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Math Help - Need help re-arranging an equation!

  1. #16
    Member rowe's Avatar
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    Re: Need help re-arranging an equation!

    Remember the rule \frac{a^2}{b^2} = {\left(\frac{a}{b}\right)}^2? well, if you square both sides of the equation, you'll remove the square root.

    Then, you can rearrange to remove the fraction, and solving for z should be easier.
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  2. #17
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    Re: Need help re-arranging an equation!

    Rowe! Your back! I thought you might have gone to bed. Not sure what time zone you are in. Anyway! do I end up with: x^2=(MR/m)^2*((r^2)^2/((1+r^2)^(2+2zr)^2))
    or am I completely off track again?
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  3. #18
    Member rowe's Avatar
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    Re: Need help re-arranging an equation!

    Almost, if my calculations are correct (this is quite a tedious one!) you should have x^2 = \left(\frac{MR}{m}\right)^2 \cdot \frac{r^4}{{(1-r^2)}^2 + (2zr)^2}. You want to try and isolate z, start by getting rid of that fraction by multiplying both sides by {(1-r^2)}^2 + (2zr)^2. Once that's "on the flat", simplify / expand where necessary so you can subtract the (2zr)^2 term on its own side of the equation and work from there.
    Last edited by rowe; June 19th 2012 at 08:36 AM.
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  4. #19
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    Re: Need help re-arranging an equation!

    Quote Originally Posted by rowe View Post
    x = \frac{MR}{m}\cdot\frac{r^2}{\sqrt{{(1-r^2)}^2+{(2zr)}^2}}
    Looks like this is the final/accepted version

    Let k = SQRT[(1 - r^2)^2 + (2zr)^2] ; then:

    x = MRr^2 / (mk)
    k = MRr^2 / (mx) ; square both sides:

    k^2 = M^2 R^2 r^4 / (m^2 x^2) ; substitute back in:

    (1 - r^2)^2 + (2zr)^2 = M^2 R^2 r^4 / (m^2 x^2) ; rearrange:

    4 z^2 r^2 = M^2 R^2 r^4 / (m^2 x^2) - (1 - r^2)^2

    z^2 = [M^2 R^2 r^4 / (m^2 x^2) - (1 - r^2)^2] / (4r^2) ; sooooooooooo:

    z = SQRT{[M^2 R^2 r^4 / (m^2 x^2) - (1 - r^2)^2] / (4r^2)}

    Can be slightly simplified to:
    z = SQRT[M^2 R^2 r^4 - (m^2 x^2)(1 - r^2)^2] / (2rmx)
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