Differential Equations: Population Dynamics

[Kelsey]

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
wher c is a constant and K is the carrying cappacity.

At what value of P does P grow fastest?

I already know
c=.15
K=1000
P(t)= 1000/(e^(e^(-.15t+ln(ln(2)))))
the limit of P(t) when t -> infinity is 1000

please help?

 

[soroban]

Hello, Kelsey!

Another model for a growth function for a limited pupulation is given by the Gompertz function,
which is a solution of the differential equation: .\(\displaystyle \dfrac{dP}{dt}\:=\:cP\ln\left(\dfrac{K}{P}\right)\)
where \(\displaystyle c\) is a constant and \(\displaystyle K\) is the carrying capacity.

At what value of \(\displaystyle P\) does \(\displaystyle P\) grow fastest?

I already know: .\(\displaystyle c\,=\,0.15,\;K\,=\,1000\)

Note that \(\displaystyle \dfrac{dP}{dt}\) is the growth of \(\displaystyle P.\)
. . Hence, we want to maximize \(\displaystyle \dfrac{dP}{dt}.\)


We have: .\(\displaystyle \dfrac{dP}{dt} \:=\:0.15P\ln\left(\frac{1000}{P}\right) \:=\:0.15P\big[\ln 1000 - \ln P\big] \)

Then: .\(\displaystyle \dfrac{d^2P}{dt^2} \:=\:0.15\bigg[(\ln1000 - \ln P) + P\left(-\dfrac{1}{P}\right)\bigg] \:=\:0 \)

. . . . . \(\displaystyle \ln\left(\dfrac{1000}{P}\right) - 1 \:=\:0 \quad\Rightarrow\quad \ln\left(\dfrac{1000}{P}\right) \:=\:1 \)

. . . . . \(\displaystyle \dfrac{1000}{P} \:=\:e \quad\Rightarrow\quad P \:=\:\dfrac{1000}{e}\)