anonymous
  • anonymous
ques
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
You need help?
anonymous
  • anonymous
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 ??
anonymous
  • anonymous
sorry that's \[\vec N \frac{d \kappa}{ds}\] not \[N \frac{d \kappa}{ds}\] in the last step

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anonymous
  • anonymous
I'm sorry, i dont know this.... @Mathmath333 @jasmineann @acxbox22 @xapproachesinfinity @divu.mkr Do you guys know this?
anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
@Mehek14
anonymous
  • anonymous
@Kainui
anonymous
  • anonymous
@triciaal
triciaal
  • triciaal
no what was the original question?
anonymous
  • anonymous
one sec
anonymous
  • anonymous
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\[\rho=\frac{1}{\kappa}\]
anonymous
  • anonymous
@triciaal
triciaal
  • triciaal
|dw:1433093340278:dw|
triciaal
  • triciaal
I don't know what torsion is either
anonymous
  • anonymous
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
triciaal
  • triciaal
@Nishant_Garg I don't know @amistre64 please take this
triciaal
  • triciaal
@skullpatrol can you help?
anonymous
  • anonymous
thx anyway
amistre64
  • amistre64
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
anonymous
  • anonymous
yeh, also here's some more equations I found \[|\vec B|=|\vec T|=|\vec N|=1\]\[\kappa=|\frac{d \vec T}{ds}|\]\[|\kappa|=\kappa\]
amistre64
  • amistre64
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?
anonymous
  • anonymous
Yeh, I think it's curvature at the specific point when we talk about \[\frac{d \vec T}{ds}=\kappa \vec N\]
amistre64
  • amistre64
and yes, B,T,N are conventionally defined as unit vectors that form an orthonormal basis ...
anonymous
  • anonymous
hmm yeh I did the order of the operations wrong in the first post
anonymous
  • anonymous
the cross product will come first
amistre64
  • amistre64
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
anonymous
  • anonymous
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?
amistre64
  • amistre64
sure
triciaal
  • triciaal
@amistre64 thanks I'm not good with vectors
anonymous
  • anonymous
\[\vec T .(\kappa \vec N \times(\kappa (\tau \vec B-\kappa \vec T)+\vec N \frac{d \kappa}{ds}))\]\[\vec T .(\kappa \vec N \times (\kappa \tau \vec B-\kappa^2 \vec T+\vec N \frac{d \kappa}{ds}))\]\[\vec T .(\kappa^2 \tau \vec N \times \vec B-\kappa^3 \vec N \times \vec T+\vec N \times \vec N \frac{d \kappa}{ds})\] \[\vec T .(\kappa^2 \tau \vec T+\kappa^3 \vec B+0)\]\[\kappa^2 \tau \vec T . \vec T+\kappa^3 \vec T . \vec B\]\[\kappa^2 \tau.1+0\]\[\kappa^2 \tau=\frac{\tau}{\rho^2}\] @amistre64
amistre64
  • amistre64
it would take me a few hours of review to most likely verify that :)

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