# [Calculus] Newton's Law of Cooling

• Mar 1st 2009, 04:38 PM
Moderatelyconfused
[Calculus] Newton's Law of Cooling
In my assignment, my Calculus teacher was murdered.

The three suspects are:
An English Teacher (no alibi from 9:15-10:00AM)
The Principal (no alibi from 10:05-10:50AM)
A History Teacher (no alibi from 10:55-11:40AM)

The temperature of the body was 87 degrees at 11:55AM, and 80 degrees at 12:45PM. The room was 75 degrees.

My job is to identify the murderer and support my answer with a written paragraph and some math.

I know I need to use (dy/dk)=K(y-Ts), e^(kt+c)=y-Ts, and y=Ce^(kt)+Ts
Assuming that Y = the temperature of an object, t = time, Ts = Room Temperature, and C = y - Ts. I have an example in my notes, but I am having trouble following the different equations and still get lost at the various transitions between e and ln.

• Mar 1st 2009, 05:34 PM
arpitagarwal82
Just at death the body temperature was 98 degrees Fahrenheit (approx Normal body temperature)

It was 87 degree at 11:55
and 75 degree at 12:45
Consider y=Ce^(kt)+Ts

Ts = 75 degree

let $\displaystyle t_1$ be time of death. temp 98
$\displaystyle t_2$be 11:55 temp 87
$\displaystyle t_3$ be 12:45 temp 80

So $\displaystyle 98 = Ce^(kt_1) + 75$
$\displaystyle 87 = Ce^(kt_2) +75$
$\displaystyle 80 = Ce^(kt_3) + 75$

i.e.

$\displaystyle 23 = Ce^(kt_1 )$ ----------eq1
$\displaystyle 12 = Ce^(kt_2)$----------eq2
$\displaystyle 5 = Ce^(kt_3)$ ----------eq3

Now do eq3 -eq2 and eq2 -eq1 and use t_3 -t_2 = 50 minutes

So a bit calculation will give you
$\displaystyle t_1 = t_2 - 50 (\ln 23/11)(\ln 12/5)$ ($\displaystyle t_2$ is in minutes here)

Find $\displaystyle t_1$ and you can take the guess of murderer on basis of $\displaystyle t_1$