no use for derivatives here.
90% of capacity is 7200 fish
So, just solve for t when P(t) = 5000 or 7200
When a reservoir is created by a dam,50 fish are introduced into the reservoir ,which has an estimated carrying capacity of 8000 fish.
A logistic model of the population is
P(t)= 400,000 / 50 + 795e^-0.5t , where t is measured in years.
How long does it take for the population to reach 5000 fish and 90% of the carrying capacity?
Should I use the derivatives to calculate it?
4 answers
Oo..I see..but what to do with e?
come on, guy. You must have solved exponential problems before, no?
5000 = 400,000 / (50 + 795e^-0.5t)
50 + 795e^-.5t = 80
795e^-.5t = 30
e^-.5t = 30/795
-.5t = log(30/795)
t = -2log(30/795) = 6.55
5000 = 400,000 / (50 + 795e^-0.5t)
50 + 795e^-.5t = 80
795e^-.5t = 30
e^-.5t = 30/795
-.5t = log(30/795)
t = -2log(30/795) = 6.55
Sorry,totally forgot about that..
Thanks^^
Thanks^^