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Particularly annoying

  1. Aug 1, 2008 #1
    (dx/dt)^2 = (Mgh/A)*(2x-1)*(1-x)^2
    where x=cos (theta)
    and M,g,h and A are all constants

    the solution is

    sec(theta) = 1 + sech(sqrt(Mght/A))
    if u can do it then could u please list your full answer cos i can get quite far into it and then the sech thing just confuses me

    i think it involves using the weierstrass elliptic p function

    HELP?!?!?!
     
  2. jcsd
  3. Aug 1, 2008 #2

    tiny-tim

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    Hi latentcorpse! :smile:

    Since they give you the answer :wink::

    Hint: substitute y = 1/x - 1, and you get … ? :smile:
     
  4. Aug 1, 2008 #3
    i forgot to mention that x = cos (theta)....my bad
     
  5. Aug 1, 2008 #4
    also i can't get your substitution to work mainly because i dont see how that helps with the sech - i have little experience with sech so you might need to explain a bit more deeply how your getting that answer to work.

    cheers guys
     
  6. Aug 1, 2008 #5
    also i missed out the t.
    ignor that first post.
    here we go:

    (dx/dt)^2 = (Mgh/A)*(2x-1)*(1-x)^2

    the solution is

    sec(theta) = 1 + sech((sqrt(Mght/A))t)

    where x = cos theta
     
  7. Aug 2, 2008 #6

    tiny-tim

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    Substitute y = 1/x - 1, and show us what you get.

    That'll give you something which you can conveniently take the square-root of.

    Just try it! :smile:
     
  8. Aug 2, 2008 #7
    ok, if i try y=1/x - 1, then x=1/(y+1) and dy = -x^(-2) dx so dx = a x^(2) dy

    this means dx/dt = sqrt(Mgh/A) (2x-1)^(0.5) (1-x)

    becomes

    - [(y+1)^(-2) dy]/[(((2/y+1)-1)^(0.5))(1-(1/y+1))] = (mgh/A)^(0.5) t

    which gives me the argument of the sech term on the right side but i cant get the rest out!!!
     
  9. Aug 2, 2008 #8

    tiny-tim

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    Hi latentcorpse! :smile:

    (have a square-root: √ :smile:)
    Yes, but why have you stopped there? :confused:

    What is (2/y+1)-1?

    And what is 1-(1/y+1)? :smile:
     
  10. Aug 3, 2008 #9
    well if y=1/x - 1 then

    x = 1/(y+1)

    and so y+1 = 1/x

    then

    (2/(y+1))-1 = 2x-1 = 2*cos(theta) - 1

    and 1 - 1/(y+1) = 1 - x = 1 - cos(theta)

    and then cos(theta)=1/(y+1), we have -sin(theta) d(theta) = -(y+1)^(-2) dy, giving

    -[sin(theta) d(theta)]/[((2cos(theta)-1)^(0.5))(1-cos(theta)] = (Mgh/A)^(0.5) t

    but how does this help us to integrate? i still can't see how the LHS works?
     
  11. Aug 3, 2008 #10

    tiny-tim

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    Hi latentcorpse! :smile:

    Forget theta … it isn't in the question, and it isn't really in the answer (sectheta is just 1/x).

    And forget x … the whole point of substituting y for x is to get away from x!

    In terms of y:

    (2/y+1)-1 = … ?

    And 1-(1/y+1) = … ? :smile:
     
  12. Aug 3, 2008 #11
    ok well i can't see any way of rearranging them other than factorising y+1

    so that

    (2/(y+1)-1)^(0.5) = (y+1)^(0.5)[1-y]^(0.5)
    and,
    1-1/(y+1)=(y+1)[(y+1)y]

    and then substituting in and collecting these(y+1) terms with that one already on the numerator we get

    -dy/[(1-y)^(0.5) (y+1)^(0.5) y]

    now what?

    also i see that sec(theta) is just 1/x but i think the reason im finding this so hard is because im looking so hard for where the sech term is going to come out
     
  13. Aug 3, 2008 #12

    tiny-tim

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    Hi latentcorpse! :smile:
    (hmm … what happened to that √ I gave you? :confused:)

    Now write that as -dy/y√(1- y2). :smile:

    Then make the obvious substitution. :smile:
     
  14. Aug 3, 2008 #13
    ok
    my bad about the root sign and damn i just realised i forgot to copy it again

    so the obvious substitution would be some trigonometric function because of that
    (1-y^2)^(0.5) term

    now by y=1/x - 1 i get y=sec(theta) - 1 but that mucks things up

    so i tried just y = cos(theta)


    giving dy=-sin(theta) dtheta

    then our LHS rearranges to

    [sin(theta) dtheta]/[cos(theta) sin(theta)]

    which is just the integral of sec(theta) dtheta

    and the integral of sec is ln( sec(theta) + tan(theta) ) and surely i want something with a sech^(-1) in it so that i can get my sech(sqrt(Mgh/A)t) term on the RHS.

    ???
     
