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Math Help - Exponential Growth and Decay

  1. #1
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    Exponential Growth and Decay

    The temperature of an ingot of silver is 60 degrees Celcius above room temperature right now. Twenty minutes ago it was 70 degrees Celcius above room temperature. How far above room temperature will the silver be 15 minutes from now?

    I'm not too sure how to start this problem. I believe the temperatures in coordinate points are (0,60) and (-20,70) and that Newton's Law is used to solve this problem [Ce^(kt) + T] Any help is appreciated. Thanks!
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    Hello, turtle!

    The temperature of an ingot of silver is 60C above room temperature right now.
    Twenty minutes ago, it was 70C above room temperature.
    How far above room temperature will the silver be 15 minutes from now?

    I believe the temperatures in coordinate points are (0,60) and (-20,70) .Right!
    and that Newton's Law is used to solve this problem [Ce^(kt) + T]

    I may be wrong, but I don't believe that "T" is required.

    We have: .d .= .Ce^{kt}


    From (0,60), we have: .Ce^{k0} .= .60 . . C = 60

    . . Hence, the function (so far) is: .d .= .60e^{kt}


    From (-20,70), we have: .60e^{-20k) .= .70

    Then: .e^{-20k} .= .7/6

    Take logs: . ln(e^{-20k}) .= .ln(7/6)

    . . (-20k)ln(e) .= .ln(7/6)

    . . k .= .(-1/20)ln(7/6) . .-0.0077

    Hence, the function is: .d .= .60e^{-0.0077t}


    When t = 15: . d .= .60e^{(-.0077)(15)} . .53.5C

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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by turtle View Post
    The temperature of an ingot of silver is 60 degrees Celcius above room temperature right now. Twenty minutes ago it was 70 degrees Celcius above room temperature. How far above room temperature will the silver be 15 minutes from now?

    I'm not too sure how to start this problem. I believe the temperatures in coordinate points are (0,60) and (-20,70) and that Newton's Law is used to solve this problem [Ce^(kt) + T] Any help is appreciated. Thanks!

    I am not familiar with this coordinate points stuff, I know the differential equations method of doing it. It's a bit longer. Have you done differential equations? If so, this is an alternate method, if not, forget about this reply.

    We begin with T' = -k(T - Tamb), where T' is the rate of change of the temperature of the object, k is a constant, T is the temperature of the object at time t, and Tamb is the ambient (or surrounding) temp, which in this case, as is typical, is the room temp

    So T' = -k(T - Tamb)
    => T'/(T - Tamb) = -k
    => ln(T - Tamb) = -kt + C
    => T - Tamb = Ae^(-kt)
    => T = Ae^(-kt) + Tamb

    Take T for 20 minutes ago as initial temp.
    => T(0) = Tamb + 70 = A + Tamb => A = 70
    => T = 70e^(-kt) + Tamb
    T(20) = 60 + Tamb = 70e^(-20k) + Tamb
    => 70e^(-20K) = 60
    => e^(-20k) = 6/7
    => lne^(-20k) = ln(6/7)
    => -20k = ln(6/7)
    => k = - ln(6/7)/20 = 0.0077

    => T = 70e^(-0.0077t) + Tamb ............finally, we have everything we need to tackle the problem. Everything I did up to this point was to find the
    unknowns in my formula

    => T 15 min from now is given by
    T(35) = 70e^{-0.0077(35)} + Tamb
    = 70e^(-0.2695) + Tamb = 53.46 + Tamb

    So the temp 15 mins from now will be 53.46 degrees celcius above room temp.

    I got the same answer as the great Soroban, so he was right in assuming that T is not required. My method is a lot longer, but it eliminates the need to assume anything--you just follow the method.
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