Q. 75 - Calculus

Question

There are four squirrels currently living in Linda’s attic. If she does nothing to evict these squirrels, the number of squirrels in her attic after t days will be given by the formula

$S\left(t\right)=\frac{12+5.5t}{3+0.25t}.$

(a) Verify that there are four squirrels in Linda’s attic at time t = 0.

(b) Determine the number of squirrels in Linda’s attic after 30 days, 60 days, and one year.

(c) Approximate $\underset{t\to \infty }{\lim }S\left(t\right)$ with a table of values. What does this limit mean in real-world terms?

(d) Graph S(t) with a graphing utility, and use the graph to verify your answer to part (c).

Step-by-Step Solution

Answer

Part (a) It is verified that there are four squirrels in Linda’s attic at time t = 0. 

Part (b) The number of squirrels in Linda's attic after 30 days is $17,$ after 60 days is $19,$ and after one year is $21.$

Part (c) After approximating $\underset{t\to \infty }{\lim }S\left(t\right)$ the value is $22.$

Part (d) The answer is verified and the graph of S(t)  with a graphing utility is

Part (a) Step 1. Given Information.

The given formula is $S\left(t\right)=\frac{12+5.5t}{3+0.25t}.$

Part (a) Step 2. Verifying that there are four squirrels in Linda’s attic at time t = 0 .

To verify that there are four squirrels in Linda’s attic at time t = 0. 

Take the limit to the given formula.

$$\begin{gathered} \underset{t\to 0}{\lim }S\left(t\right)=\underset{t\to 0}{\lim }\frac{12+5.5t}{3+0.25t} \\ \underset{t\to 0}{\lim }S\left(t\right)=\frac{12+5.5\left(0\right)}{3+0.25\left(0\right)} \\ \underset{t\to 0}{\lim }S\left(t\right)=\frac{12}{3} \\ \underset{t\to 0}{\lim }S\left(t\right)=4 \end{gathered}$$

Hence it is proved that there are four squirrels in Linda's attic at the time $t=0.$

Part (b) Step 1. Determining the number of squirrels in Linda’s attic after 30 days.

To determine the number of squirrels in Linda’s attic after 30 days.

Take the limit $t\to 30$ to the given formula.

$$\begin{gathered} \underset{t\to 30}{\lim }S\left(t\right)=\underset{t\to 30}{\lim }\frac{12+5.5t}{3+0.25t} \\ \underset{t\to 30}{\lim }S\left(t\right)=\frac{12+5.5\left(30\right)}{3+0.25\left(30\right)} \\ \underset{t\to 30}{\lim }S\left(t\right)=\frac{12+165}{3+7.5} \\ \underset{t\to 30}{\lim }S\left(t\right)=\frac{177}{10.5} \\ \underset{t\to 30}{\lim }S\left(t\right)\approx 16.86 \\ \underset{t\to 30}{\lim }S\left(t\right)\approx 17 \end{gathered}$$

Part (b) Step 2. Determining the number of squirrels in Linda’s attic after 60 days.

To determine the number of squirrels in Linda’s attic after 60 days.

Take the limit $t\to 60$ to the given formula.

$$\begin{gathered} \underset{t\to 60}{\lim }S\left(t\right)=\underset{t\to 60}{\lim }\frac{12+5.5t}{3+0.25t} \\ \underset{t\to 60}{\lim }S\left(t\right)=\frac{12+5.5\left(60\right)}{3+0.25\left(60\right)} \\ \underset{t\to 60}{\lim }S\left(t\right)=\frac{12+330}{3+15} \\ \underset{t\to 60}{\lim }S\left(t\right)=\frac{342}{18} \\ \underset{t\to 60}{\lim }S\left(t\right)=19 \end{gathered}$$

Part (b) Step 3. Determining the number of squirrels in Linda’s attic after one year.

To determine the number of squirrels in Linda’s attic after one year.

Take the limit $t\to 365$ to the given formula.

$$\begin{gathered} \underset{t\to 365}{\lim }S\left(t\right)=\underset{t\to 365}{\lim }\frac{12+5.5t}{3+0.25t} \\ \underset{t\to 365}{\lim }S\left(t\right)=\frac{12+5.5\left(365\right)}{3+0.25\left(365\right)} \\ \underset{t\to 365}{\lim }S\left(t\right)=\frac{12+2007.5}{3+91.25} \\ \underset{t\to 365}{\lim }S\left(t\right)=\frac{2019.5}{94.25} \\ \underset{t\to 365}{\lim }S\left(t\right)\approx 21.43 \\ \underset{t\to 365}{\lim }S\left(t\right)\approx 21 \end{gathered}$$

Part (c) Step 1. Approximating lim t → ∞ S ( t ) with a table of values.

Let's choose a sequence of values approaching $t=\infty$ to approximate $\underset{t\to \infty }{\lim }S\left(t\right)$ with a table of values.

t	10	100	1000	10000
S(t)	12.18	20.07	21.79	21.98

Thus, the values of S(t) approach to 22 as $t\to \infty .$

Hence, $\underset{t\to \infty }{\lim }S\left(t\right)=22.$

The limit means in the real world is that the limit tells us that a function approaches as that function's inputs get nearer and nearer to some number.

Part (d) Step 1. Sketch of a graph S(t) by using a graphing utility.

To graph S(t) by using a graphing utility, insert the function and adjust the window.

The graph is

From the graph, we can depict that for $t\to \infty$ the value of the function tends to $22.$

Hence the answer is verified.