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Chapter 1: Problem 51

A patient's temperature was \(97.0^{\circ} \mathrm{F}\) at 8: 05 A.M. and \(101.0^{\circ} \mathrm{F}\) at 12: 05 P.M. If the temperature change with respect to elapsed time was linear throughout the day, what would the patient's temperature be at 3: 35 P.M.?

Short Answer

Expert verified
Answer: The patient's temperature would be approximately 103.8°F at 3:35 P.M.

Step by step solution

01

Convert given times to a common format

First, convert the given times to minutes elapsed since midnight to make calculations easier. 8:05 A.M. is 485 minutes since midnight, and 12:05 P.M. is 725 minutes since midnight.
02

Calculate the slope of the temperature change

Using the given temperatures, calculate the slope (rate of change) by subtracting the two temperature values and dividing by the time difference. Let \(x\) represent the time in minutes since midnight and \(y\) represent the temperature in Fahrenheit. Slope, denoted by \(m\): \[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{101 - 97}{725 - 485} = \frac{4}{240}\]
03

Calculate the temperature at 3:35 P.M.

To find the patient's temperature at 3:35 P.M., first convert the time to minutes past midnight. 3:35 P.M. is 895 minutes since midnight. Now we will use the slope-point equation of the line: \[y - y_1 = m(x - x_1)\] Using the coordinates (485, 97) and the slope, 1/60, we have: \[y - 97 = \frac{1}{60}(x - 485)\] Plug in 895 for x to find the patient's temperature at 3:35 P.M.: \[y - 97 = \frac{1}{60}(895 - 485)\] \[y - 97 = \frac{1}{60}(410)\] \[y - 97 = 6.83\] \[y = 103.83\] So the patient's temperature would be approximately \(103.8^{\circ} \mathrm{F}\) at 3:35 P.M.

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