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

Answer:

S(3)=22

Step-by-step 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).

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

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

When t=0, S(t)=15

$15=30+Ce^{-k*0}\\C=15-30\\C=-15$

When t = 2, S(t)=20

$20=30-15e^{-2k}\\20-30=-15e^{-2k}\\-10=-15e^{-2k}\\e^{-2k}=\dfrac23\\ Take the natural log of both sides \\-2k=\ln \dfrac23\\k=-\dfrac{\ln(2/3)}{2}$

Therefore:

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