2010, 05:03 PM
WahooMan
Logistic Growth
q(t) = a / 1 + be^-kt

where b = a / q(initial) - 1 and k > 0 is constant.

Assuming a > q(initial) > 0, prove that the function q is everywhere increasing by analyzing its first derivative. Do not use specific values for any of the constants.

I found the derivative of q(t) to be

q'(t) = (-a(be^-kt)-bk) / ((1+be^-kt)^2)

First of all, is that right? And if it is, how do I prove that the function is everywhere increasing without using specific values for any of the constants?
• Apr 10th 2010, 05:11 PM
mr fantastic
Quote:

Originally Posted by WahooMan
q(t) = a / 1 + be^-kt

where b = a / q(initial) - 1 and k > 0 is constant.

Assuming a > q(initial) > 0, prove that the function q is everywhere increasing by analyzing its first derivative. Do not use specific values for any of the constants.

I found the derivative of q(t) to be

q'(t) = (-a(be^-kt)-bk) / ((1+be^-kt)^2)

First of all, is that right? And if it is, how do I prove that the function is everywhere increasing without using specific values for any of the constants?

Your derivative is wrong. Fix it and then use the fact that a> 0, k > 0 and a > q(initial) > 0.
• Apr 10th 2010, 05:26 PM
WahooMan
Quote:

Originally Posted by mr fantastic
Your derivative is wrong. Fix it and then use the fact that a> 0, k > 0 and a > q(initial) > 0.

Is it

q'(t) = (-a(be^-kt)) / ((1+be^-kt)^2)

?
• Apr 10th 2010, 05:29 PM
mr fantastic
Quote:

Originally Posted by WahooMan
Is it

q'(t) = (-a(be^-kt)) / ((1+be^-kt)^2)

?

No.

Instead of fumbling around in the dark and posting the result, please show all the details of your calculation so that your mistake(s) can be pointed out.
• Apr 10th 2010, 05:40 PM
WahooMan
Quote:

Originally Posted by mr fantastic
No.

Instead of fumbling around in the dark and posting the result, please show all the details of your calculation so that your mistake(s) can be pointed out.

q(t) = a / 1 + be^-kt

Quotient Rule: (Derivative of Top(Bottom) - Top(Derivative of Bottom)) / ((Bottom)^2)

Derivative of Top: 0
Derivative of Bottom: 0 + be^-kt (I think this is where my problem is)

q'(t) = (0(1+be^-kt)-a(be^-kt)) / ((1+be^-kt)^2)
= (-a(be^-kt)) / ((1+be^-kt)^2)
• Apr 10th 2010, 05:47 PM
mr fantastic
Quote:

Originally Posted by WahooMan
q(t) = a / 1 + be^-kt

Quotient Rule: (Derivative of Top(Bottom) - Top(Derivative of Bottom)) / ((Bottom)^2)

Derivative of Top: 0
Derivative of Bottom: 0 + be^-kt (I think this is where my problem is)

q'(t) = (0(1+be^-kt)-a(be^-kt)) / ((1+be^-kt)^2)
= (-a(be^-kt)) / ((1+be^-kt)^2)

Just to be clear: Are you saying that you do not know how to differentiate $1 + b e^{-kt}$ ? If that is the case, your best course of action is to go back to your class notes or textbook and review differentiation of basic exponential functions.
• Apr 10th 2010, 05:55 PM
WahooMan
Quote:

Originally Posted by mr fantastic
Just to be clear: Are you saying that you do not know how to differentiate $1 + b e^{-kt}$ ? If that is the case, your best course of action is to go back to your class notes or textbook and review differentiation of basic exponential functions.

q'(t) = (-a(-bke^-kt)) / ((1+be^-kt)^2)

That's it, isn't it? I'm feeling pretty good about that one.
• Apr 10th 2010, 06:01 PM
mr fantastic
Quote:

Originally Posted by WahooMan
q'(t) = (-a(-bke^-kt)) / ((1+be^-kt)^2)

That's it, isn't it? I'm feeling pretty good about that one.

Yes. And the numerator simplifies to $abk e^{-kt}$. Now go back and read the second sentence in post #2.
• Apr 10th 2010, 06:03 PM
WahooMan
Quote:

Originally Posted by mr fantastic
Yes. And the numerator simplifies to $abk e^{-kt}$. Now go back and read the second sentence in post #2.

The function will always be positive. Thanks.
• Apr 10th 2010, 06:15 PM
mr fantastic
Quote:

Originally Posted by WahooMan
The function will always be positive. Thanks.

Do you understand why b > 0 ....?