IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

danny_007

Please help me with this puzzle given to me by my calculus professor:

Suppose that you come into your professor's office to ask some 
questions shortly before 9:00 a.m. on Friday. You find him lying on 
the floor of his office in a pool of chalk dust, dead. You quickly 
call the police and their investigators take several measurements over 
the next hour, including:

  1) the body temperature at 9:00 a.m. - 80 degrees
  2) the body temperature at 10:00 a.m. - 78 degrees
  3) room temperature - 70 degrees (constant)

You quickly realize that the police believe you to be a prime suspect, 
so you need an alibi. You know that you were studying until midnight, 
but you aren't sure if that is enough information. You need to know 
the time of death!
 
Determine the time of death by creating an exponential model. Use the 
following statement: the difference between body temperature and room 
temperature changes at a rate proportional to that difference. How 
good is your alibi?

The only thing I have figured to do is use the exponential formula 
y = Ce^kt, but I'm not sure which numbers to plug in and what 
information that will give me. Please help!

danny_007

as many salutes as u want for the right answer

djdylan2000

Its basically Newtons law of cooling, right?

sandy1990

use newtons law with laplace correction

danny_007

i know that but please help me in the last paragraph i have written which numbers to put yaar

danny_007

ramkumar_november

may be this is a clue....

" the body temperature at the time of death must be 37 degrees which is normal human body temperature......"

so at t =0 temp = 37 degrees

newton's law of cooling states that :

$\frac{dT}{dt}\;=\;-K(T-S)$

where S is the room temperature... which is 70 degrees

now derive the expression by integrating

$\int\frac{dT}{T-70}\;=\int \;-Kdt$

and then use the data to find the value of K

chimanshu_007

edited...

see below

anchitsaini

http://mathforum.org/library/drmath/view/53409.html

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