Question

ques

Answer 1

You need help?

Answer 2

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 ??

Answer 3

sorry that's
\[\vec N \frac{d \kappa}{ds}\]
not
\[N \frac{d \kappa}{ds}\]
in the last step

Answer 4

I'm sorry, i dont know this.... @Mathmath333 @jasmineann @acxbox22 @xapproachesinfinity @divu.mkr Do you guys know this?

Answer 5

@ganeshie8

Answer 6

@Mehek14

Answer 7

@Kainui

Answer 8

@triciaal

Answer 9

no what was the original question?

Answer 10

one sec

Answer 11

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\[\rho=\frac{1}{\kappa}\]

Answer 12

@triciaal