Murdered Professor

Date: 06/26/97 at 14:22:58
From: Brett Wilson
Subject: Exponential puzzle

Dr. Math,

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!

Thank you.

Date: 06/26/97 at 18:06:55
From: Doctor Toby
Subject: Re: Exponential puzzle

The t in that exponential formula is the time since the time of death.
That's hard to work with when you don't know the time of death. So use 
the formula y = Ce^k(t-T), where T is the time of death. In this 
formula, t is the actual time, which you can work with. 

There are three unknown constants in this formula, C, k, and T. You 
have three data points, y = 28.6 degress when t = T, y = 10 degrees 
when t = 9:00, and y = 8 degrees when t = 10:00. Now you have enough 
to solve for C, k, and, most importantly, T.

Note that y is the difference between the temperature and the base 
temperature of 70 degrees, not the temperature itself. If you like, 
you can use the equation y-Y = Ce^k(t-T), where now y is the actual 
temperature and Y is the base temperature, and add the fourth datum 
that y = 70 degrees when t = infinity. Then you immediately get 
Y = 70 degrees.

Anyway, if you use the datum that y = 28.6 degrees when t = T, you 
learn that C = 28.6 degrees right away. So the hard part is solving 
the two remaining equations for the two remaining constants, k and T.
(But that's not really too hard. The really hard part will be finding 
another alibi!)

-Doctor Toby,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 06/26/97 at 18:12:28
From: Doctor Anthony
Subject: Re: exponential puzzle

You should start with the differential equation that is actually 
quoted in the question. The equation, using T for temperature and 
t for time is:

     dT/dt = -k(T-70)     (k is the constant of proportionality)

 dT/(T-70) = -k*dt    

Integrating both sides:

  ln(T-70) = -kt + const.     

     T-70  =  Ae^(-kt)     where A = e^(const.) = constant  

At t = 9, T = 80 and at t = 10,  T = 78. 

So, 10 = Ae^(-9k) and 8 = Ae^(-10k)  

Dividing these we have, 10/8 = e^(-9k+10k) = e^(k)

This gives k = ln(5/4) = 0.223

To find A, use 10 = Ae^(-9 x .223) = 0.1342A

So A = 10/.1342  =  74.5

The full equation is then: T = 70 + 74.5e^(-.223t)

Normal body temperature is 98.6, so now we require the value of t for 
when T = 98.6:

      98.6 - 70 = 74.5e^(-.223t)

       28.6/74.5 = e^(-.223t)

        .3839 = e^(-.223t)    

    ln(.3839) = -.223t

       -.9574 = -.223t

            t = .9574/.223  
 
            t =  4.293

This corresponds to 4:17 a.m., so you need to think of someone who can 
prove what you were doing at 4:17 a.m.
 
-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 07/04/97 at 22:52:21
From: Doctor Marko
Subject: Re: exponential puzzle

Dear Brett,

You are on a good track.  The "proper" way to do this is to set up a
differential equation from the statement: The difference between body
temperature and the room temperature changes at a rate proportional to 
that difference.  But the basic equation (y = Ce^kt) is right.

From this statement you can infer that the only two important 
quantities are time (t) and the difference between the two 
temperatures, body-room which we can call T.  Additionally, the 
statement tells you that the rate of change (also known as the time 
derivative) of T is proportional to T itself. Thus, you would write 
dT/dt = -k*T, where dT/dt stands for the time derivative and k is the 
constant of proportionality. The reason for the minus sign is that 
T is always positive but decreasing, and thus, T > 0, but dT/dt < 0.

Now you can solve this differential equation in many ways (I hope you 
have studied this in class) and the standard way is the separation 
of variables. So you would write:

dT/T = -k*dt

You can now integrate both sides and get:

Log(T) = -k*t + c1 (c1 is a constant from the indefinite integral)  

The next step is exponentiating both sides of the equation:

e^(Log(T)) = e^(-k*t+c1)

By exponential and log rules, e^(Log(T)) = T, and 
e^(-k*t+c1) = e^(-k*t)*e^c1. Now e^c1 is just another constant, call 
it C, so the final equation becomes:

T = C*e^(-k*t)

Now you want to plug your information:

Room temp = 70
At 9 a.m., body temp = 80 => t = 9, T = 10 => 10 = C*e^(-k*9)
At 10 a.m., body temp = 78 => t = 10, T = 8 => 8 = C*e^(-k*10)

So dividing one equation by the other you get:

8/10 = e^(-10k)/e^(-9k) or .8 = e^(-k), or Log(.8) = -k.

Then you pull out your trusty calculator and -0.2231 = -k. We can go 
back and figure out C by: 

10 = C*e^(-0.2231*9) => 10 = C*0.1343 => C = 74.4766.  

Then approximately, T = 74.48*e^(-0.223*t). The final question: When 
did the professor die?  Well, normal body temperature is 98, so when 
professor died T = 28.  Plug it in:

     28 = 74.48*e^(-0.223*t) 
 0.3759 = e^(-0.223*t)
-0.9783 = -0.223*t 
      t = 4.39 

The professor died at bit before 4:30am.  So, if the student can prove 
that s/he was in bed at that time, everything will be fine.  

I hope this cleared things a bit.  I know I was quite verbose, but 
this is an important scheme of solving problems, and chances are that 
you will have many such puzzles (possibly on an exam as well!) so 
learn it well.  Good luck,

-Doctor Marko,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/