"P(t)= \\frac{14,250}{1+29e^{-0.62t}}."
75% = 3 / 4. By what number do you divide 14250 so that you get 3/4 of it? By 4/3. So,
"1+29e^{-0.62t} = \\frac{4}{3}"
"29e^{-0.62t} = \\frac{1}{3}"
"e^{-0.62t} = \\frac{1}{87}"
ln(1/87) = -0.62t Rounded to the nearest tenth t=ln(1/87)/(-0.62)=7.2.
the carrying capacity is the upper bound of the logistic function. If you express a logistic function as "\\frac{c}{1 + b^x}" the carrying capacity is c.
In this problem, the carrying capacity is 14,250
It is the population at time zero, when t=0:
"14250\/(1+29*e^0)=14250\/(1+29*1)=475"
The answer to this question should really be tuned in to your classroom discussions, because many explanations are possible. Basically, the difference between "P(t)=P_0 \\ e^{Kt} ," and the logistic function is that the first is unlimited exponential growth and the second is growth in the realistic conditions of limited resources. One virtue of the logistic model is that it is realistic, because we live in the reality of limited resources.
Comments
Leave a comment