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Math Help - Find the limit as t approaches infinity of T(t).

  1. #1
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    Find the limit as t approaches infinity of T(t).

    I have this problem and it seems pretty easy. I think I'm just overthinking it.

    At time t = 0 minutes, the temperature of a cup of coffee is 180 degrees Fahrenheit. Left in a room whose temperature is 70 degrees Fahrenheit, the coffee cools so that its temperature function T(t), also measure in degrees Fahrenheit, satisfies the differential equation: dT/dt = -1/2(T) + 35.

    Find \lim_{t \to \infty} T(t) . Explain what this means in the context of the problem.

    What is it asking me to do when it says " \lim_{t \to \infty} T(t) ?" Is it asking me to set up an equation and solve (how would I do that?), or just explain what's happening? Also, can someone briefly explain what is happening? I'm stumped. Thanks for any help.

    the limit as t approaches infinity of T(t)
    Last edited by maddawg579; April 4th 2010 at 10:46 AM.
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  2. #2
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    Quote Originally Posted by maddawg579 View Post
    I
    What is it asking me to do when it says "Find the limit as t approaches infinity of T(t)?" Is it asking me to set up an equation and solve (how would I do that?), or just explain what's happening? Also, can someone briefly explain what is happening? I'm stumped. Thanks for any help.

    the limit as t approaches infinity of T(t)
    It's asking you to solve the differential equation, obtain the function T(t), and then take it's limit as t\to\infty right?

    You can solve \frac{dT}{dt}=-1/2 T+35 by separating the variables: \frac{dT}{35-1/2 T}=dt

    See if you can get to \ln(35-1/2T)=-t/2+c

    then plug in the initial values T(0)=180 to determine what c is, then take exp of both sides, (or take exp of both side then figure out what c is) to get it to the form T(t), then take the limit \lim_{t\to\infty} T(t).
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  3. #3
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    So if I plug in 180 for T, I'd get \ln(35-90)=-180/2+c?

    Then solve for c, which becomes 94.828, I think. Unless I skipped a step.

    Then do I take the e of both sides? Is that what you mean by exp?
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  4. #4
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    Dawg, you're getting the function name T confused with the variable name t. Keep them straight. So if I have \ln(35-1/2 T)=-t/2+c and c is an arbitrary constant from the integration, then if I add, subtract, multiply anything to it, it's still a constant. Then doing a little algegra, I get:

    35-1/2T=ce^{-t/2}

    (still just an arbitrary constant c even though I took exponentials.

    little more:

    T(t)=70-ce^{-t/2}

    Now substitute the initial conditions T(0)=180 to figure out that c=-110 to get:

    T(t)=70+110 e^{-t/2}

    Now, you take that limit of that as t\to\infty.
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  5. #5
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    Ok, that makes sense. I'm new to calc, so bear with me.

    I can see you took the e of both sides, so the ln cancels with the e on left side, and on the right side the -t/2 becomes the exponent of e. But why is c multiplied to that?

    After that I can see you moved 35 over and divided by -1/2 to get 70. I get that and I understand the rest.
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  6. #6
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    \ln(35-1/2 T)=-t/2+c

    so taking exponents of both sides:

    35-1/2 T=e^{-t/2+c}=e^c e^{-t/2}

    but e^c is still just some arbitrary constant if c is so I can replace it with just the arbitrary constant c and get:

    35-1/2 T=ce^{-t/2}
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  7. #7
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    Ok, thanks so much. That all makes perfect sense. You were a huge help.
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  8. #8
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    Wasn't it obvious from the start that \lim_{t\to\infty} T(t)= 70 degrees?

    Mathematically, that " -\frac{1}{2} multiplying the T in the differential equation tells you that the solution will involve an exponential with negative exponent that will go to 0 as t goes to infinity.

    Physically, the coffee cools down to room temperature.
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  9. #9
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    Oh I knew it had to equal 70 degrees in the end. I just needed to know the steps to get there.
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