The full quesiton is "The population of a certain species of fish introduced into a lake is described by the logistic equation G(t) = (12,000) / 1+39e^(-1.2t) where G(t) is the population after t years. Find the point at which the growth rate of this population begins to decline."
To solve- I know you should take the derivate and set equal to zero. I have (3.97, 9000) as the answer but not sure it is correct and I cannot recreate the answer.
What is the derivative of (12,000) / 1+39e^(-1.2t)
3 answers
g = 12000/(1+e^(-1.2t))
= 12000 * (1+e^(-1.2t))^-1
g' = (12000)(-1)((1+e^(-1.2t))^-2 * (-1.2 e^(-1.2t))
= 14400e^(-1.2t)/((1+e^(-1.2t))^2
Now, g' is the growth rate. It starts to decline when g'' changes from positive to negative.
g'' = 17280*e^1.2t*(1 - e^1.2t)/((1+e^(-1.2t))^3
g'' = 0 where 1 = e^1.2t, or t=0
Hmmm. There's something wrong here. g' is always positive, meaning that the population is always growing.
However, g'' is negative for t>0, meaning that the growth rate is always slowing down. In fact, as t grows large, g(t) approaches 12000.
In fact the graph of g(t) displays this behavior, with fastest growth at t=0. Is there a typo somewhere?
= 12000 * (1+e^(-1.2t))^-1
g' = (12000)(-1)((1+e^(-1.2t))^-2 * (-1.2 e^(-1.2t))
= 14400e^(-1.2t)/((1+e^(-1.2t))^2
Now, g' is the growth rate. It starts to decline when g'' changes from positive to negative.
g'' = 17280*e^1.2t*(1 - e^1.2t)/((1+e^(-1.2t))^3
g'' = 0 where 1 = e^1.2t, or t=0
Hmmm. There's something wrong here. g' is always positive, meaning that the population is always growing.
However, g'' is negative for t>0, meaning that the growth rate is always slowing down. In fact, as t grows large, g(t) approaches 12000.
In fact the graph of g(t) displays this behavior, with fastest growth at t=0. Is there a typo somewhere?
Steve,
The problem has a 39 in front of the "e".
The problem has a 39 in front of the "e".
5 answers