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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: letsride on September 08, 2010, 06:36:08 pm

Title: solve for T
Post by: letsride on September 08, 2010, 06:36:08 pm: express T in terms of t

this was an anaylis type q based on newtons law of cooling

btw Te, To, k, P, T, t are terms just like x,y,z

(http://i54.tinypic.com/2rdyjrr.jpg)

im kinda stuck w/ the absolute sign

the answer i think was T = Te + (P/k) + (Te-To-(P/k))e^-kt
Title: Re: solve for T
Post by: letsride on September 08, 2010, 06:53:25 pm: i know it gets to e^kt = |[-k(To-Te)+P]/[-k(T-Te)+P]|

w/ the absolute sign, can't you just take it as +-e^kt, however the answer provided throws away the - ???
Title: Re: solve for T
Post by: brightsky on September 08, 2010, 07:26:08 pm: If it helps, always try to think of the absolute values signs as $\sqrt{(...)^2}$ .
Title: Re: solve for T
Post by: letsride on September 08, 2010, 07:34:48 pm: don't understand how the answer is T = Te + (P/k) + (Te-To-(P/k))e^-kt unless you ignore the absolute :s
Title: Re: solve for T
Post by: Mao on September 09, 2010, 02:18:34 am: Here:

$ \begin{align*} t & = \log_e \left| \frac{-k (T_o - T_e) + P}{-k (T - T_e) + P} \right| \\ \pm e^{t} & = \frac{-k (T_o - T_e) + P}{-k (T - T_e) + P} \\ -k (T - T_e) + P & = \pm \left( -k (T_o - T_e) + P \right) e^{-t} \\ T - T_e & = \frac{P}{k} \pm \left( -(T_o - T_e) + \frac{P}{k} \right) e^{-t} \\ T & = T_e + \frac{P}{k} \pm \left( T_e - T_o + \frac{P}{k} \right) e^{-t} \\ \end{align*} $

And then, there will be some other constraints you'll use to eliminate negative option from the $\pm$