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Math Help - Newton's Law of Cooling

  1. #1
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    Newton's Law of Cooling

    Newton's Law of Cooling can be written as the DE \frac{dT}{dt}=k(T-T_{a}) for some value of k.

    If T-T_{a} > 0 is k positive or negative? Briefly explain why.

    Find the GS of the DE \frac{dT}{dt}=k(T-T_{a}). You may assume T-T_{a} > 0 and the final solution should give T explicitly in terms of k, t, T_{a} and a constant of integration.

    Could someone give me a helping hand with this? I'm new to differential equations and this question was just thrown at me unexpectedly.
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  2. #2
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    Quote Originally Posted by Konidias View Post
    Newton's Law of Cooling can be written as the DE \frac{dT}{dt}=k(T-T_{a}) for some value of k.

    If T-T_{a} > 0 is k positive or negative? Briefly explain why.
    What do "T" and " T_a" mean? If T is the variable temperature and T_a is a constant temperature, then T- T_a> 0 means temperature T is greater than the the temperature T_a and T will decrease to T_a. So dT/dt must be negative and that means that k must be negative.
    By the way, although it wasn't asked, if T< T_a, T- T_a< 0 so T increases to T_a. dT/dt is positive now which means that, since T- T_a is negative, k must still be negative. The "k" in Newton's law of cooling is always negative.

    Find the GS of the DE \frac{dT}{dt}=k(T-T_{a}). You may assume T-T_{a} > 0 and the final solution should give T explicitly in terms of k, t, T_{a} and a constant of integration.

    Could someone give me a helping hand with this? I'm new to differential equations and this question was just thrown at me unexpectedly.
    "Separate" it: \frac{dT}{T- T_a}= kdt. Integrate both sides. You will get a logarithm on the left so solving for T will give an exponential of kt.
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     <br />
\frac{dT}{dt}=k(T-T_{a})<br />

    \int \frac{dT}{T-T_{a}}=\int kdt

     <br />
ln|T-T_{a}|+C=kt<br />

     <br />
T-T_{a}+C=e^{kt}<br />

     <br />
T=e^{kt}+T_{a}+C<br />

    Is this correct?
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