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anonymous

  • 4 years ago

1. Newton’s law of cooling states that the rate of cooling of an object is proportional to the difference between its temperature and the ambient temperature. (a) Assuming this, formulate an initial value problem that models the cooling of a cup of coffee. (b) Suppose that the ambient temperature is 20◦ C, the coffee is initially at a temperature of 90◦ C, and is initially cooling at a rate of 8◦/minute. How long will it take for the coffee to cool to 75◦.

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  1. anonymous
    • 4 years ago
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    i believe the answer is1.875 minutes

  2. anonymous
    • 4 years ago
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    can someone instruct me how to get the answer ...

  3. JamesJ
    • 4 years ago
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    Let \( T_A \) be the ambient temperature and T the temperature of the object cooling. What the problem tells you is \[ \frac{dT}{dt} \ \ \ \alpha \ \ - (T - T_A) \] hence you can write down the ODE with a constant of proportionality.

  4. JamesJ
    • 4 years ago
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    Now. Use the facts given to you in part (b) to calculate that constant. Then, solve the ODE and apply the initial values given to obtain the solution to this problem. Finally, find the time t such that T = 75 C.

  5. JamesJ
    • 4 years ago
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    So, does what I wrote down above make sense and do you see what to do next?

  6. anonymous
    • 4 years ago
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    not really .. i dont think that we're supposed to use the second part of the question to figure out part a.. and i'm not sure where the alpha came from

  7. JamesJ
    • 4 years ago
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    alpha means "is proportional to"

  8. JamesJ
    • 4 years ago
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    The change in temperature of the object, like a cup of hot water, is proportional to the difference of its temperature to the ambient temperature. The negative sign is there because the temperature is going down towards the ambient temperature if it currently above it; or changing upwards, if T < T_a.

  9. anonymous
    • 4 years ago
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    hmmm.. i wish i understood, thanks anyways though

  10. JamesJ
    • 4 years ago
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    Let's try this one more time. You're told that: "Newton’s law of cooling states that the rate of cooling of an object is proportional to the difference between its temperature and the ambient temperature." Let's write T(t) for the temperature of the cooling object. This law of cooling is telling us something about \[ \frac{dT}{dt} \] so far so good?

  11. anonymous
    • 4 years ago
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    ok.

  12. anonymous
    • 4 years ago
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    ok i think i may have misread the question ... it's asking for an inital problem and not initial formula .. i'm not really sure what it's asking.

  13. JamesJ
    • 4 years ago
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    Now, what is the change proportional to? "proportional to the difference between its temperature and the ambient temperature" Let \( T_A \) be the ambient temperature, then the difference between the temperature T and the ambient temperature is \[ T - T_A.\] Hence \[ \frac{dT}{dt} \ \text{ is proportional to } -(T-T_A) \] (We write \[ \frac{dT}{dt} \ \ \alpha \ -(T-T_A) \] this is standard notation; I'm sure you've seen it before, but it isn't very important for this problem.)

  14. JamesJ
    • 4 years ago
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    What is important is the negative sign. It must be negative because if \( T > T_A \) then T should be going down. I.e., the object is cooling and its temperature T should be getting closer to the ambient temperature. So far, so good? Be honest.

  15. anonymous
    • 4 years ago
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    Ok if I'm being honest, I just started a differential calculus class and I don't even know what we're learning .. I'm not even sure what the objective of the course is. I sound stupid, but promise just takes a little time ahah .. Ok, so for the first part all we're trying to say is that the 2 are proportional to eachother? we dont need an equal sign or anything then in the answer right?

  16. JamesJ
    • 4 years ago
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    We do. We now introduce a constant of proportionality. What that equation means in concrete terms is \[ \frac{dT}{dt} = -k(T-T_A) \] for some positive constant k. This is the ODE we want.

  17. anonymous
    • 4 years ago
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    ODE means original differential equation?

  18. JamesJ
    • 4 years ago
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    ordinary differential equation.

  19. anonymous
    • 4 years ago
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    lmao alright. you must think i'm a retard.

  20. JamesJ
    • 4 years ago
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    I'm going to refer you now to this lecture where this equation is solved. http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-3-solving-first-order-linear-odes/ I *strongly* recommend you watch it, take notes, rewind, watch it again. We all take a while to understand new concepts. This example of Newton's Law of Cooling is a classic.

  21. anonymous
    • 4 years ago
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    ok thank you.

  22. JamesJ
    • 4 years ago
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    after you've watched it, if you still have doubts, come find me.

  23. anonymous
    • 4 years ago
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    ok .. will do, thanks

  24. JamesJ
    • 4 years ago
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    btw, the solution method he shows you in the lecture is more powerful than what you need. But no harm done, because it is a good general method. I can still show you a simpler method for you equation in particular: separation of variables. You may know it already.

  25. JamesJ
    • 4 years ago
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    but more than "it's a good general method", this is going to be the next thing you learn in your class anyway.

  26. anonymous
    • 4 years ago
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    kind of... i'm working on b now, and for solving ... are you able to walk me through the steps to getting the solution?

  27. anonymous
    • 4 years ago
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    if your still there ... i got T=Ta +_ e^(-kt+c)

  28. JamesJ
    • 4 years ago
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    Right, \[ T(t) = T_a + Ae^{-kt} \] For some constant A ( = e^c ) Now, in part b, we can solve for A. We have that T(0) = 90. Therefore \[ 90 = T(0) = T_a + A \] hence, \[ A = 90 - T_a \] We also know that \( T_a = 20 \), thus \( A = 90 - 20 = 70 \) and the solution is \[ T(t) = 20 + 70e^{-kt} \] ok?

  29. anonymous
    • 4 years ago
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    yes thank you so much!

  30. JamesJ
    • 4 years ago
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    Now, one more thing. You need to find k. How are yo going to do that? You're given one more piece of information in (b) "[the coffee] is initially cooling at a rate of 8◦/minute". Translate that into the mathematics.

  31. anonymous
    • 4 years ago
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    so in math terms thats .. -8t?

  32. JamesJ
    • 4 years ago
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    "Cooling at a rate of 8 degrees/minute" That must mean dT/dt at t = 0 is 8 degrees/minute Now dT/dt = k(T_a - T) You're told that for t = 0 dT/dt = -8 degrees/minute We also have that k(T_a - T) = k(20-90) = -70k Thus -70k = -8 You can now solve for k.

  33. anonymous
    • 4 years ago
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    ohhh .. ok, and then once i've solved for k i can solve for t?

  34. JamesJ
    • 4 years ago
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    Yes, then finally you can find the t for which T(t) = 75

  35. anonymous
    • 4 years ago
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    aren't you just wonderful, thank you so so so much!

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