Finding a population growth function

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jjm5119

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Another model for a growth function for a limited pupulation is given by the Gompertz function, which is a solution of the differential equation dP/dt=c ln(K/P)*P where c is a constant and K is carrying the capacity.

a) solve this differential equation for c=.2, k=5000, and initial population P(0)=500
Find P(t)

b)At what value of P does P grow fastest?

So i know that P(t)=P(0)e^(kt)
So i subbed in value of P(0) and k and came up with P(t)=500*e(5000t) and that answer is not correct, obviously because I didn't incorporate the constant c at all. Basically I'm pretty much lost. Any help would be appreciated.
 
You can re-write the D.E. as \(\displaystyle \frac{d}{dt} ln(ln(P/K)) = -c\).
 
jjm5119 said:
Another model for a growth function for a limited pupulation is given by the Gompertz function, which is a solution of the differential equation

dP/dt=c ln(K/P)*P where c is a constant and K is carrying the capacity.
Click to expand...
If your problem is

dP/dt=c ln(K/P)*P

dP/(p*ln(p/k) = -c dt

d[ln(p/k)]/ln(p/k) = -c/k * dt

ln[M*ln(p/k)] = -ct/k <<<<< Ln(M) is the constant of integration

Now continue.....
 
This doesn't really make sense. Could you please explain what to solve for or do from there? You don't have to give me any more work, just an explanation will do. Thanks a lot!
 
This is how I have seen the Gompertz curve equation. It is a logistics equation.

From \(\displaystyle \frac{dP}{dt}=P(a-bln(P))\) we separate variables and get

\(\displaystyle \frac{-1}{b}ln|a-bln(p)|=t+c_{1}\)

Solving for P, we get \(\displaystyle P=e^{\frac{a}{b}}e^{-ce^{-bt}}\)

Now, use your initial conditions to find c.
 
you're confusing me by using a and b. is a=c and b=k or is a=k and b=c
 
I'm sorry. That's just how I am used to doing it.

If you separate variables, you get:

\(\displaystyle \frac{dP}{Pln(\frac{5000}{P})}=\frac{1}{5}dt\)

Now integrate and get:

\(\displaystyle -ln(ln(\frac{5000}{P}))=\frac{1}{5}t+C\)

Now, solve for P:

\(\displaystyle P=5000e^{-e^{\frac{-t}{5}-C}}\)

Now, use P(0)=500 and find C.

This gives you \(\displaystyle P=5000(\frac{1}{10})^{e^{\frac{-t}{5}}}\)
 
genius. how am i to go about finding the value of P that makes P grow fastest?
 
jjm5119 said:
genius. how am i to go about finding the value of P that makes P grow fastest?
Click to expand...
If you don't know how to do this, then it may possibly be you do not have the prerequisite for this class.

Tell us - how does one find fastest growth rate of a given function (with one independant variable) - say f(t)?
 
Wouldn't you just take the first derivative and set it equal to zero like so?

d/dp(cln(k/p)p)

= clnk - clnp - c

clnk - clnp - c = 0

lnk - lnp - 1 = 0

p = e^(lnk - 1)

p = k + 1/e

Does anybody know if this is right?
 
I don't know what happened to the replies after this post.

Anyway, looks good to me....
 
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