algebraic manipulation

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January 24th 2010, 12:58 PM
madmartigano

algebraic manipulation

I need to find the relationship between A and B, where

$A = \frac{{{L_2} - {L_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}<br />$ and $B = \frac{{{V_2} - {V_1}}}{{{V_1}\left( {{t_2} - {t_1}} \right)}}$

I'm assuming I need to make ${V_1} = {L_1}{W_1}{H_1}\,\,\,{\rm{and}}\,\,\,{V_2} = {L_2}{W_2}{H_2}$

So basically, how do I get

$\frac{{{L_2} - {L_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}$ out of $\frac{{{L_2}{W_2}{H_2} - {L_1}{W_1}{H_1}}}{{{L_1}{W_1}{H_1}\left( {{t_2} - {t_1}} \right)}}$ ?

I've only gotten so far:

${W_1}{H_1}B = \frac{{{L_2}{W_2}{H_2} - {L_1}{W_1}{H_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}$

and I can't figure out how to extract ${{L_2} - {L_1}}$ from the numerator on the right side of the equation.

Is this even possible? Am I going about this problem incorrectly?
January 24th 2010, 01:43 PM
Moo

Hello,

Your initial question is to get a relation between A and B, right ?

If so, just look at their formulas. They both have $(t_2-t_1)$ in it.

So, for example, write that $t_2-t_1=\frac{V_2-V_1}{V_1B}$ and substitute it in A.

If it's not what you're looking for, can you be more precise about what you want ? I can't see what you want to do with the stuff you've written down in the rest of your message (Worried)
January 24th 2010, 02:13 PM
Archie Meade

Or, if we divide A by B, we can invert B and multiply

$\frac{A}{B}=\frac{L_2-L_1}{L_1(t_2-t_1)}\ \frac{V_1(t_2-t_1)}{V_2-V_1}=\frac{V_1(L_2-L_1)}{(V_2-V_1)L_1}$
January 24th 2010, 02:22 PM
madmartigano

Thank you. I suppose I was just overthinking the problem. Got it now :)