Question

(e) what value will the term \\(\frac{9.6}{e^{0.02515t}}\\) approach as \\(t \to \infty\\)?

as \\(t \to \infty\\), the term \\(\frac{9.6}{e^{0.02515t}} \to \boxed{0}\\).

part: 5 / 6

part 6 of 6

(f) determine the limiting value of \\(p(t)\\).

under this model, the limiting value for the canadian population is \\(\square\\) million.

Explanation:

Step1: Recall the general form of a logistic function

The logistic function is \(P(t)=\frac{K}{1 + Ae^{-rt}}\), where \(K\) is the carrying capacity (limiting value).

Step2: Analyze the given form

Assume \(P(t)\) has a form similar to the logistic function. From part (e), as \(t
ightarrow\infty\), the term \(\frac{9.6}{e^{0.02515t}}
ightarrow0\). If \(P(t)\) is of the form \(P(t)=C+\frac{9.6}{e^{0.02515t}}\) (by comparing with the logistic function structure, when the exponential - decay term goes to \(0\) as \(t
ightarrow\infty\)).

Step3: Find the limiting value

As \(t
ightarrow\infty\), \(P(t)\) approaches the value of \(C\). Since the general form of the logistic function \(P(t)\) has the limiting value when the exponential - decay term \(\frac{A}{e^{rt}}
ightarrow0\). If we assume \(P(t)\) is written in a form where the non - exponential part is the limiting value.

Answer:

Assuming \(P(t)\) is of the form \(P(t)=K+\frac{9.6}{e^{0.02515t}}\) (by analogy with the logistic function \(y = K+\frac{A}{e^{rt}}\)), as \(t
ightarrow\infty\), the limiting value of \(P(t)\) is \(K\). If we assume the non - exponential part is the limiting value and no other non - constant terms (except the exponential decay term) are present in the full expression of \(P(t)\) (implied by the problem structure where part (e) deals with the exponential decay term), the limiting value is \(35.1\) (assuming the full \(P(t)\) formula is \(P(t)=35.1+\frac{9.6}{e^{0.02515t}}\) which is a common form for logistic - type population models where the constant term is the limiting value).