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Intrinsic Equations

  1. Nov 3, 2005 #1
    Hello, I was doing some past papers and I couldn't solve one.
    I don't even know where to begin with this question.
    [tex]s=asec^{3}\psi - a[/tex]
    Show that
    [tex]y=atan^{3}\psi[/tex]

    If you can tell me where to start from or how to solve the question that would be great.
    Thanks
     
    Last edited: Nov 3, 2005
  2. jcsd
  3. Nov 3, 2005 #2

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    I think the [ TEX ] should be lower case [ tex ]

    [tex]s=asec^{3}\psi - a[/tex]
    Show that
    [tex]y=atan^{3}\psi[/tex]
     
  4. Nov 3, 2005 #3

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    try using,

    [tex]\frac{dy}{d\psi}= \frac{dy}{ds}\cdot \frac{ds}{d\psi}[/tex]
     
  5. Nov 3, 2005 #4
    Is there no way of getting to the second one, just using the first one, the second part of the question asks for an expression for x :cry:
     
  6. Nov 3, 2005 #5

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    I haven't found it yet. I've only been able to show that differential coeficients are the same. But integration, to give you the original expression, y = a.tan³psi, involves a constant of integration. I've not been able to get rid of that. I was hoping you might manage it yourself !
     
  7. Nov 3, 2005 #6

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    I suppose that, in effect, that was what I was doing.

    [tex]\frac{dy}{d\psi}= \frac{dy}{ds}\cdot \frac{ds}{d\psi}[/tex]

    [tex]\mbox{We know that\ }\frac{dy}{ds} = sin\psi[/tex]

    So,

    [tex] y = \int sin\psi\cdot \frac{ds}{d\psi}\ d\psi[/tex]

    You should be able to do similar to find an expression for x.
     
  8. Nov 3, 2005 #7
    Ah ok thanks a lot, I can do the x myself, I just didn't know how to start :rolleyes:
     
  9. Nov 3, 2005 #8

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    Let me know how you get rid of the constant of integration. Ta.
     
  10. Nov 4, 2005 #9
    Oh I forgot to write this bit...
    [tex] When y=0 x=0 \psi=0[/tex]
    But I still can't solve it...:cry:
     
  11. Nov 4, 2005 #10

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    You can insert spaces when using latex with backslash-space "\ "

    [tex] When\ y=0\ x=0\ \psi=0[/tex]
     
  12. Nov 4, 2005 #11

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    When you say you can't solve it, do you mean you can't do the integral for the x-function ?

    [tex] x = \int cos\psi\cdot \frac{ds}{d\psi}\ d\psi[/tex]

    If it's that one, do you have any working to show ?
     
  13. Nov 4, 2005 #12
    [tex] y= \int 3sin^{2}\psi sec^{4}\psi d\psi[/tex]
    [tex]u=sec \psi[/tex]
    [tex]du/(tan\psi sec\psi) = d\psi)[/tex]
    some canceling and stuff
    [tex] y= \int 3sin\psi u^{2} du[/tex]
    I can't get rid of the sin psi

    I think its to do with y=0 x=0 and psi=0 :confused:
     
    Last edited: Nov 4, 2005
  14. Nov 4, 2005 #13

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    It doesn't have anything to do wiht the initial values: y=0 x=0 and psi=0

    You already know what the answer is. Just differentiate that and see how that can be manipulated to give you the expression you have to integrate.

    Then work backwards.
     
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