Newtons Cooling problem/ possible error in book.

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I have a book called REA 's problem solvers: Differential Equations Year 2004 and the problem 12-14 on page 244 might have problems with it and I want to ask to see if anyone else thinks so.

The problem is basically Newton's cooling where they ask to find

T - T_o = (T_i - T_o) e^(-kt) (1)

where T_i = T when t = 0.

Given dT/dt = -k(T - T_o) (2)

where T_o is the temperature of the surrounding medium.

The part that seems incorrect to me is where they integrate after rearranging (2) to get

LN(T - T_o) = -kt + LN(C)

1. Shouldn't the left side be -LN(T_o - T) after integration?
2. How do they get LN(C) instead of just C and then solving for t = 0, C = -LN(T_o - T)?
 

Answers and Replies

  • #2
dextercioby
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To put C or ln C is the same, the constant is arbitrary. Ln C is more convenient. As for the first question, you're basically computing

[tex] \int \frac{dT}{T-T_0} [/tex]
 
  • #3
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To put C or ln C is the same, the constant is arbitrary. Ln C is more convenient. As for the first question, you're basically computing
Would C = T - T_o instead? That would make more sense.
And then LN C (or LN(T - T_o) ) at t = 0 would be the integrating factor?

[tex] \int \frac{dT}{T-T_0} [/tex]
Right, I know I'm computing [tex] \int \frac{dT}{T-T_0} [/tex] but is it correct in the book or am I right. I used Derive 6 and I got a different answer after integration.

EDIT: Nevermind on the integration, it is correct. I made a mistake in Derive 6. But I'm still not sure about the constant C.
 
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  • #4
Integral
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... But I'm still not sure about the constant C.
C is a constant, so is lnC, or C2 or any other operation done to C. The constant is arbitrary, so the exact form of it is not important. Though the form can make succeeding steps more transparent.
 
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https://www.youtube.com/watch?v=xq-JiYMQydc
 

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