Chapter 3.4, Problem 135E

a.

To determine

To find: the graph of the function $T=20\left[1+7(2^{-h})\right]$ .

Explanation of Solution

Given information :

Given function $T=20\left[1+7(2^{-h})\right]$ .

Graph:

  

b.

To determine

the horizontal asymptote of the function $T=20\left[1+7(2^{-h})\right]$ . And its meaning in the context of the problem.

Answer to Problem 135E

Horizontal asymptote is $h$ -axis .

Explanation of Solution

Given information :

Given function $T=20\left[1+7(2^{-h})\right]$ .

The horizontal asymptote is the of $T(h)$ for large $h$ . That is,

  $\begin{array}{l}\underset{h\to \infty }{\lim }T(h)\\ =\underset{h\to \infty }{\lim }20\left[1+7(2^{-h})\right]\\ =20\underset{h\to \infty }{\lim }\left[1+7(2^{-h})\right]\\ =20\left[1+7\underset{h\to \infty }{\lim }\frac{1}{2^h}\right]\\ =20+\left(20\times 7\times \frac{1}{2^{\infty }}\right)\\ =20+\left(20\times 7\times \frac{1}{0}\right)\\ =\infty \end{array}$

Hence the horizontal asymptote is at $h=\infty$ that is the whole $h$ -axis. That is the temperature will decrease to $h$ -axis but will not meet. That is the temperature will never zero degree Celsius.

c.

To determine

To find: the time $T$ to the object cool to ${80}^{\circ }C$ .

Answer to Problem 135E

The time required $29.33$ years.

Explanation of Solution

Given information :

Given function $T=20\left[1+7(2^{-h})\right]$ .

When $T(h)={80}^{\circ }C$ we get,

  $\begin{array}{l}T=20\left[1+7(2^{-h})\right]=80\\ \Rightarrow 1+7(2^{-h})=4\\ \Rightarrow 7(2^{-h})=3\\ \Rightarrow 2^{-h}=\frac{3}{7}\\ \Rightarrow \frac{1}{2^h}=\frac{3}{7}\\ \Rightarrow 2^h=\frac{7}{3}\\ \Rightarrow h\log (2)=\log (7/3)\\ \Rightarrow h=\frac{\log (7/3)}{\log (2)}=1.22\text{ (Correct up to two decimal places)}\end{array}$

So the time required $1.22$ hours.