Another model for a growth function for a limited pupulation is given by the Gom

Question

Another model for a growth function for a limited pupulation is given by the Gompertz function, which is a solution of the differential equation [rac{dP}{dt} = c ln left( rac{K}{P} ight) P] where (c) is a constant and (K) is the carrying capacity.

(a) Solve this differential equation for (c=0.1), (K=1000), and initial population (P_0=300).
(P(t)=) .

(b) Compute the limiting value of the size of the population.
(displaystyle lim_{t o infty} P(t) =) .

(c) At what value of (P) does (P) grow fastest?
(P=

Explanation / Answer

dP/dt = c ln(K/P)P dP/(ln(K/P)P) = cdt dP/( (ln K - ln P)P ) = cdt put (lnK - lnP) = x => -(1/P)dP = dx dP/( (ln K - ln P)P ) = cdt => -(1/x)dx = cdt integrating both sides => x = ke^-ct ( k is some constant ) => lnK - lnP = ke^-ct putting t = 0 => lnK - lnPo = k => k = ln(K/Po) so, lnK - lnP = ln(K/Po)e^-ct lnP = lnK - ln(K/Po)e^-ct P = e^(lnK - ln(K/Po)e^-ct) P = e^lnK*{e^-ln(K/Po)}^e^-ct P = K*(Po/K)^(e^-ct) a) P(t) = 1000*(0.3)^(e^-0.1t) b) lim t->+inf P(t) = 1000*(0.3)^(0) = 1000 c) dP/dt = c ln(K/P)P d^2P/dt^2 = c ln(K/P) + c (P/K)(-K/P^2)P d^2P/dt^2 = c ln(K/P) - c for max rate, d^2P/dt^2 = 0 c ln(K/P) - c = 0 ln(K/P) = 1 K/P = e => P = K/e = 1000/e => P = 367.879 or 368 (aprrox)

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