Calc problem

[crazeinc]

A murder victim was discovered at noon in an apartment in which the thermostat was set at 78*F. The coroner immediately measured the body temperature to be 93*F. Two hours later the temperature was measured again, and it had dropped one degree. Approximately when (to the nearest minute) had the murder been committed? Note: You will need to use the fact that the body temperature is 98.6*F.

Using Newtons Law of Cooling: y(t)=(y(original)-T)e^(kt)+T here's the solution I came up with. I was wondering if somebody could tell me if I'm on the right track?

93=(98.6-78)e^(tk)+78
15=20.6e^(kt)
(ln .73)/t=k

92=(98.6-78)e^((t+2)k)+78
14=20e^(kt)
(ln .68)/(t+2)=k

(ln .73)/t = (ln .68)/(t+2)
t=8.87 (not sure how to solve this, used my calculator)

noon - 8.87 hours = 3.13 = 3:08AM

 

[crazeinc]

actually, i don't think that's right at all...

 

[crazeinc]

Then again, I thought it was wrong because t would be different for both, but that's why I added 2 to it in the second equation.

 

[TuxDave]

Are you sure you have the equation correct? Shouldn't it be e^(-kt)?

 

[TuxDave]

So given this new equation:


93=(98.6-78)e^(-kt)+78
15=20.6e^(-kt)

92=(98.6-78)e^(-k(t+2))+78
14=20.6e^(-k(t+2))

15/14=e^(-kt+kt+2k)
15/14=e^(2k)
k=something

etc....

 

[crazeinc]

my equation is correct, the k isn't negative in the formula. It is negative though if you solve for k in (ln .73)/t=k or (ln .68)/(t+2)=k if you plug 8.87 in for t which leads me to believe this may be the right answer.

 

[crazeinc]

bump

 

[agnitrate]

not a calculus problem

 

[aplefka]

Looks like Algebra 2 to me.

Edit: It is Algebra 2... growth and decay. I'd do it but I'm too tired.

 

[crazeinc]

great contribution guys...

 

[xirtam]

It's an ordinary differential equation. Not algebra 2, I don't think. Just a second, I suck at these.

 

[bigj3347]

algebra two for sure . .. i'd help if i actually remember how.

 

[pray4mojo]

I'm also thinking of calculus when I read the problem but that's probably because I have to take the AP exam in about a week. I would do it but it's 1AM and I going to sleep...

 

[xirtam]

I got 2:48 AM.

dT/dt=k*(78-T)

Integrating yields

ln(T-78)=-k*t + C

T(0)=93 degrees, ambient 78 degrees
T(2)=92 degrees, ambient 78 degrees

ln(93-78) = -k*0+C ... C=ln(15)
ln(92-78) = -2k + ln(15)... k=ln(15/14)/2

ln(98.6-78) = -tD * ln(15/14)/2 + ln(15)

tD=-9.197 (check this, I didn't use a fancy calculator)

= approximately 9 hours and 12 minutes before noon, so I get 2:48 am, time of death.