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Chapter 5

Q. 5.3.11

Modeling room temperature

Ten minutes after a furnace is turned on, the temperature in a room reaches 74° and the furnace turns off. It takes two minutes for the room to cool to 70° and two minutes for the furnace to bring it back to 74° as shown in the following table.

\begin{array}{l|c|c|c|c|c|c|c|c|c} \text { Time (min) } & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\ \hline \text { Temperature }\left({ }^{\circ} \mathrm{F}\right) & 74 & 72 & 70 & 72 & 74 & 72 & 70 & 72 & 74 \end{array}

Assuming that the temperature (after time 10) is a sine function of the time, find the function and graph it.

Step-by-Step

Verified Solution

Because the temperature ranges from 70° to 74°, the amplitude of the sine curve is 2. Because the temperature goes from its maximum of 74° back to 74° in 4 minutes, the period is 4. Since the period is 2π/B, we get B = π/2. Now concentrate on one cycle of the function. Starting at (13, 72) the temperature increases to its maximum of 74°, decreases to its minimum of 70°, and then ends in the middle at (17, 72). We choose this cycle because it duplicates the behavior of y = sin(x) on its fundamental cycle [0, 2π]. So shifting y = sin(x) to the right 13 and up 72 gives us y=2 \sin \left[\frac{\pi}{2}(x-13)\right]+72. Its graph is shown in Fig. 5.58. Changing this equation with any right or left shift by a multiple of 4 gives an equivalent equation. For example, y=2 \sin \left[\frac{\pi}{2}(x-1)\right]+72 is an equivalent equation.

Modeling room temperature Ten minutes after a furnace is turned on, the temperature in a room reaches 74° and the furnace turns off. It takes two minutes for the room to cool to 70° and two minutes for the furnace to bring it back to 74° as shown in the following table. Time (min) 10 11 12 13 14