Chapter 7, Problem 54EP

a.

To determine

To find:The average temperature to the nearest Fahrenheit between $t=6\text{ and }14$ .

Answer to Problem 54EP

The average temperature is $87$ .

Explanation of Solution

Given information:

  $F\left(t\right)=80-10\cos \left(\frac{\pi t}{12}\right)$

Calculation:

The formula to find the average temperature is given by,

  $\frac{1}{b-a}{\displaystyle {\int }_a^{b}F}(t)dt$

The given function is,

  $\begin{array}{c}F(t)=80-10\cos \left(\frac{\pi t}{12}\right)\\ a=6\\ b=14\end{array}$

Substituting the values in the average equation to get,

  $\begin{array}{c}Avg=\frac{1}{b-a}{\displaystyle {\int }_a^{b}F}(t)dt\\ =\frac{1}{14-6}{\displaystyle {\int }_6^{14}8}0-10\cos \left(\frac{\pi t}{12}\right)dt\\ =\frac{1}{8}{\displaystyle {\int }_6^{14}\left(80-10\cos \left(\frac{\pi t}{12}\right)\right)}dt\end{array}$

Solving the integral,

  $\begin{array}{c}Avg=\frac{1}{8}{\displaystyle {\int }_6^{14}\left(80-10\cos \left(\frac{\pi t}{12}\right)\right)}dt\\ =\frac{1}{8}{\displaystyle {\int }_6^{14}8}0{\Vert }^2-\frac{1}{8}{\displaystyle {\int }_6^{14}1}0\cos \left(\frac{\pi t}{12}\right)dt\\ =Av{g}_1-Av{g}_2\end{array}$

To find the value of $Av{g}_1$

  $\begin{array}{c}Av{g}_1=\frac{1}{8}{\displaystyle {\int }_6^{14}8}0dt\\ ={\left.\frac{1}{8}(80t)\right|}_6^{14}\\ =\frac{1}{8}[80(14)-80(6)]\\ =\frac{1}{8}(1120-480)\\ =\frac{1}{8}(640)\\ Av{g}_1=80\end{array}$

To find the value of $Av{g}_2$

  $Av{g}_2=\frac{1}{8}{\displaystyle {\int }_6^{14}1}0\cos \left(\frac{\pi t}{12}\right)dt$

Using substitution method,

  $\begin{array}{c}u=\frac{\pi t}{12}\\ du=\frac{\pi }{12}dt\\ \frac{12}{\pi }du=dt\end{array}$

To find the new boundaries:

At $t=14$ the upper boundary is,

  $\begin{array}{c}u=\frac{\pi (14)}{12}\\ =\frac{7\pi }{6}\end{array}$

At $t=6$ the lower boundary is,

  $\begin{array}{c}u=\frac{\pi (6)}{12}\\ =\frac{1\pi }{2}\end{array}$

Now,

  $\begin{array}{c}Av{g}_2=\frac{1}{8}{\displaystyle {\int }_{\frac{\pi }{2}}^{\frac{7\pi }{6}}1}0\cos (u)\frac{12}{\pi }du\\ ={\left.\frac{1}{8\pi }120\sin (u)\right|}_{\frac{\pi }{2}}^{\frac{7\pi }{6}}\\ =\frac{15}{\pi }\left(\sin \frac{7\pi }{6}-\sin \frac{\pi }{2}\right)\\ =\frac{15}{\pi }\left(-\frac{1}{2}-1\right)\\ =\frac{15}{\pi }\left(-\frac{3}{2}\right)\\ =-\frac{45}{2\pi }\\ =-7.16\end{array}$

Hence, the average temperature is given by,

  $\begin{array}{c}Avg=Av{g}_1-Av{g}_2\\ =80-\left(-7.16\right)\\ =87.16\\ \approx 87\end{array}$

Therefore, the average temperature is $87$ .

b.

To determine

To find: For what values of $t$ was the air conditioner cooling the house.

Answer to Problem 54EP

The value of $t$ is $5.231\le t\le 18.77$ .

Explanation of Solution

Given information:

  $F\left(t\right)=80-10\cos \left(\frac{\pi t}{12}\right)$

Calculation:

Let's interpret the situation and formulate the inequality first, then solve for $t$ .

