How to satisfy a logistic growth equation?

Joyce Sullivan

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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.
 
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.

does that mean u swallow or spit...........:hysterical:
 
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

does that mean u swallow or spit...........:hysterical:
No, you stick your head between your legs and kiss your *** goodbye!
 
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.
 
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.

now i would like to see that one.
 
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