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anonymous
 one year ago
ques
anonymous
 one year ago
ques

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let \[\vec r(s)\] 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
 one year ago
Best ResponseYou've already chosen the best response.0sorry that's \[\vec N \frac{d \kappa}{ds}\] not \[N \frac{d \kappa}{ds}\] in the last step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry, i dont know this.... @Mathmath333 @jasmineann @acxbox22 @xapproachesinfinity @divu.mkr Do you guys know this?

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0no what was the original question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Prove 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\[\rho=\frac{1}{\kappa}\]

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433093340278:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0I don't know what torsion is either

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm 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
 one year ago
Best ResponseYou've already chosen the best response.0@Nishant_Garg I don't know @amistre64 please take this

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0@skullpatrol can you help?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2i 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
 one year ago
Best ResponseYou've already chosen the best response.0yeh, also here's some more equations I found \[\vec B=\vec T=\vec N=1\]\[\kappa=\frac{d \vec T}{ds}\]\[\kappa=\kappa\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2and 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
 one year ago
Best ResponseYou've already chosen the best response.0Yeh, I think it's curvature at the specific point when we talk about \[\frac{d \vec T}{ds}=\kappa \vec N\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2and yes, B,T,N are conventionally defined as unit vectors that form an orthonormal basis ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm yeh I did the order of the operations wrong in the first post

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the cross product will come first

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2not 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
 one year ago
Best ResponseYou've already chosen the best response.0The 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?

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0@amistre64 thanks I'm not good with vectors

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\vec T .(\kappa \vec N \times(\kappa (\tau \vec B\kappa \vec T)+\vec N \frac{d \kappa}{ds}))\]\[\vec T .(\kappa \vec N \times (\kappa \tau \vec B\kappa^2 \vec T+\vec N \frac{d \kappa}{ds}))\]\[\vec T .(\kappa^2 \tau \vec N \times \vec B\kappa^3 \vec N \times \vec T+\vec N \times \vec N \frac{d \kappa}{ds})\] \[\vec T .(\kappa^2 \tau \vec T+\kappa^3 \vec B+0)\]\[\kappa^2 \tau \vec T . \vec T+\kappa^3 \vec T . \vec B\]\[\kappa^2 \tau.1+0\]\[\kappa^2 \tau=\frac{\tau}{\rho^2}\] @amistre64

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2it would take me a few hours of review to most likely verify that :)
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