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Math Help - Transposition of Formulae - Can someone check my answers

  1. #1
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    Transposition of Formulae - Can someone check my answers

    Hi,

    I've been asked the following question:

    The temperature of an electrical device at a time 't' is given by the function:

    {T}_{2} = {T}_{1}(1-{e}^{\frac{-t}{\tau}})

    Where {T}_{1} is the temperature at time t = 0 and \tau is a constant

    Find a) {T}_{1} when {T}_{2} is 60 degrees Celsius, t = 50 seconds and \tau = 90 seconds

    b) The time t, for {T}_{2} to be \frac{1}{3} the value of {T}_{1}


    I believe the answer for (a) to be:

    \frac{60}{1-{e}^{\frac{-50}{90}}} = {T}_{1} = 140.763593

    However for question (b) I've gotten:

    46.92119767 = 140.763593(1-{e}^{\frac{-t}{90}})

    \frac{46.92119767}{140.763593} = 1-{e}^{\frac{-t}{90}}

    (-1)\frac{46.92119767}{140.763593} = (-1)1-{e}^{\frac{-t}{90}}

    -\frac{46.92119767}{140.763593} = -1+{e}^{\frac{-t}{90}}

    -\frac{46.92119767}{140.763593} + 1 = -1+1+{e}^{\frac{-t}{90}}

    -\frac{46.92119767}{140.763593} + 1 = {e}^{\frac{-t}{90}}

    ln(-\frac{46.92119767}{140.763593} + 1) = ln({e}^{\frac{-t}{90}})

    ln(-\frac{46.92119767}{140.763593} + 1) = \frac{-t}{90}

    ln(-\frac{46.92119767}{140.763593} + 1)(90) = -t

    (-1)ln(-\frac{46.92119767}{140.763593} + 1)(90) = (-1)-t

    (-1)ln(-\frac{46.92119767}{140.763593} + 1)(90) = t

    t = 36.49185973 Seconds

    Can someone please let me know if I've gotten these correct?

    Thanks.
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  2. #2
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    They look fine, though your LaTeXing needs work...
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  3. #3
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    Brill thanks.

    Yeh my LaTeXing kinda sucks.. my handwritings even worse.
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  4. #4
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    You don't need to use the " T_1" from the first problem. When T_2= \frac{1}{3}T_1, the equation becomes \frac{1}{3}T_1= T_1(1- e^{-t/\tau}). Divide both sides by T_1: \frac{1}{3}= 1- e^{-t/90} so e^{-t/90}= \frac{2}{3}. -t/90= ln(2/3), t= -90 ln(2/3).
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