chapter4

According to newtons law of cooling the body will

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According to Newton’s law of cooling, the body will radiate heat energy into the room at a rate proportional to the difference in temperature between the body and the room. If T(t) is the body temperature at time t, then for some constant of proportionality k, T'(t)=k[T(t)-70] This is a separable differential equation and is written as 101

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kdt dT 70 T 1 = - Upon integrating both sides, one gets ln|T-70|=kt+c Taking exponential, one gets |T-70|=e kt+C =Ae kt where A = e C . Then T-70= ± Ae kt = Be kt Then T(t)=70 + Be kt Constants k and B can be determined provided the following information is available: Time of arrival of the police personnel, the temperature of the body just after his arrival, temperature of the body after certain interval of time. Let the officer arrived at 10.40 p.m. and the body temperature was 94.4 degrees. This means that if the officer considers 10:40 p.m. as t=0 then T(0)=94.4=70+B and so B=24.4 giving T(t)=70 + 24.4 e kt. Let the officer makes another measurement of the temperature say after 90 minutes, that is, at 12.10 a.m. and temperature was 89 degrees. This means that T(90)=89=70+24.4 e 90k Then 102
, 4 . 24 19 k 90 e = so = 4 . 24 19 ln 90 k and = 4 . 24 19 ln 90 1 k The officer has now temperature function + = 4 . 24 19 ln 90 t e 4 . 24 70 ) t ( T In order to find when the last time the body was 98.6 (presumably the time of death), one has to solve for time the equation + = = 4 . 24 19 ln 90 t e 4 . 24 70 6 . 98 ) t ( T To do this, the officer writes = 4 . 24 19 ln 90 t e 4 . 24 6 . 28 and takes logarithms of both sides to obtain = 4 . 24 19 ln 90 4 . 24 6 . 28 ln t Therefore, the time of death, according to this mathematical model, was ) 4 . 24 / 19 ln( ) 4 . 24 / 6 . 28 ln( 90 t = which is approximately –57.0.7 minutes. 103

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The death occurred approximately 57.07 minutes before the first measurement at 10.40 p.m. , that is at 9.43 p.m. approximately 4.10 Drug Distribution (Concentration) in Human Body To combat the infection to human a body appropriate dose of medicine is essential. Because the amount of the drug in the human body decreases with time medicine must be given in multiple doses. The rate at which the level y of the drug in a patient’s blood decays can be modeled by the decay equation ky dt dy - = where k is a constant to be experimentally determined for each drug. If initially, that is, at t=0 a patient is given an initial dose y p , then the drug level y at any time t is the solution of the above differential equations, that is, y(t)=y p e -kt Remark: 4.10.1. In this model it is assumed that the ingested drug is absorbed immediately which is not usually the case. However, the time of absorption is small compared with the time between doses. Example 4.14: A representative of a pharmaceutical company recommends that a new drug of his company be given every T hours in doses of quantity y 0 , for an extended period of time. Find the steady state drug in the patient’s body.
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