The park service that administers a state park estimates that there are 495 deer in the park. They decide to remove deer according to the differential equation dP/dt = −0.1P.

a. Show that the solution to the differential equation dP/dt = −0.1P is P = 495e^−0.1t , where t is measured in years and P is the population of deer. Use it to find the deer population in 5 years to the nearest deer.

b. After this 5-year period, no human intervention is taken and the deer population grows again. From that time, the deer population increases directly proportional to 650−P, where the constant of proportionality is k. Find an equation for the deer population P(t) in terms of t and k for this 5-year period.

c. Using the growth model from part b), 1 year later the deer population is 350. Find k.

d. Using the growth model from part b) and the value of k from part c), find lim t→∞ P(t)

The deer population in 5 years is approximately 294 deer.

A rise in the resultant quantity for a given quantity is referred to as exponential growth, and a decrease in the resultant quantity for a given quantity is referred to as exponential decay. Over time, exponential functions monitor constant growth. Examples from the actual world include radioactive decay, compound interest, and bacterial proliferation.

a. To solve the differential equation dP/dt = −0.1P, we can separate the variables and integrate:

dP/P = −0.1 dt

Integrating both sides gives:

ln|P| = −0.1t + C

where C is the constant of integration. To solve for C, we can use the initial condition that P = 495 when t = 0:

ln|495| = −0.1(0) + C

C = ln|495|

Substituting this value of C back into the equation gives:

ln|P| = −0.1t + ln|495|

Simplifying using logarithm properties, we get:

ln|P/495| = −0.1t

Exponentiating both sides gives:

|P/495| = e^−0.1t

Taking the absolute value off and multiplying both sides by 495, we get:

P = 495e^−0.1t

To find the deer population in 5 years, we can substitute t = 5 into the equation:

[tex]P = 495e^{-0.1(5)} = 294[/tex] deer (rounded to the nearest deer).

Therefore, the deer population in 5 years is approximately 294 deer.

b. To find an equation for the deer population P(t) in terms of t and k for the period after 5 years, we can use the differential equation:

dP/dt = k(650−P)

Separating the variables and integrating, we get:

∫dP/(650−P) = ∫k dt

Using partial fraction decomposition and integrating, we get:

ln|650−P| = −kt + C

where C is the constant of integration. To solve for C, we can use the initial condition that P = 294 when t = 5:

ln|650−294| = −k(5) + C

C = ln|356| + 5k

Substituting this value of C back into the equation gives:

ln|650−P| = −kt + ln|356| + 5k

Simplifying using logarithm properties, we get:

ln|(650−P)/356| = −kt

Exponentiating both sides gives:

[tex]|(650-P)/356| = e^{-kt[/tex]

Taking the absolute value off and multiplying both sides by 356, we get:

[tex]650-P = 356e^{-kt[/tex]

Solving for P gives:

[tex]P = 650-356e^{-kt[/tex]

Therefore, the equation for the deer population P(t) in terms of t and k for the period after 5 years is P = 650−356e^−kt.

c. Using the growth model from part b), if the deer population is 350 one year later, we can substitute t = 1 and P = 350 into the equation and solve for k:

350 = 650−356e^−k

Solving for [tex]e^{-k[/tex] gives:

[tex]e^{-k = (650-350)/356 = 3/2[/tex]

Taking the natural logarithm of both sides gives:

−k = ln(3/2)

Solving for k gives:

k = −ln(3/2) ≈ 0.4055

Therefore, k ≈ 0.4055.

d. Using the growth model from part b) and the value of k from part c), the long-term behavior of the deer population as t → ∞ is determined by the limit:

lim P(t) as t → ∞

We can use the fact that [tex]e^{-kt[/tex]approaches 0 as t →

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