Results 1 to 7 of 7

Math Help - Differential equations and a dead body.

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Differential equations and a dead body.

    Let's assume we find a dead body in a room with a constant temperature of 24 degrees.

    At 8a.m, the body has a temperature of 28 degrees. At 9a.m it has a temperature of 26 degrees. Assuming normal body temperature is 37 degrees, when was the person killed?
    This is what I have so far:

    \frac{d \theta}{dt}=-k(\theta-\theta_0)
    I need to find k.

    \frac{d \theta}{dt_1}=-k(28-24)=-4k
    \frac{d \theta}{dt_2}=-k(26-24)=-2k

    Both these equations give:
    \theta=-4kt_1+C
    \theta=-2kt_2+C

    Hence:
    28=-4k(8)+C \Rightarrow \ 28=-32k+C
    26=-2k(9)+C \Rightarrow \ 26=-18k+C

    2=-14k \Rightarrow k=-\frac{1}{7}

    However, my book says k should be ln \ 2. What's going wrong?
    Last edited by mr fantastic; March 16th 2009 at 12:19 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276
    The problem is that you don't have
    \frac{d\theta}{dt_1}=-k(28-24)=-4k
    \frac{d\theta}{dt_2}=-k(26-24)=-2k
    (these are notationally and conceptually incorrect), but
    \left.\frac{d\theta}{dt}\right|_{t=t_1}=-k(28-24)=-4k
    \left.\frac{d\theta}{dt}\right|_{t=t_2}=-k(26-24)=-2k.
    And you cannot turn these into values for \theta like you did.
    Instead, you need to solve the general differential equation
    \frac{d\theta}{dt}=-k(\theta-\theta_0) to get \theta(t) in terms of k and \theta_0, and then plug in your two pairs (t,\theta) to obtain two equations that can be solved for the unknowns k and \theta_0.

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    \frac{d \theta}{dt}=-k(\theta-\theta_0)

    \frac{d \theta}{dt}=-k \theta+k\theta_0

    \frac{d \theta}{dt}+k \theta=k \theta_0

    R=e^{\int k dt}=e^{kt}

    e^{kt} \frac{d \theta}{dt}+ke^{kt} \theta=k \theta_0 e^{kt}

    \frac{d}{dt} (e^{kt} \theta)=k \theta_0e^{kt}

    e^{kt} \theta= \theta_0 e^{kt}+C

    \theta=\theta_0+Ce^{-kt}

    Ah, this looks promising!

    So we have:

    28=24+Ce^{-8k} and 26=24+Ce^{-9k}.

    28=24+Ce^{-8k}\Rightarrow C=4e^{8k}

    Therefore:

    2=4e^{8k}e^{-9k} \Rightarrow \ 2=4e^{-k}

    \frac{1}{2}=e^{-k}

    -ln \ 2=-k \Rightarrow \ k=ln \ 2

    Thanks!

    One more thing, when I substitute my value for k back into the equation and work everything out I get t=-2.28 hours. Which time do I add this to?

    Do I do 8:00-2.28 or 9-2.28?
    Last edited by Showcase_22; March 17th 2009 at 04:27 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276
    Neither. You did t=8 for 8:00 AM and t=9 for 9:00 AM, thus t=0 is midnight, and t=-2.28 is 2.28 hours before midnight, or about 9:43 PM the previous night.

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    hmm, okay then. Let me see if i'm doing this correctly:

    C=4e^{8ln\ 2}=2^{10}

    \theta=24+2^{10}e^{-ln 2 \ t}

    \theta=24+2^9e^t

    37=24+2^9e^t

    13=2^9e^t

    t=ln\frac{37}{2^9}=-2.627 which is approximately -2 hours 38 minutes (sorry, I mistyped it in the previous post:s)

    Hence the time should be 9:22pm the previous day.

    Unfortunately my book has done 9:00-2 hours 38 minutes giving 6:22 am.

    Does this mean my book is wrong?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276
    Well, first I can see that when you went from \theta=24+2^{10}e^{-(\ln{2})t} to the next line, you made an error:
    e^{-(\ln{2})t} is not \frac{1}{2}e^{t}. Instead, you have
    e^{-(\ln{2})t}=\left(e^{\ln{2}}\right)^{-t}=2^{-t},
    so you should go from
    \theta=24+2^{10}e^{-(\ln{2})t} to \theta=24+2^{10-t} (which, rechecking, still gives theta=28 at t=8 and theta=26 at t=9).
    Letting 37=24+2^{10-t}
    we get
    13=2^{10-t}
    10-t=\log_2(13)
    t=10-\log_2(13)

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Thanks Kevin!

    I actually managed to get the right answer now, HUZZAH!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Dead Link in FAQ to a Latex thread
    Posted in the LaTeX Help Forum
    Replies: 8
    Last Post: December 21st 2010, 02:54 PM
  2. dead on the field
    Posted in the Math Puzzles Forum
    Replies: 2
    Last Post: October 14th 2009, 02:55 AM
  3. Replies: 2
    Last Post: May 18th 2009, 03:49 AM
  4. Replies: 0
    Last Post: April 1st 2007, 11:30 AM

Search Tags


/mathhelpforum @mathhelpforum