Two bacterial populations, Q(t) and Z(t) are assumed to grow according to Euler exponential growth model, with parameters a and b. Suppose that the two populations are grown together in a beaker. W(t) = Q(t) / (Q(t) + Z(t)) to be the fraction of the total population of Q type. Using differential equations for Q and Z, show that W(t) satisfies a logistic growth equation.

Let Q(t) = Q(0)exp(at) Z(t) = Z(0)exp(bt) then W(t) = Q(0)exp(at)/(Q(0)exp(at) + Z(0)exp(bt)) = 1/[1 + c*exp((b-a)t)] where c = Z(0)/Q(0) This gives W(t)*(1 + c*exp((b-a)t)) = 1 and W(t) = 1 - W(t)*c*exp((b-a)t) and c*exp((b-a)t) = (1 - W(t))/W(t) Taking derivatives, we get W'(t) = -W'(t)*cexp((b-a)t) - W(t)*c*(b-a)*exp((b-a)t) or W'(t)(1 + c*exp((b-a)t)) = -W(t)*(b-a)*c*exp((b-a)t) or W'(t)/W(t) = -W(t)*(b-a)*(1 - W(t))/W(t) or W'(t) = -(b-a)*W(t)*(1-W(t)) This is the logistic differential equation No, you stick your head between your legs and kiss your *** goodbye!

Al No that's the difference between like and love. Joyce, you joined this board today and this is your first post. Jesus what a waste of time posting meaningless drivel.

Not only can I blind it into an alley dock, I can do with a road tractor and a 53' trailer slid all the way back during rush hour across three lanes of city traffic while smoking a cigar and singing along with Jimmy Buffett and not rub my tires.