  15. Aug 3, 2008 #14

    tiny-tim

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    Yes, when you have a √(1 - y2), the usual substitution is either sin or cos, because sin2 + cos2 = 1.

    But you can also use either sech or tanh, because sech2 + tanh2 = 1 also.

    As you say, using cos gives you ln (sec + tan), which is really awkward.

    So try sech intead of cos, and see what you get! :biggrin:
     
  16. Aug 3, 2008 #15
    ok
    well if y = sech(theta), dy = -sech(theta) tanh(theta) dtheta

    and 1-y^2 = [tanh(theta)]^2 and so the denominator is sech(theta) tanh(theta)

    then -dy/[y(1-y^2)^(0.5)] becomes

    [sech(theta) tanh(theta) dtheta]/[sech(theta) tanh(theta)] but then everything cancels and we just have the integral of dtheta do we not?
     
  17. Aug 3, 2008 #16

    tiny-tim

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    Yes!! :biggrin:

    And the integral of dtheta is … ? :smile:
     
  18. Aug 3, 2008 #17
    yeah but the integral of dtheta is theta
    and that's equal to my right hand side of sqrt(Mgh/A) t
    but if u look back to my first post that's not the solution i was aiming for

    we're trying to get

    sec(theta) = 1 + sech(sqrt(Mgh/A)t)
     
  19. Aug 3, 2008 #18
    actually might be able to do a part without you!! yay!

    is this because if we take sech of both sides

    but then sech(theta) = y = sec(theta) -1 and so it all rearranges

    fantastic!
    cheers for all your help m8!

    legend!
     
  20. Aug 3, 2008 #19

    tiny-tim

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    D'oh! :smile:

    That's a different theta!! :biggrin:

    Try it with phi instead of theta … :wink:

    EDIT: ah … you beat me to it … well done!! :smile:
     
  21. Aug 4, 2008 #20
    Hi there!

    Here's my solution:

    [tex](\frac{dx}{dt})^2=\frac{mgh}{A}(2x-1)(1-x)^2[/tex]
    [tex]\frac{dx}{dt}=\sqrt{\frac{mgh}{A}(2x-1)(1-x)^2}[/tex]
    [tex]\frac{dx}{dt}=(1-x)\sqrt{\frac{mgh}{A}(2x-1)}[/tex]
    [tex]\frac{mgh}{A}=K[/tex]
    [tex]\frac{dx}{(1-x)\sqrt{K(2x-1)}}=dt[/tex]
    [tex]\displaystyle{\frac{1}{\sqrt{K}}\int\frac{dx}{(1-x)\sqrt{2x-1}}}=\displaystyle{\int}dt[/tex]

    [tex]2x-1=z^2 <=> x=\frac{z^2+1}{2}[/tex]
    [tex]2dx=2zdz <=> dx=zdz[/tex]

    [tex]\displaystyle{\int}\frac{1}{(1-x)\sqrt{2x-1}} dx}=\displaystyle{\int}\frac{z}{(1-\frac{z^2+1}{2})z} dz=\displaystyle{2\int}\frac{1}{1-z^2} dz=\displaystyle{2\int}(\frac{1}{2(1-z)}-\frac{1}{2(1+z)}) dz=\displaystyle{\int}(\frac{1}{(1-z)}-\frac{1}{(1+z)}) dz=\ln{\frac{1-z}{1+z}[/tex]

    [tex]z=\sqrt{2x-1}[/tex]

    [tex]\ln{\frac{1-z}{1+z}=\ln{\frac{\sqrt{2x-2}}{\sqrt{2x}}=\ln\sqrt{\frac{2-2x}{2x}}=\displaystyle{\frac{1}{2}}\ln{2\frac{1-x}{x}}=\displaystyle{\frac{1}{2}}\ln{\frac{1-x}{x}}+c[/tex]


    [tex]\displaystyle{\frac{1}{\sqrt{K}}\int\frac{dx}{(1-x)\sqrt{2x-1}}}=\displaystyle{\int}dt[/tex]
    [tex]\frac{1}{2\sqrt{K}}\ln{\frac{1-x}{x}}+c=t[/tex]
    [tex]\ln{\frac{1-x}{x}}=2\sqrt{K}(t-c)[/tex]
    [tex]\displaystyle{\frac{1-x}{x}}=e^{2\sqrt{K}(t-c)}[/tex]
    [tex]1-x=xe^{2\sqrt{K}(t-c)}[/tex]
    [tex]x(1+e^{2\sqrt{K}(t-c)})=1[/tex]
    [tex]x=\displaystyle{\frac{1}{1+e^{2\sqrt{K}(t-c)}}[/tex]
    [tex]\cos{\theta}=\displaystyle{\frac{1}{1+e^{2\sqrt{K}(t-c)}}[/tex]
    [tex]\sec{\theta}=1+e^{2\sqrt{K}(t-c)}[/tex]
    [tex]\sec{\theta}=1+e^{2\sqrt{\frac{mgh}{A}}(t-c)}[/tex]

    which is very similar to the solution given :) - s.o. has to prove the last term
     
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