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Complex equation (for a 9th grader :>)

  1. May 19, 2010 #1
    Warning: This is mixed with physics, though all the physics work is done. Only formatting the equation is needed.
    1. The problem statement, all variables and given/known data
    L is needed from this equation:
    L-l= 1/2 a(T-τ)2
    2. Relevant equations

    3. The attempt at a solution
    L= 1/[STRIKE]2[/STRIKE] [STRIKE]2[/STRIKE]l/t2 (([STRIKE]2[/STRIKE]Lt2)/[STRIKE]2[/STRIKE]l-2*√((2Lt2)/2l)*τ+τ2 ) + l

    (the 2's at 1/2 2l/t^2 and 2lt^2/2l are removed right?)
    Completely lost here. I think it would be better to attach a picture no?
    Last edited: May 19, 2010
  2. jcsd
  3. May 19, 2010 #2


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    Welcome to PF!

    Hi Avathacis ! Welcome to PF! :wink:

    Sorry, but that's really difficult to read :redface: … can you type it again, putting in spaces, and using the X2 tag just above the Reply box (for the 2)? :smile:
  4. May 19, 2010 #3
    How does it look? :> I can still attach a screenshot of MS Word.
  5. May 19, 2010 #4


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    Hi Avathacis! :smile:
    ah, that's better! :biggrin:

    (i assume that last * is supposed to be a + ? :wink:)

    that's ok (and yes, those 2s do cancel), but

    i] where's the "-l" gone?

    ii] it's probably easier to leave the ()2 unexpanded.

    ok, fix that, and then I suggest you get rid of the fractions by multiplying both sides of the equation by powers of t and l.

    (after that, I'm not sure … it looks as if it's going to turn into a nasty quartic equation, but let's wait and see if something easy comes to light)
  6. May 19, 2010 #5
    Err, what -l? Nvm, got your point. Editing.
    Btw, thanks a lot for the help. (I fixed the * :p)
  7. May 19, 2010 #6


    Staff: Mentor

    This is somewhat confusing. Is l lower case L? Also, you have T, t, and τ (Greek letter tau). Are all three of them needed?

    Also, I don't see the need to replace a with 2l/t2.

    The original equation can be rewritten as
    [tex]L - l = \frac{a(T - \tau)^2}{2} = \frac{a(T^2 - 2\tau T + \tau^2)}{2} = \frac{a(2L/a - 2\tau \sqrt{2L/a} + \tau^2}{2}[/tex]

    [tex]\Rightarrow L - l = \frac{a(2L/a - 2\tau \sqrt{2L/a} + \tau^2}{2}[/tex]

    This equation is quadratic in form. If you let u^2 = L, it will be a quadratic.
  8. May 19, 2010 #7
    I) Yes all 3 are needed. There are 3 times.
    II) l is lower case L.
    III) a is unknown, so it is swapped for 2 known ..err.. letters. It is the same with T.
    Also what are we going to do with -l?
  9. May 19, 2010 #8


    Staff: Mentor

    But there is no need to replace a that I can see, since it doesn't involve L. If you need to replace it, you can do that later, after you have solved for L.

    If you make the substitution I suggested, you will have a u^2 term, a u term and everything else. l (ell) will be in amongst the "everything else."
  10. May 20, 2010 #9
    I still can't find a way to solve it with your equation :(. Sorry.
  11. May 20, 2010 #10
    Also, my attempt at the solution might be wrong. Try redoing it yourselves :p.
  12. May 20, 2010 #11


    Staff: Mentor

    Here's the equation at the end of post 6.
    [tex]L - l = \frac{a(2L/a - 2\tau \sqrt{2L/a} + \tau^2)}{2}[/tex]
    [tex]\Rightarrow 2L - 2l = 2L - 2a\tau \sqrt{2L/a} + a\tau^2[/tex]

    I didn't realize it before but the 2L terms drop out. Move the term with L in it to one side, and move everything else to the other side. Square both sides to get L out of the radical.
  13. May 21, 2010 #12
    Oh, right! This is the "answer". It's pretty much done but i will need to reformat it again after i make a=2l/t2.
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