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Math Help - 2 questions

  1. #1
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    2 questions

    The first is a problem:
    A casserole is removed from the oven and placed on a bench in the kitchen. The casserole's rate of cooling is proportional to the difference between its temperature and the temperature of its surroundings. When removed from the oven, the casserole is at a temperature of 80 deg Celcius, and it cools to 65 deg in 10 minutes. If the temperature of the kitchen is kept at a constant 20 deg, find the exact time to cool to 50 deg.
    All temps are in Celcius.

    I sort of got d(temp)/dt = k(temp - 20)
    hence dt/d(temp) = 1/k(temp-20)
    so t = 1/k int(1/(temp - 20)) d(temp)
    but i dont know how i should find the k constant and the C constant of integration to get the time.

    Second question:
    Find the exact value of k if int_1/4->1/2( 1/(sqrt(1-4x^2)) ) dx = kPI

    i sort of got [ asin(x/2) ]1/4 -> 1/2 = kPI
    but i think i ended up with something incorrect. I dont have the solutions at hand so im not sure if im right or not.
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  2. #2
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    for the first question can someone confirm the values of k and c to be -.029 and 142.3 respectively?
    I also got the answer to be 24.09 minutes. is that correct?

    I worked out the second question, i basically got the integral wrong.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by scorpion007 View Post
    The first is a problem:
    A casserole is removed from the oven and placed on a bench in the kitchen. The casserole's rate of cooling is proportional to the difference between its temperature and the temperature of its surroundings. When removed from the oven, the casserole is at a temperature of 80 deg Celcius, and it cools to 65 deg in 10 minutes. If the temperature of the kitchen is kept at a constant 20 deg, find the exact time to cool to 50 deg.
    All temps are in Celcius.

    I sort of got d(temp)/dt = k(temp - 20)
    hence dt/d(temp) = 1/k(temp-20)
    so t = 1/k int(1/(temp - 20)) d(temp)
    but i dont know how i should find the k constant and the C constant of integration to get the time.
    You have:

    d(temp)/dt -k temp =-20 k

    Now multiply through by exp(-kt) to get:

    exp(-kt) d(temp)/dt - k exp(-kt) temp = -20 k exp(-kt)

    Now the left hand side is the derivative of exp(-kt) temp, so:

    d/dt[exp(-kt) temp] = -20 k exp(-kt)

    hence integrating we get:

    exp(-kt) temp = 20 exp(-kt) + const.

    At t=0, temp=80, so:

    80 = 20 + const,

    or const = 60.

    So we have:

    temp = 20 + 60 exp(-kt).

    We are told when t=10 min, temp=65 Celsius so:

    65 = 20 +60 exp(-10k),

    or (if I have done this right):

    k = (-1/10) ln(45/60) ~= 0.028768.

    Now we need to find t so that temp=50, at such a time we have:

    50 = 20 + 60 exp(-0.028768*t),

    or:

    exp(-0.028768*t) = 0.5,

    -0.028768*t = ln(0.5),

    t = -ln(0.5)/0.028768 = 24.094 min

    RonL
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  4. #4
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    great, i got it right! although my method was quite different to yours :/
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