Question

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t), where t is the time in years. When t = 0, the population was 15, and when t = 2, the population increased to 20. Find the population when t = 3.

Answer 1

S(3)=22

Explanation:

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t).

`(dS)/(dt)=k(30-S(t))\\ (dS)/(dt)+kS(t)=30k\\$The integrating factor: e^(\int k dt)=e^(kt)\\$Multiply all through by the integrating factor\\ (dS)/(dt)e^(kt)+kS(t)e^(kt)=30ke^(kt)`

`(Se^(kt))'=30ke^(kt) dt\\$Integrate both sides\\ Se^(kt)=(30ke^(kt))/(k)+C$ (C a constant of integration)\\Se^(kt)=30e^(kt)+C\\$Divide both sides by e^(kt)\\S(t)=30+Ce^(-kt)`

When t=0, S(t)=15

`15=30+Ce^(-k*0)\\C=15-30\\C=-15`

When t = 2, S(t)=20

`20=30-15e^(-2k)\\20-30=-15e^(-2k)\\-10=-15e^(-2k)\\e^(-2k)=\frac23\\$Take the natural log of both sides$\\-2k=\ln \frac23\\k=-(\ln(2/3))/(2)`

Therefore:

`S(t)=30-15e^{(\ln(2/3))/(2)t}\\$When t=3$\\S(t)=30-15e^{(\ln(2/3))/(2) * 3}\\S(3)=21.8 \approx 22`