Chapter 8, Problem 54RE

(a)

To determine

To find the average temperature, to the nearest degree Fahrenheit, between $t=16\text{ and }t=14$

Answer to Problem 54RE

Average temperature $\approx {87}^{\circ }$ F

Explanation of Solution

Given:

The temperature outside a house during a 24-hour period is given by

  $F\left(t\right)=80-10\cos \left(\frac{\pi t}{12}\right),0\le t\le 24$

Concept Used:

Application of integrals

Calculation:

Average Temperature

  $\begin{array}{l}=\frac{1}{14-6}{\displaystyle \underset{6}{\overset{14}{\int }}\left(80-10\cos \left(\frac{\pi t}{12}\right)\right)}dt\\ =\frac{1}{8}\left[80{\displaystyle \underset{6}{\overset{14}{\int }}dt}-10{\displaystyle \underset{6}{\overset{14}{\int }}\cos \left(\frac{\pi t}{12}\right)}dt\right]\\ =\frac{1}{8}\left[80\left(t\right){|}_6^{14}-10\times \frac{12}{\pi }\left(\sin \left(\frac{\pi t}{12}\right)\right){|}_6^{14}\right]\\ =\frac{1}{8}\left[80\left(14-6\right)-\frac{120}{\pi }\left(\sin \left(\frac{14\pi }{12}\right)-\sin \left(\frac{6\pi }{12}\right)\right)\right]\\ =\frac{1}{8}\left[640-\frac{120}{\pi }\left(\sin \left(\frac{7\pi }{6}\right)-\sin \left(\frac{\pi }{2}\right)\right)\right]\\ =\frac{1}{8}\left[640-\frac{120}{\pi }\left(-\frac{1}{2}-1\right)\right]\\ =\frac{1}{8}\left[640-\left(\frac{120}{\pi }\times -\frac{3}{2}\right)\right]\\ =\frac{1}{8}\left[640-\left(\frac{-180}{\pi }\right)\right]\\ \approx \frac{1}{8}\left[640+57.2956\right]\\ \approx \frac{1}{8}\times 697.2956\\ \approx 87.162\approx 87°F\end{array}$

Conclusion:

Average temperature $\approx 87°F$

(b)

To determine

To determine values of $t$ at which air conditioner was cooling the house.

Answer to Problem 54RE

The value of t is $5.231\le t\le 18.770$ .

Explanation of Solution

Given:

Air conditioner cooled the house whenever the outside temperature was at or above 78 degrees Fahrenheit.

Concept Used:

Concept of inequality

Calculation:

  $\begin{array}{l}\left[80-10\cos \left(\frac{\pi t}{12}\right)\right]-78\ge 0\\ 2-10\cos \left(\frac{\pi t}{12}\right)\ge 0\\ 10\cos \left(\frac{\pi t}{12}\right)\le 2\\ \cos \left(\frac{\pi t}{12}\right)\le \frac{1}{5}\\ taking{\cos }^{-1}\text{ both sides}\\ {\cos }^{-1}\cos \left(\frac{\pi t}{12}\right)\le {\cos }^{-1}\frac{1}{5}\\ \Rightarrow \frac{\pi t}{12}=1.369\text{ or 4}\text{.914}\\ \Rightarrow \text{t=1}\text{.369}\times \frac{12}{\pi }\text{ or 4}\text{.914}\times \frac{12}{\pi }\\ \Rightarrow 5.231\le t\le 18.770\end{array}$

Conclusion:

The value of t is $5.231\le t\le 18.770$ .

(c)

To determine

To find total cost, to the nearest cent, to cool the house for this 24-hour period

Answer to Problem 54RE

Total cost, to the nearest cent, to cool the house for this 24-hour period= $\$5.10$

Explanation of Solution

Given:

Rate at which the cost of cooing the house accumulates = $\$0.05$

Concept Used:

Applications of integral

Calculation:

Rate at which the cost of cooing the house accumulates

  $\begin{array}{l}=0.05{\displaystyle \underset{5.231}{\overset{18.770}{\int }}\left(80-10\cos \left(\frac{\pi t}{12}\right)-78\right)}dt\\ =0.05{\displaystyle \underset{5.231}{\overset{18.770}{\int }}\left(2-10\cos \left(\frac{\pi t}{12}\right)\right)}dt\\ =0.05\left[2{\displaystyle \underset{5.231}{\overset{18.770}{\int }}dt-10{\displaystyle \underset{5.231}{\overset{18.770}{\int }}\cos \left(\frac{\pi t}{12}\right)dt}}\right]\\ =0.05\left[2\left(t\right){|}_{5.231}^{18.770}-\frac{120}{\pi }\left(\sin \left(\frac{\pi t}{12}\right)\right){|}_{5.231}^{18.770}\right]\\ \text{putting the limits to get answer}\\ 0.05{\displaystyle \underset{5.231}{\overset{18.770}{\int }}\left(80-10\cos \left(\frac{\pi t}{12}\right)-78\right)}dt=5.096\end{array}$

Conclusion:

Total cost, to the nearest cent, to cool the house for this 24-hour period= $\$5.10$