Q. 88 - Calculus

Question

In Exercises 87–89, suppose that Leila is a population biologist with the Idaho Fish and Game Service. Wolves were introduced formally into Idaho in 1994, but there were some wolves in the state before that. Leila has been assigned the task of estimating the rate at which the number of wolves in Idaho increased naturally, before the animals were introduced. The only information she has is a population model, which indicates that the wolf population currently satisfies the formula

                             $w\left(t\right)=835\left(1-e^{-.006t}\right),$

where $t$ is the number of years since $1994$.

To approximate the rate of increase of wolves per year at the beginning of $1994$, Leila decides to take the limit of $\frac{w\left(t\right)}{t}$ as $t\to 0$. Why does this approach make sense, and what is the value of that limit? Is there another way she could find the same number?

Step-by-Step Solution

Answer

Since the average rate of change of wolves satisfies the mean value theorem, Leila's reasoning makes sense. The value of the limit $\underset{t\to 0}{\lim }\frac{w\left(t\right)}{t}$ is, $5.01$.

Step 1 . Given information

The Wolf population currently is,

$w\left(t\right)=835\left(1-e^{-.006t}\right),$

where $t$ is the number of years since $1994$.

The average rate of change of wolves from $1994$  to a given time $t$ is given by $\frac{w\left(t\right)}{t}$.

Step 2 . To decide whether the approach makes sense or not , use mean value theorem.

From the mean value theorem,

$\frac{\Delta w}{\Delta t}=\frac{w\left(t\right)-w\left(0\right)}{t-0}$

        $=\frac{w\left(t\right)-0}{t}$   [ Since, $w\left(0\right)=0$]

         $=\frac{w\left(t\right)}{t}$

The value of the limit is,

$\underset{t\to 0}{\lim }\frac{w\left(t\right)}{t}=\underset{t\to 0}{\lim }\frac{835\left(1-e^{-0.006t}\right)}{t}$  [ in the form of $\frac{0}{0}$]

                $=\underset{t\to 0}{\lim }\frac{835\left(0-\left(e^{-0.006t}(-.006)\right)\right)}{1}$  [ Using L'Hopital's rule]

                  $$\begin{gathered} =\underset{t\to 0}{\lim }\left(835·0.006e^{-0.006t}\right) \\ =835\times 0.006e^0 \\ =5.01\times 1 \\ =5.01 \end{gathered}$$

Thus, Leila's reasoning makes sense.