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April 20, 2014

April 20, 2014

Posted by **Chris** on Thursday, May 5, 2011 at 4:02pm.

T(t)=T0+(T1-T0)e^-kt

The police discover the body of a murder victim. Critical to solving the crime is determining when the murder was committed. The coroner arrives at the murder scene at 12:00pm. She immediately takes the temperature of the body and finds it to be 94.6 degrees F. She then takes the temperature 1 hour later and finds it to be 93.4 degrees F. The temperature of the room is 70 degrees F. When was the murder committed?

- College Algebra Application -
**MathMate**, Friday, May 6, 2011 at 12:32pmLet t=0 at time of death.

We have two temperatures, 94.6 at t=t1, and 93.4 at t=t1+1.

Also, normal body temperature is 98.6, so T1=98.6.

T0=ambient temperature = 70.

Substitute in given formula:

T(t)=T0+(T1-T0)e^-kt

94.6=70+(98.6-70)e^(k*t1)...(1)

93.4=70+(98.6-70)e^(k*(t1+1))...(2)

We try to solve for t1 and k.

Subtract (2) from (1):

Rearrange (1) and (2)

e^(k*t1)=0.86014 ...(1a)

e^(k*(t1+1))=0.81818 ...(2a)

Divide (2a) by (1a)

e^(k*(t1+1)-k*t1)=0.81818/0.86014

e^k=0.95122

take natural log on both sides,

k=-0.0500

Substitute k in (1a):

e^(-0.05t1)=0.86014

take natural log on both sides

-0.05t1 = -0.15066

t1=-0.15066/(-0.05)

=3.01

So the victim was dead 3 hours before the coroner arrived, or at 9 a.m.

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