  $\begin{array}{c}80-10\cos \left(\frac{\pi t}{12}\right)\ge 78\\ -10\cos \left(\frac{\pi t}{12}\right)\ge 78-80\\ -10\cos \left(\frac{\pi t}{12}\right)\ge -2\\ \cos \left(\frac{\pi t}{12}\right)\ge \frac{-2}{-10}\\ \left(\frac{\pi t}{12}\right)\ge \arccos \frac{1}{5}\\ t\ge \arccos \frac{1}{5}\left(\frac{12}{\pi }\right)\\ 5.231\le t\le 18.77\end{array}$

Therefore, the value of t is $5.231\le t\le 18.77$ .

c.

To determine

To find: The total cost to the nearest cent to cool the house for $24$ -hour period.

Answer to Problem 54EP

The total cost was about $\$5.10$ .

Explanation of Solution

Given information:

  $F\left(t\right)=80-10\cos \left(\frac{\pi t}{12}\right)$

Calculation:

Let's first analyze the problem,

  $\begin{array}{c}\text{Temperature with payment }=F(t)-78\\ \cos t=0.05\end{array}$

Integrating the term to get,

  $\text{ Total cost}=k{\displaystyle {\int }_a^b(F(}t)-78)dt$

Substituting the value,

  $\begin{array}{c}\text{Total cost }=0.05{\displaystyle {\int }_{5.251}^{18\pi }\left(80-10\cos \left(\frac{\pi t}{12}\right)-78\right)}dt\\ =0.05{\displaystyle {\int }_{5.251}^{18\pi }\left(2-10\cos \left(\frac{\pi t}{12}\right)\right)}dt\\ =0.05{\displaystyle {\int }_{5.251}^{18\pi }\left(2-10\cos \left(\frac{\pi t}{12}\right)\right)}dt\\ =0.05{\displaystyle {\int }_{5.251}^{18\pi }2}dt-0.05{\displaystyle {\int }_{5.251}^{18\pi }1}0\cos \left(\frac{\pi t}{12}\right)dt\\ ={\text{ Total cost }}_1-{\text{ Total cost }}_2\end{array}$

To find ${\text{Total cost }}_1$:

  $\begin{array}{c}{\text{Total cost }}_1=0.05{\displaystyle {\int }_{5.231}^{18.77}2}dt\\ ={\left.0.05(2t)\right|}_{5.231}^{18.77}\\ =0.05[2(18.77)-2(5.231)]\\ =0.05(37.54-10.462)\\ =0.05(27.078)\\ =1.3539\end{array}$

To find ${\text{Total cost }}_2$:

  $\text{ Total }{\mathrm{cost}}_2=0.05{\displaystyle {\int }_{5.231}^{18.77}1}0\cos \left(\frac{\pi t}{12}\right)dt$

Integrating by substitution method,

  $\begin{array}{c}u=\frac{\pi t}{12}\\ du=\frac{\pi }{12}dt\\ \frac{12}{\pi }du=dt\end{array}$

To find the new boundaries of integration,

At $t=18.77$ , the upper boundary is

  $u=\frac{\pi (18.77)}{12}$

At $t=5.231$ , the lower boundary is

  $u=\frac{\pi (5.231)}{12}$

Now,

  $\begin{array}{c}{\text{Total cost }}_2=0.05{\displaystyle {\int }_{\frac{5.231\pi }{12}}^{\frac{18.77\pi }{12}}1}0\cos (u)\frac{12}{\pi }du\\ ={\left.\frac{0.05}{\pi }(120\sin (u))\right|}_{\frac{\pi }{2}}^{\frac{7\pi }{6}}\\ =\frac{6}{\pi }\left(\sin \frac{18.77\pi }{12}-\sin \frac{5.231\pi }{12}\right)\\ =\frac{6}{\pi }(-0.97975-0.9798)\\ =\frac{6}{\pi }(-1.95955)\\ =-3.742\end{array}$

The total cost is given by,

  $\begin{array}{c}\text{Total cost }={\text{ Total cost }}_1-{\text{ Total cost }}_2\\ =1.3539-\left(-3.742\right)\\ \approx 5.10\end{array}$

Therefore, the total cost to cool the house for $24$ -hour periodis $\$5.10$ .