Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

theEric

  • one year ago

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 (\(70^\circ\text F\)).

  • This Question is Closed
  1. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    *

  2. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    hM?

  3. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Hm?*

  4. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I figure that I probably need to find out what \(k\) is...

  5. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @sauravshakya

  6. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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...

  7. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I haven't done anything like it in class, that I recall.

  8. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    But I don't have \(T(t)\) to do integration...

  9. Saeeddiscover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You can say nothing about 8:00 AM here. Because We don't know how the temperature function is changing over time.

  10. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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...

  11. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I know that initial value problems require one point...

  12. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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

  13. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thank you very much! I'm working on digesting it now.

  14. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(A\) is an arbitrary constant like \(C\), right?

  15. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    yes A = e^C we don't care about C per se.

  16. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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?

  17. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thank you!

  18. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    you can use non zero values of t. But obviously t=0 makes it especially easy to find A

  19. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Right, thank you very much! :)

  20. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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!

  21. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.