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Math Help - Need help with 3 Calculus Extra Credit Problems

  1. #1
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    Need help with 3 Calculus Extra Credit Problems

    I'm in intro to calculus and I need help setting this equation up:

    Question 1-(Additional Topics in Integration)-

    Newton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium. A cold drink is removed from a refrigerator on a hot summer day and placed in a room where the temperature is 80F. Express the temperature of the drink as a function of time (minutes) if the temperature of the drink was 40F when it left the refrigerator and was 50F after 20 minutes in the room.

    Thanks!
    Last edited by mr fantastic; November 30th 2008 at 07:54 PM. Reason: Re-edited by Mr F. Original post restored.
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  2. #2
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    Newton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium.

    \frac{dT}{dt} = k(T-A)

    A = room temperature (a constant)
    T = temperature of the cold drink at any time t in minutes
    k = proportionality constant

    separate variables and integrate.
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  3. #3
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    Ok, with your help and help from another problem, I have this. Is it the next step?

    dT/dt= -k(T-TM)

    Where T(t) is the temperature of the drink, and TM is the temperature of the surrounding solution.

    So:

    dT/dt= -(T-80)

    This doesn't seem right, I don't know how to incorporate the other numbers and variables. Grrr calculus
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  4. #4
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    \frac{dT}{dt} = k(T-80)

    separate variables ...

    \frac{dT}{T-80} = k \, dt

    integrate ...

    \ln|T-80| = kt + C_1

    |T-80| = e^{kt + C_1}

    |T-80| = e^{C_1} \cdot e^{kt}

    T - 80 = C_2 \cdot e^{kt}

    T = 80 + C_2 \cdot e^{kt}

    The calculus is done, so I'm stopping at this point. The rest is algebra ... you were given two temperatures at two different times. With that info, you can determine the constants C_2 and k and finalize the temperature as a function of time.
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  5. #5
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    I greatly appreciate your help. I have:

    Initial value (40 degrees at time 0)
    T= 80 + Ce^kt
    40= 80 + Ce^k*0
    40= 80 + C
    C= -40
    How do I find k?

    Second value (50 degrees at time 20)
    T= 80 + Ce^kt
    50= 80 + Ce^k*20
    -30= Ce^20k
    Again finding k has stumped me.

    What does k represent and how do i find it?
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  6. #6
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    Quote Originally Posted by Sm10389 View Post
    I greatly appreciate your help. I have:

    Initial value (40 degrees at time 0)
    T= 80 + Ce^kt
    40= 80 + Ce^k*0
    40= 80 + C
    C= -40
    good.

    How do I find k?
    Second value (50 degrees at time 20)
    how about substituting in -40 for C ? then find k with the second value.
    T= 80 + Ce^kt
    do it.
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  7. #7
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    you my friend are a genius.

    so....

    Initial value (40 degrees at time 0)
    T= 80 + Ce^kt
    40= 80 + Ce^k*0
    40= 80 + C
    C= -40

    Second value (50 degrees at time 20)
    T= 80 + Ce^kt
    50= 80 + Ce^k*20
    -30= -40e^20k
    3/4= e^20k
    ln3/4=20k
    k= ln(3/4)/80

    so would this be my final answer?

    T= 80 + -40e^ln((3/4)/20)t
    Last edited by Sm10389; November 21st 2008 at 08:42 PM.
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  8. #8
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    skeeter

    how does that look
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  9. #9
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    Quote Originally Posted by Sm10389 View Post
    how does that look
    check it out yourself ... graph the result in your graphing utility and see if the given info matches up.
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  10. #10
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    i just did, and it is not correct. where did i goof up?
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  11. #11
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    so would this be my final answer?

    T= 80 + -40e^ln((3/4)/20)t
    looks like you have ...

    k = \ln\left(\frac{\frac{3}{4}}{20}\right)

    ... which it ain't.

    should be ...

    k = \frac{\ln\left(\frac{3}{4}\right)}{20}
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  12. #12
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    Thank you, I had that too, I just did not use the parentheses correctly in my calculator.

    The next step of the problem is to calculate it if the drink were warmer then room temperature (>80).

    I know how to set it up like the last one, but we would only be given one point (0, 85) for example.

    How would I calculate k here?
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  13. #13
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    k will remain the same because the rate of heat transfer will remain the same

    ... however, you'll have to recalculate C_2.
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  14. #14
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    QUESTION #2

    More from the Introduction to Differential Equations-

    Investment plan- an investor makes regular deposits totaling D dollars each year into an account that earns interest at the annual rate r compounded continuously.

    A: Explain why the account grows at the rate ( dV/dt = rV + D ) where V(t) is the value of the account 2 years after the initial deposit. Solve this differential equation to express V(t) in terms of r and D.

    I came up with this:
    V(t)= (C/r)*e^rt - (D/r)

    I am sure it is correct. This is the next part:
    Amanda wants to retire in 20 years. To build up a retirement fund, she makes regular annual deposits of $8,000. If the prevailing interest rate stays constant at 4% compounded continuously, how much will she have in her account at the end of the 20 year period?

    I know how to do everything but:
    Find C
    Figure out how compounding continuously would affect the equation.

    Please help me.
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