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Sonprelis August 6th, 2014 03:02 AM

Gompertz equation - differential equation
 
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Hello everyone!

Could someone please help me on this one?

How do you get from the first equation to the second (see uploaded image)?
Is the second equation the first derivative of the differential equation? In this case, how would you thus find the second derivative of this differential equation?

Thank you very much!! ;)

ZardoZ August 6th, 2014 03:24 AM

You should have in mind that N is changing with respect to t, so it is N(t).

$\displaystyle \frac{\mathbb{d}}{\mathbb{d}t}\left[N(t)\ln\left(\frac{K}{N(t)}\right)\right]=r\cdot\frac{\mathbb{d}}{\mathbb{d}t}\left[N(t)\ln\left(\frac{K}{N(t)}\right)\right]=r\left[N'(t)\ln\left(\frac{K}{N(t)}\right)+N(t)\ln'\left( \frac{K}{N(t)}\right)\right]$

Sonprelis August 6th, 2014 03:35 AM

Quote:

Originally Posted by ZardoZ (Post 202368)
You should have in mind that N is changing with respect to t, so it is N(t).

$\displaystyle \frac{\mathbb{d}}{\mathbb{d}t}\left[N(t)\ln\left(\frac{K}{N(t)}\right)\right]=r\cdot\frac{\mathbb{d}}{\mathbb{d}t}\left[N(t)\ln\left(\frac{K}{N(t)}\right)\right]=r\left[N'(t)\ln\left(\frac{K}{N(t)}\right)+N(t)\ln'\left( \frac{K}{N(t)}\right)\right]$



Thank you very much for your answer, I am still somewhat confused though. I've never really studied differential equations and have chosen to do so for a mathematical exploration. Could you thus please expand on the explanation? Why and how is it that it goes from one form to the other?
I would be infinitely grateful for your help! :)

ZardoZ August 6th, 2014 03:37 AM

Do you know anything about differentiation rules?

$\displaystyle \left(f(t)g(t)\right)'=f'(t)g(t)+f(t)g'(t)$

in our case $f(t)=N(t)$ and $g(t)=\ln\left(\frac{K}{N(t)}\right)$, the ' implies $\frac{d}{dt}$.

Sonprelis August 6th, 2014 09:50 AM

Now it's clear, thank you :D

However, I still wanted to know... when we apply the d/ dt to the equation rN ln(K/N),
what is it exactly that we're doing?
Are we finding the first derivative of the equation? Can this be done with a differential equation?

The original equation is indeed a differential one:

dN / dt = rN ln (K/N)

and then we have to apply the d/dt to it:

d/dt [rN ln (K/N)]

ZardoZ August 6th, 2014 10:02 AM

first derivative.
 
Rate of change with respect to t (probably time here)!

Sonprelis August 6th, 2014 10:07 AM

Quote:

Originally Posted by ZardoZ (Post 202406)
Rate of change with respect to t (probably time here)!


...and how would the second derivative look like for the following function?

dN / dt = rN ln (K/N)

The first derivative we now know to be:

r[dN/dt ln(K/N) + N d/dt ln(K/N)]


Thank you very much for your kind help!! :);)


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