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Algebraic Manipulation

  1. May 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Simplify so that there is no square root function. Leave it in terms of S

    S = xc^2 / (1 - 2Gy/rc^2)^1/2

    [tex] s = \frac{xc^2}{(1-\frac{2Gy}{rc^2})^{1/2}}[/tex]

    Attempt

    I've tried too many things, it would be too cumbersome to copy all of it down.
     
  2. jcsd
  3. May 24, 2012 #2

    rock.freak667

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    Homework Helper

    Try squaring both sides and making the fraction simpler. Not sure which term you want as the subject though.
     
  4. May 24, 2012 #3
    I've done that, but ultimately we need it to be an explicit function of E, so you will still have to take the square root of the right side and it doesn't seem to be possible to simplify it.
     
  5. May 24, 2012 #4

    Mark44

    Staff: Mentor

    You've been around here long enough to know the drill. You don't need to copy all of it down, but you do need to show some effort. I cut you some slack this time.

    E? There's no E in this problem.
     
  6. May 25, 2012 #5
    Sorry, mistype. I meant S.

    Sorry about that.
     
  7. May 26, 2012 #6

    HallsofIvy

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    Staff Emeritus
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    What, exactly, do you mean by "no square root function"? You can, of course, immediately replace the square root by a 1/2 power, as you have done. Why is that not a correct answer? You can, of course, proceed to combine the fraction and simplify but you will still have 1/2 powers whatever you do.
     
  8. May 26, 2012 #7
    Thanks for your time.

    The 1/2 power is the equivalent of the square root so that defeats the purpose.
     
  9. May 26, 2012 #8

    Mark44

    Staff: Mentor

    Is the goal to get rid of the square root in the denominator?
     
  10. May 26, 2012 #9
    Yes, but within the condition that equation is left as an explicit function of E.
     
  11. May 26, 2012 #10
    Multiply the quotient on the right-hand side of the equation by a convenient form of 1; in this case, [tex]\frac{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}}{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}}[/tex]

    This technique is called rationalizing the denominator.
     
    Last edited: May 26, 2012
  12. May 26, 2012 #11

    SammyS

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    There's that "E" again !

    Perhaps ...
    Are you supposed to be solving for y ?​
     
  13. May 26, 2012 #12
    [tex]s = \frac{xc^2}{(1-\frac{2Gy}{rc^2})^{1/2}} \frac{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}}{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}} [/tex]

    [tex] s = xc^2 \frac{\sqrt{1 - \frac{2Gy}{rc^2}}} {1-\frac{2Gy}{rc^2}}[/tex]
    [tex] s = xc^2 \sqrt{1 - \frac{2G}{rc^2}} (1 - \frac{rc^2}{2Gy})[/tex]
    Whoops.

    No, for S.
     
    Last edited: May 26, 2012
  14. May 26, 2012 #13
    There is no longer a square root in the denominator, so you have succeeded in eliminating the square root in the denominator. However, I am afraid the last step is incorrect because

    [tex]\frac{1}{1 - \frac{2Gy}{rc^2}} \neq 1 - \frac{rc^2}{2Gy}[/tex]

    The penultimate step is the last correct step.
     
  15. May 26, 2012 #14
    True, I've rewrote it below.

    [tex] s = xc^2 \sqrt{1 - \frac{2G}{rc^2}} \frac{1}{1 - \frac{2Gy}{rc^2}}[/tex]
     
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