# Differential equation- find max slope

1. Oct 11, 2010

### hover

1. The problem statement, all variables and given/known data

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 the carrying capacity.

At what value of P does P grow fastest?

2. Relevant equations

c = .05
K=1000
P_0 = 500 (initial condition)

P(t) = 1000/e^(e^(-.05t-.3665)) (this is the specific solution)

3. The attempt at a solution

I think it is asking to find what the max value of dP/dt. However I think to do this, I need to find the derivative of dP/dt which is d^2P/dt^2 and set it equal to zero. This will let me find t when the slope of P is at its max and then I plug t back into P. Is this correct?

If it is, I am in for one hell of a derivative...

Thanks!

2. Oct 11, 2010

### HallsofIvy

Why solve for P? You were asked for a value of P for which P grows fastest which mean its derivative is a maximum. You can find a maximum of a function by setting its derivative to 0. Here, the function is dP/dt so differentiate that:
$$\frac{d}{dt}\frac{dP}{dt}= \frac{d^2P}{dt^2}$$$$= C\frac{d(P(ln(K)- ln(P))}{dt}= 0$$

3. Oct 11, 2010

### hover

Ok. Solving that derivative I get t = -7.33 and in the end P has a max slope at P= 367.879. These are right.

Thanks!