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love history [14]
3 years ago
13

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.
Mathematics
1 answer:
pishuonlain [190]3 years ago
8 0

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

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Explanations:

Part (a)

In this case, the vertex is the highest point, so the vertex of function 1 is (-3, -1)

If the parabola was flipped so that it opened upward, instead of downward, then the vertex would be the lowest point.

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Part (b)

We could graph this function to see where the highest or lowest point is, but there's another way we could find the vertex using algebra.

The given equation is in the form y = ax^2 + bx + c where in this case

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----------------------------

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