# Math Help - Need some help with a hard question

1. ## Need some help with a hard question

Ok the question is

Coroners estimate the time of death from body temperature using the simple rule that a body cools about 1C in the first hour after death and about 1/2C for each addictional hour.

Assuming an air temperature of 20C and a living body temperature of 37C, the temperature, T(t), in degrees Celcius is given by T(t) = 20 + 17e^-kt, where t=0 is the instant that death occurred..

a) For what value of k will the body cool by 1C in the first hour?
b) Using the value of k found in part a) after how many hours will the temperature of the body be decreaing at a rate of 1/2 C per hour?
c) Using the value of k found in part a) Show that 24h after death, the coroner's rule gives approx the same temp as the forumula.

I know I need an equation to do this but Im not sure what it is, I understood my other log study questions but this one stumps me. Any help would be appreciated greatly.

2. Originally Posted by Skor
Ok the question is

Coroners estimate the time of death from body temperature using the simple rule that a body cools about 1C in the first hour after death and about 1/2C for each addictional hour.

Assuming an air temperature of 20C and a living body temperature of 37C, the temperature, T(t), in degrees Celcius is given by T(t) = 20 + 17e^-kt, where t=0 is the instant that death occurred..

a) For what value of k will the body cool by 1C in the first hour?
b) Using the value of k found in part a) after how many hours will the temperature of the body be decreaing at a rate of 1/2 C per hour?
c) Using the value of k found in part a) Show that 24h after death, the coroner's rule gives approx the same temp as the forumula.

I know I need an equation to do this but Im not sure what it is, I understood my other log study questions but this one stumps me. Any help would be appreciated greatly.
a) You have:

$T(t) = 20 + 17e^{-kt}$

Now if we measure time in hours we have if $1$ is the time to cool by 1 degree C

$
36=20+17 e^{-k}
$

or:

$
e^{-k}=16/17
$

or:

$
k=-\ln(16/17)\approx 0.060625
$

b) The rate of fall of temprature is:

$\frac{dT}{dt}=-k \times 17 e^{-kt}$

If the temprature is falling at a rate of 1/2 a degree per hour we have:

$-k \times 17 e^{-kt}=-1/2$

so:

$e^{-kt}=0.5/(k \times 17)$

or:

$-kt=\ln(0.5/(k \times 17))$

so:

$t=-(1/k)\ln(0.5/(k \times 17))\approx 11.28 \mbox{ hours}$

RonL