Newtons cooling law DE, mathematical approach

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Discussion Overview

The discussion revolves around applying Newton's law of cooling to determine the time of death based on temperature readings at different times. Participants explore the mathematical approach to solving the differential equation involved, including integration techniques and the use of initial conditions.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant presents a differential equation based on Newton's law of cooling and seeks guidance on integrating it to find the time of death.
  • Another participant suggests a substitution method to simplify the integration process.
  • A third participant confirms the initial approach and proposes an alternative representation of the temperature function, introducing a new variable.
  • There is a discussion about determining the constants 'C' and 'k' using the known temperature values at specific times.
  • A participant expresses confusion about finding 'k' without an initial condition, prompting further clarification from others.
  • Participants confirm that using 10 AM as a reference point for 't=0' is valid and that a negative value for time before this point is expected.
  • One participant concludes with a calculated time of death based on the discussion, indicating a specific time prior to 10 AM.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical approach to solving the problem, but there is some uncertainty regarding the initial conditions and the determination of the constant 'k'. The discussion remains focused on exploring these mathematical aspects without reaching a consensus on the final interpretation of the results.

Contextual Notes

Participants note the lack of explicit initial conditions for determining 'k', which may affect the integration process. The discussion also highlights the dependence on the chosen reference time for calculations.

Pengwuino
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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?
 
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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??
 
[tex]\int\frac{{dT}}{{(T - 78)}} = - k\int dt\Rightarrow\ln (T - 78) = - kt+C[/tex]
 
Last edited:
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.
 
I got to the second line benorin stated and now I am stuck again... I feel I need to find 'k' but I don't have an initial condition to work with, only a time change...
 
arildno said:
[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.

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?
 
Yup! :smile:

And yes, you'll get a negative value.
 
Ahhhh there we go! -2.75 hours later... so 7:14am!
 

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