Chapter 5: Applications of Integration
Section 5.6: Differential Equations
A medical examiner (M.E.) notes the temperature of the body of a deceased person is 85°F, and the environment in which the body has been located is 65°F. Being careful not to alter the surrounding temperature, the M.E. waits 15 minutes and again checks the body's temperature, finding it to be 80°F. Using Newton's law of cooling, what estimate can the M.E. make for the time of death of the deceased?
Starting with the model ut=Us+A ek t developed in Example 5.6.9, write ut=65+A ek t and form the two equations u0=85 and u15=80 from the data points t,u=0,85 and t,u=15,80. The solution of the equations 65+A=85 and 65+A e15 k=80 is A=20,k=ln3/4/15≐−0.0192.
To estimate the time of death, solve ut=98.6, where the average body-temperature is taken as 98.6°F.
65+20 eln3/4 t/15
20 eln3/4 t/15
So, the approximate time of death is 27 minutes prior to the arrival of the M.E.
Define the function ut=65+A ek t
Context Panel: Assign Function
ut=65+A ⅇk t→assign as functionu
Form and solve two algebraic equations for A and k
Write the two equations arising from the given data points, and press the Enter key.
Context Panel: Solve≻Solve
Obtain ut and determine the time of death
Expression palette: Evaluation template
Equate ut to 98.6.
Press the Enter key.
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