Chapter 3.4, Problem 151E

a.

To determine

To use : the graphing utility to graph the function

Explanation of Solution

Given information : The question is

The model for the data $T=20\left[1+72\left(2^{-h}\right)\right]$ and the table shown below

  

Graph : Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $20\left[1+72\left(2^{-h}\right)\right]$ which is required to graph.

Step 4: Press $\boxed{\text{GRAPH}}$ button to graph the function.

The graph is obtained as:

  

Interpretation :

From the above graph the red line shows the function $20\left[1+72\left(2^{-h}\right)\right]$

b.

To determine

To calculate : the horizontal asymptote of the function

Answer to Problem 151E

The horizontal asymptote is x-axis component to $\infty$

Explanation of Solution

Given information : The question is

The model for the data $T=20\left[1+72\left(2^{-h}\right)\right]$

Calculation :

The horizontal asymptote is at $T={20}^{\circ }$

If the object is placed in the room for long time the temperature of the object falls to room temperature

So, the horizontal asymptote of the function is given by substituting the value of x-axis component to $\infty$

c.

To determine

To calculate : when the temperature of the object is ${100}^{\circ }C$

Answer to Problem 151E

In $0.81hr$ the temperature of the object is ${100}^{\circ }C$

Explanation of Solution

Given information : The question is

The model for the data $T=20\left[1+72\left(2^{-h}\right)\right]$

Calculation :

  $\begin{array}{l}T=100\\ 100=20\left[1+72\left(2^{-h}\right)\right]\\ 5=1+7\left(2^{-h}\right)\\ 4=7\left(2^{-h}\right)\\ \frac{4}{7}=\left(2^{-h}\right)\\ \ln \frac{4}{7}=\ln \left(2^{-h}\right)\\ h=\frac{\ln \frac{4}{7}}{-\ln 2}\\ h=0.81hr\end{array}$

d.

To determine

How long does it take to cool to ${80}^{\circ }C$ one-half the initial temperature

Answer to Problem 151E

It takes $1.22hr$ to cool to ${80}^{\circ }C$ one-half the initial temperature

Explanation of Solution

Given information : The question is

The model for the data $T=20\left[1+72\left(2^{-h}\right)\right]$

Calculation :

  $\begin{array}{l}T=80\\ 80=20\left[1+72\left(2^{-h}\right)\right]\\ 4=1+7\left(2^{-h}\right)\\ 3=7\left(2^{-h}\right)\\ \frac{3}{7}=\left(2^{-h}\right)\\ \ln \frac{3}{7}=\ln \left(2^{-h}\right)\\ h=\frac{\ln \frac{3}{7}}{-\ln 2}\\ h=1.22hr\end{array}$