Page 29 - Advanced engineering mathematics

1.1 Terminology and Separable Equations 9

To solve for T , take the exponential of both sides of this equation to get
|T − 68|= e kt+c = Ae kt
c
where A = e . Then
kt
kt
T − 68 =±Ae = Be ,
so
kt
T (t) = 68 + Be .
Now the constants k and B must be determined. Since there are two constants, we will need two
pieces of information. Suppose the lieutenant arrived at 9:40 p.m. and immediately measured the
body temperature, obtaining 94.4 . It is convenient to let 9:40 p.m. be time zero in carrying out
◦
measurements. Then
T (0) = 94.4 = 68 + B,
so B = 26.4. So far,
kt
T (t) = 68 + 26.4e .
To determine k, we need another measurement. The lieutenant takes the body temperature again
at 11:00 p.m. and finds it to be 89.2 . Since 11:00 p.m. is 80 minutes after 9:40 p.m., this means
◦
that
80k
T (80) = 89.2 = 68 + 26.4e .
Then
21.2
e 80k = ,
26.4
so
21.2

80k = ln .
26.4
Then
1 21.2
k = ln .
80 26.4
The temperature function is now completely known as
T (t) = 68 + 26.4e ln(21.2/26.4)t/80 .
The time of death was the last time at which the body temperature was 98.6 (just before it began
◦
to cool). Solve for the time t at which
T (t) = 98.6 = 68 + 26.4e ln(21.2/26.4)t/80 .
This gives us
30.6
= e ln(21.2/26.4)t/80 .
26.4
Take the logarithm of this equation to obtain
30.6 t 21.2

ln = ln .
26.4 80 26.4
According to this model, the time of death was
80ln(30.6/26.4)
t = ,
ln(21.2/26.4)
which is approximately −53.8 minutes. Death occurred approximately 53.8 minutes before
(because of the negative sign) the first measurement at 9:40 p.m., which was chosen as time
zero in the model. This puts the murder at about 8:46 p.m.

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October 14, 2010 14:9 THM/NEIL Page-9 27410_01_ch01_p01-42