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theEric Group Title

Hi! Any help with this ODE word problem will be appreciated! I'll simplify it to the important data: We are keeping track of the temperature of coffee. Well, we have the temperature at two times and want a third. I would like to be able to conceptualize what's going on, if anyone can help! 7:30am \(200^\circ\text F\) 7:45am \(120^\circ\text F\) 8:00am \(?^\circ\text F\) The rate of change of heat is a simplified version of Newton's Law of Cooling, that \(\dfrac{dT}{dt}=k(T-T_r)\) where \(k\) is a proportionality constant and \(T_r\) is the room temperature (\(70^\circ\text F\)).

  • one year ago
  • one year ago

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  1. ganeshie8 Group Title
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    *

    • one year ago
  2. theEric Group Title
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    hM?

    • one year ago
  3. theEric Group Title
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    Hm?*

    • one year ago
  4. theEric Group Title
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    I figure that I probably need to find out what \(k\) is...

    • one year ago
  5. ganeshie8 Group Title
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    @sauravshakya

    • one year ago
  6. theEric Group Title
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    Can I use \(\Large{\int\limits_{7.5}^{7.75}}\normalsize\dfrac{dT}{dt}dt=\Large{\int\limits_{7.5}^{7.75}}\normalsize k(T-T_r)\ dt=200-120\)? I'll think on that...

    • one year ago
  7. theEric Group Title
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    I haven't done anything like it in class, that I recall.

    • one year ago
  8. theEric Group Title
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    But I don't have \(T(t)\) to do integration...

    • one year ago
  9. Saeeddiscover Group Title
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    You can say nothing about 8:00 AM here. Because We don't know how the temperature function is changing over time.

    • one year ago
  10. theEric Group Title
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    That's what I was thinking as well... But we know the rate-of-change function, two points on the \(T\) vs \(t\) curve, which is one solution to the rate-of-change function...

    • one year ago
  11. theEric Group Title
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    I know that initial value problems require one point...

    • one year ago
  12. phi Group Title
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    I think I would solve the equation then use the given data to find the constants \[ \frac{dT}{dt}= k(T -T_r) \\ \frac{dT}{(T -T_r)}= k \ dt \\\ln(T -T_r)= k \ t + C \] where t is time in minutes from 7:30 make each side the exponent of e \[ T- T_r= A e^{k\ t} \] now use 7:30 t=0 T= 200 to find A and 7:45 t=15 to find k

    • one year ago
  13. theEric Group Title
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    Thank you very much! I'm working on digesting it now.

    • one year ago
  14. theEric Group Title
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    \(A\) is an arbitrary constant like \(C\), right?

    • one year ago
  15. phi Group Title
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    yes A = e^C we don't care about C per se.

    • one year ago
  16. theEric Group Title
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    This is somewhat related, @phi , but do we always need to look at things like \(t=0\) for initial value problems? Or can we use non-zero values for the independent variable?

    • one year ago
  17. theEric Group Title
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    Thank you!

    • one year ago
  18. phi Group Title
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    you can use non zero values of t. But obviously t=0 makes it especially easy to find A

    • one year ago
  19. theEric Group Title
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    Right, thank you very much! :)

    • one year ago
  20. theEric Group Title
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    In case it is of any viewer's concern: \(A=130^\circ\text F\) \(k=-0.0637\ [s^{-1}]\) and the coffee will be about \(89.2^\circ\text F\) at 8:00am. :) The problem was about a professor with an 8am class, who never drinks coffee unless it is at \(90^\circ\text F\) or cooler. Good news for him - he can drink the coffee in class! I might show off and solve for the time \(t\) at which he can take his first sip. :P But not until I get the real problems done. Thank you!

    • one year ago
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