# Modeling with exponential and logarithmic functions help?

1. Apr 30, 2011

### camp3r101

Modeling with exponential and logarithmic functions help???

1. The problem statement, all variables and given/known data
Use Newton's Lay of Cooling, T = C + (T0 - C)e-kt, to solve this exercise. At 9:00 A.M., a coroner arrived at the home of a person who had died during the night. The temperature of the room was 70 degrees F, and at the time of death the person had a body temperature of 98.6 degrees F. The coroner took the body's temperature at 9:30 A.M., at which time it was 85.6 degrees F, and again at 10:00 A.M., when it was 82.7 degrees F. At what time did the person die?

2. Relevant equations
T = C + (T0 - C)e-kt
If you do not know what the variable's mean...these are their meanings:
T = temperature of a heated object
C = constant temperature of the surrounding medium (the ambient temp)
T0 = initial temperature of the heated object
k = negative constant associated with the cooling object
t = time (in minutes)

3. The attempt at a solution
I tried solving for k by doing:
Steps (I plugged all the values in to their corresponding places):

85.6 = 70 + (98.6 - 70)e-k(30)
15.6 = 28.6e-30k
0.5454545455 = e-30k
ln(0.5454545455) = ln(e-30k)
ln(0.5454545455) = -30k
k = (ln(0.5454545455)/(-30))
k = 0.0202045268

After getting this....i do not know what to do next...of even if I did the process of anything correctly as yet.

2. Apr 30, 2011

### HallsofIvy

Re: Modeling with exponential and logarithmic functions help???

Excellent! Writing them all out like this is a very good idea- even on your home work paper. Not every one uses the same symbols. One criticism- you say "t= time (in minutes)" but do not say "time from what?" From what you write later, t appears to be the time (in minutes) after 9:00. You should say that.

NO. You don't know what the temperature of the body was at 9:00- you are told that " at the time of death the person had a body temperature of 98.6 degrees F" and you don't know when the time of death was. If you had said "t= time (in minutes) after death until the Coroner arrived, at 9:00", you would have
85.6= 70+ (98.6- 70)e-k(t+ 30)
and
82.7= 70+ (98.6- 70)e-k(t+ 60)

That gives you two equations to solve for k and t. Subtracting t minutes from 9:00 gives you the time of death.

3. Apr 30, 2011

### camp3r101

Re: Modeling with exponential and logarithmic functions help???

Which values of k ant t do I use to find the time of death???
Do I use both; do I add both t values together and then subtract from 9:00 ??

4. Apr 30, 2011

### eumyang

Re: Modeling with exponential and logarithmic functions help???

camp3r101: while what HallsofIvy said is correct, I wouldn't do it this way, because you defined T0 as the initial temperature of the object... that is to say, the temperature at t = 0. I would define t as simply the number of minutes after death. So the equations I would write are
$$85.6 = 70 + (98.6 - 70)e^{-kt}$$
for 09:30am, and
$$82.7 = 70 + (98.6 - 70)e^{-k(t + 30)}$$
for 10:00am.

No. (The hints I'm going to give may be too much, so apologies to the mods in advance.)

Take my 1st equation and simplify it a bit:
$$15.6 = 28.6e^{-kt}$$

Take my 2nd equation and simplify:
$$12.7 = 28.6e^{-k(t + 30)}$$
$$12.7 = 28.6e^{-kt - 30k}$$

Use property of exponents:
$$12.7 = \frac{28.6e^{-kt}}{e^{30k}}$$

Notice the numerator? That equals 15.6, from the first equation, so we substitute:
$$12.7 = \frac{15.6}{e^{30k}}$$

Now I'll shut up. Solve for k, and plug this in back into
$$15.6 = 28.6e^{-kt}$$
and solve for t. Remember that this t value corresponds to 09:30am, so subtract to find the time of death.

Last edited: Apr 30, 2011
5. May 6, 2011

### camp3r101

Re: Modeling with exponential and logarithmic functions help???

DANNGGG! You guys are soo much help!! Much thanks to you!
I see what steps I didn't take. Thanks again! :)

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