Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Newtons cooling law DE, mathematical approach

  1. Jan 29, 2006 #1

    Pengwuino

    User Avatar
    Gold Member

    I have this one problem dealing with Newton's law of cooling:

    [tex]\frac{{dT}}{{dt}} = - k(T - A)[/tex]

    I'm basically trying to determine at what time someone died! The info I have is that at time of death, the temperature was 98.6 degrees, at 10 AM it was at 92 degrees, and at 2PM it was at 86 degrees. The surrounding temperature, A, was 78 degrees and constant. Unfortunately, the book does not give an example as to how this DE works.

    I have a feeling I need to do this:

    [tex]\frac{{dT}}{{(T - 78)}} = - kdt[/tex]

    And integrate it… but I'm not sure how I would do that… especially with it just being k*dt. Any suggestions?
     
  2. jcsd
  3. Jan 29, 2006 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What? you have a problem with kdt? What is the integral of a constant?

    If dT/(T-78), try a simple substitution: let u= T- 78 so du= dT. Can you integrate du/u??
     
  4. Jan 29, 2006 #3

    benorin

    User Avatar
    Homework Helper

    [tex]\int\frac{{dT}}{{(T - 78)}} = - k\int dt\Rightarrow\ln (T - 78) = - kt+C[/tex]
     
    Last edited: Jan 29, 2006
  5. Jan 29, 2006 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Well, your approach looks perfectly fine to me.
    Alternatively, you could introduce the the dependent variable u(t)=T(t)-A,
    and recognizing that u obeys the diff. eq [itex]\frac{du}{dt}=-ku[/itex]
    whereby you see that we have:
    [itex]u(t)=Ce^{-kt}\to{T}(t)=A+Ce^{-kt}[/itex]
    If you let t=0 correspond to 10 AM, then you can determine C and k from the info known at that time, along with the info known at 2P.M.
    Finally, use the info about the body temp. at time of death to determine when he died.
     
  6. Jan 29, 2006 #5

    Pengwuino

    User Avatar
    Gold Member

    I got to the second line benorin stated and now im stuck again... I feel I need to find 'k' but I don't have an initial condition to work with, only a time change...
     
  7. Jan 29, 2006 #6

    Pengwuino

    User Avatar
    Gold Member

    Oh so are you trying to say I could use 10:00AM as a t=0 and simply use the negative value (would I get a negative value?) to determine how long before 10:00am he died?
     
  8. Jan 29, 2006 #7

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Yup! :smile:

    And yes, you'll get a negative value.
     
  9. Jan 29, 2006 #8

    Pengwuino

    User Avatar
    Gold Member

    Ahhhh there we go! -2.75 hours later... so 7:14am!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Newtons cooling law DE, mathematical approach
  1. Solving DE Mathematica (Replies: 5)

  2. Mathematics Software (Replies: 1)

Loading...