7.5.21-eg-1 - k> 0 Substituting we have T t = 12 e-kt 70 1 Find the constant k Using T(1 = 76 76 = 12 e-k 70 ln ± 1 2 ² =-k and so k = ln 2 2

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Example A homicide victim was found in a room kept constant at 70 F. Measurements of the victim’s temperture were made once the police arrived and one hour later. The results were 82 F and 76 F respectively. Assuming Newton’s Law of Cooling and that the victim’s temperature was 98 . 6 F just before death, what was the time of death relative to the arrival of the police? We start with Newton’s Law of Cooling dT dt = - k ( T - T S ) , to obtain the model T ( t ) - T S = ( T 0 - T S ) e - kt where T is the victim’s temperature in F as a function of time t in hours since the police arrived, surrounding temperature T S = 70, initial temperature T 0 = 82, and constant
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Unformatted text preview: k > 0. Substituting we have T ( t ) = 12 e-kt + 70 . 1. Find the constant k . Using T (1) = 76 76 = 12 e-k + 70 . ln ± 1 2 ² =-k and so k = ln 2 . 2. Use the model to determine the answer. Using the exact value of k throughout the remainder of the problem, the model becomes T ( t ) = 12 e-(ln 2) t + 70 . We use the model and estimate the time of death: 98 . 6 = 12 e-(ln 2) t + 70 28 . 6 12 = e-(ln 2) t and so t =-ln 28 . 6 12 ln 2 ≈ -1 . 25298 implying that the time of death was about 1.25 hours prior to the arrival of the police....
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

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