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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    You need help?

  2. anonymous
    • one year ago
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    Let \[\vec r(s)\] be the position vector of a curve C where s is the arc length measured from a fixed point on the curve C consider \[\frac{d \vec r}{ds}.\frac{d^2 \vec r}{ds^2} \times \frac{d^3 \vec r}{ds^3}\] Now this is equal to \[\vec T . \kappa \vec N \times (\kappa \frac{d \vec N}{ds}+\frac{d \kappa}{ds}\vec N)\] \[\implies \vec T.\kappa \vec N \times (\kappa(\tau \vec B-\kappa \vec T)+N \frac{d \kappa}{ds})\] Now shouldn't \[\frac{d \kappa}{ds}\] be 0??Isn't curvature and torsion constant ??

  3. anonymous
    • one year ago
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    sorry that's \[\vec N \frac{d \kappa}{ds}\] not \[N \frac{d \kappa}{ds}\] in the last step

  4. anonymous
    • one year ago
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    I'm sorry, i dont know this.... @Mathmath333 @jasmineann @acxbox22 @xapproachesinfinity @divu.mkr Do you guys know this?

  5. anonymous
    • one year ago
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    @ganeshie8

  6. anonymous
    • one year ago
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    @Mehek14

  7. anonymous
    • one year ago
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    @Kainui

  8. anonymous
    • one year ago
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    @triciaal

  9. triciaal
    • one year ago
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    no what was the original question?

  10. anonymous
    • one year ago
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    one sec

  11. anonymous
    • one year ago
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    Prove that \[\frac{d \vec r}{ds}.\frac{d^2 \vec r}{ds^2} \times \frac{d^3 \vec r}{ds^3}=\frac{\tau}{\rho^2}\] where\[\rho=\frac{1}{\kappa}\]

  12. anonymous
    • one year ago
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    @triciaal

  13. triciaal
    • one year ago
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    |dw:1433093340278:dw|

  14. triciaal
    • one year ago
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    I don't know what torsion is either

  15. anonymous
    • one year ago
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    hmm here's the equations ur suppose to use \[\vec T=\frac{d \vec r}{ds}\]\[\frac{d \vec T}{ds}=\frac{d^2 \vec r}{ds^2}=\kappa \vec N\]\[\frac{d \vec B}{ds}=-\tau \vec N\]\[\frac{d \vec N}{ds}=\tau \vec B-\kappa \vec T\] \[\vec B=\vec T \times \vec N\] kappa is curvature and tau is torsion I wanted to know if curvature and torsion r constant... @triciaal

  16. triciaal
    • one year ago
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    @Nishant_Garg I don't know @amistre64 please take this

  17. triciaal
    • one year ago
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    @skullpatrol can you help?

  18. anonymous
    • one year ago
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    thx anyway

  19. amistre64
    • one year ago
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    i assume since a scalar corssed with a vector is meaningless, then the order of operations if better seen as. \[\frac{d \vec r}{ds}.\left(\frac{d^2 \vec r}{ds^2} \times \frac{d^3 \vec r}{ds^3}\right)\] just trying to get a bearing on it

  20. anonymous
    • one year ago
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    yeh, also here's some more equations I found \[|\vec B|=|\vec T|=|\vec N|=1\]\[\kappa=|\frac{d \vec T}{ds}|\]\[|\kappa|=\kappa\]

  21. amistre64
    • one year ago
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    and i dont believe curvature is a constant with respect to some generic curve. it changes with respect to where you are on the curve right?

  22. anonymous
    • one year ago
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    Yeh, I think it's curvature at the specific point when we talk about \[\frac{d \vec T}{ds}=\kappa \vec N\]

  23. amistre64
    • one year ago
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    and yes, B,T,N are conventionally defined as unit vectors that form an orthonormal basis ...

  24. anonymous
    • one year ago
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    hmm yeh I did the order of the operations wrong in the first post

  25. anonymous
    • one year ago
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    the cross product will come first

  26. amistre64
    • one year ago
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    not sure how useful this property may be for you but might was well include it in our toolbox a · (b × c) = (a × b) · c

  27. anonymous
    • one year ago
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    The question I'm doing has not taken \[\frac{d \kappa}{ds}=0\] and completed the proof so I was thinking if it varies with s Would you like to see the proof anyway?

  28. amistre64
    • one year ago
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    sure

  29. triciaal
    • one year ago
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    @amistre64 thanks I'm not good with vectors

  30. anonymous
    • one year ago
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    \[\vec T .(\kappa \vec N \times(\kappa (\tau \vec B-\kappa \vec T)+\vec N \frac{d \kappa}{ds}))\]\[\vec T .(\kappa \vec N \times (\kappa \tau \vec B-\kappa^2 \vec T+\vec N \frac{d \kappa}{ds}))\]\[\vec T .(\kappa^2 \tau \vec N \times \vec B-\kappa^3 \vec N \times \vec T+\vec N \times \vec N \frac{d \kappa}{ds})\] \[\vec T .(\kappa^2 \tau \vec T+\kappa^3 \vec B+0)\]\[\kappa^2 \tau \vec T . \vec T+\kappa^3 \vec T . \vec B\]\[\kappa^2 \tau.1+0\]\[\kappa^2 \tau=\frac{\tau}{\rho^2}\] @amistre64

  31. amistre64
    • one year ago
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    it would take me a few hours of review to most likely verify that :)

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