the squirrel population in a small self-contained forest was studied by a biologist. the biologist found that the squirrel population, P, measured in hundreds, is a function of time, t, where t is measured in weeks. the function P(t)=20/(1+3e^0.02t). the biologists study the squirrel population for 52 weeks.
a) determine the maximum population of squirrels over the first 52 weeks. (remember to check the endpoints and the critical point)
b) as time goes to infinity what will the limit of P(t) be? (you should be able to do this without a calculator)
c) how does what you discovered in b) confirm what you learned in a?
e) determine the point of inflection.
f) explain the meaning of the point of inflect in terms of the squirrel population growth.
P(t)=20/(1+3e^0.02t)
= 20(1+3e^0.02t)^-1
P ' (t) = -20(1+3e^0.02t)^-2 (.06)(e^(.02t) )(.02)
= -.024(e^(.02t) ) / (1+3e^0.02t)^2
check my steps, I should have written it out on paper
a) To determine the maximum population of squirrels over the first 52 weeks, we need to find the critical point and evaluate the function at the endpoints.
First, let's find the critical point by finding when the derivative of P(t) equals zero. The derivative of P(t) with respect to t can be found using the quotient rule:
P'(t) = [20 * (0.02) * e^(0.02t)] / (1 + 3e^(0.02t))^2
Setting this derivative equal to zero:
0 = [20 * (0.02) * e^(0.02t)] / (1 + 3e^(0.02t))^2
To simplify this equation, we can multiply both sides by (1 + 3e^(0.02t))^2:
0 = 20 * (0.02) * e^(0.02t)
Now, divide both sides by 20 and (0.02):
0 = e^(0.02t)
Since e^(0.02t) is always positive, there is no value of t that makes the equation true. Therefore, there is no critical point within the given time frame.
Next, let's evaluate P(t) at the endpoints, t = 0 and t = 52:
P(0) = 20 / (1 + 3e^(0.02 * 0)) = 20 / (1 + 3) = 20 / 4 = 5
P(52) = 20 / (1 + 3e^(0.02 * 52)) ≈ 20 / (1 + 3e^(1.04)) ≈ 20 / (1 + 3 * 2.831) ≈ 20 / (1 + 8.493) ≈ 20 / 9.493 ≈ 2.109
Comparing the population values, we can see that the maximum population of squirrels over the first 52 weeks occurs at t = 0, where P(t) = 5.
b) To find the limit of P(t) as time goes to infinity, we need to examine the behavior of the exponential term e^(0.02t).
As t approaches infinity, the exponent 0.02t also approaches infinity. When the exponent approaches infinity, e^(0.02t) also approaches infinity. Therefore, the limit of P(t) as t approaches infinity is:
lim(t→∞) P(t) = 20 / (1 + 3(∞)) = 20 / (1 + ∞) = 20 / ∞ = 0
c) The result in part b) confirms what we learned in part a) because as time goes to infinity, the squirrel population approaches zero. This means that the maximum population of squirrels occurs at the beginning (t = 0) and decreases as time goes on.
e) To determine the point of inflection, we need to find where the second derivative of P(t) equals zero. Let's find the second derivative of P(t):
P''(t) = [20 * (0.02)^2 * e^(0.02t)] / (1 + 3e^(0.02t))^3 - 2 * [20 * (0.02) * e^(0.02t)]^2 / (1 + 3e^(0.02t))^4
Setting this second derivative equal to zero:
0 = [20 * (0.02)^2 * e^(0.02t)] / (1 + 3e^(0.02t))^3 - 2 * [20 * (0.02) * e^(0.02t)]^2 / (1 + 3e^(0.02t))^4
To simplify this equation, we can multiply both sides by (1 + 3e^(0.02t))^3 and cancel out the common terms:
0 = 20 * (0.02)^2 * e^(0.02t) - 2 * [20 * (0.02) * e^(0.02t)]^2 / (1 + 3e^(0.02t))
Now, let's multiply through by (1 + 3e^(0.02t))^4 to eliminate the denominator:
0 = 20 * (0.02)^2 * e^(0.02t) * (1 + 3e^(0.02t)) - 2 * [20 * (0.02) * e^(0.02t)]^2
Simplifying further, we have:
0 = 0.2 * e^(0.02t) * (1 + 3e^(0.02t)) - 0.08 * e^(0.04t)
Dividing both sides by 0.2 * e^(0.02t), we get:
0.08 * e^(0.02t) = 1 + 3e^(0.02t)
0.08 * e^(0.02t) - 1 - 3e^(0.02t) = 0
Unfortunately, this equation does not have a simple algebraic solution. We would need to use numerical methods to approximate the point of inflection.
f) The point of inflection represents a change in the concavity of the squirrel population growth. At the point of inflection, the rate of change of the population transitions from increasing to decreasing (convex to concave) or vice versa. In terms of the squirrel population growth, the point of inflection indicates a shift in the growth rate.