1. ## 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:

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

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

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

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

I believe the answer for (a) to be:

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

However for question (b) I've gotten:

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle t = 36.49185973$ Seconds

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

Thanks.

2. They look fine, though your LaTeXing needs work...

3. Brill thanks.

Yeh my LaTeXing kinda sucks.. my handwritings even worse.

4. You don't need to use the "$\displaystyle T_1$" from the first problem. When $\displaystyle T_2= \frac{1}{3}T_1$, the equation becomes $\displaystyle \frac{1}{3}T_1= T_1(1- e^{-t/\tau})$. Divide both sides by $\displaystyle T_1$: $\displaystyle \frac{1}{3}= 1- e^{-t/90}$ so $\displaystyle e^{-t/90}= \frac{2}{3}$. $\displaystyle -t/90= ln(2/3)$, $\displaystyle t= -90 ln(2/3)$.