Solution-(7)-Newton

Solution-(7)-Newton - . 6 e-k 1 . Now solve for k : 57 . 2...

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Mathematics 115 – Fall 2011 (7) Exercise 32: p. 322 Cooling Body Solution 1 : Assume that time t = 0 corresponds to 2 A.M. Let T ( t ) be the body temperature at time t hours, let C = 10 F be the ambient temperature. We know that T (0) = 61 . 6 F, so Newton’s Law of Cooling gives us T ( t ) = 10 + (61 . 6 - 10) e - kt . To determine what time the murder was committed we need to find at what time the body temperature was 98.6 F, so we need to solve T ( t ) = 98 . 6 for t . So, we need to find k . The other piece of information we have is that at 3 A.M. the body temperature was 57.2 F. So we have T (1) = 57 . 2 and T (1) = 10 + 51
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Unformatted text preview: . 6 e-k 1 . Now solve for k : 57 . 2 = 10 + 51 . 6 e-k 47 . 2 / 51 . 6 = e-k ln(47 . 2 / 51 . 6) =-k k = . 0891 . Now solve T ( t ) = 98 . 6, that is 98 . 6 = 10 + 51 . 6 e-. 0891 t . We get t =-6 . 07, following the same arithmetic steps displayed above. This means that the murder was committed 6.07 hours before 2 A.M. , namely, about 7:56 P.M. of the previous day. Solution 2 : Assume that time t = 0 corresponds to the time of death, so T (0) = 98 . 6. Now proceed exactly as in Example 5, p. 319....
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This document was uploaded on 10/24/2011.

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