# Thread: Transposition of Formulae - Can someone check my answers

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:

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

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.

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 " $T_1$" 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 $e^{-t/90}= \frac{2}{3}$. $-t/90= ln(2/3)$, $t= -90 ln(2/3)$.