FS - N , and we just deFne the magnitude to be- . In other...

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The Frenet-Serret Formulas September 13, 2006 We start with the formula we know by the deFnition: d T ds = κ N . We also deFned B = T × N . We know that B is a unit vector, since T and N are orthogonal unit vectors. This means, just as for T , that B · d B ds = 0 . Now we di±erentiate: d ds B = d ds ( T × N ) = p d T ds P × N + T × p d N dx P . Since the derivative on T is in the same direction as N , we know that the Frst cross product is 0 . The second one we can’t say much about. We do, however, know that it is orthogonal to T , since T is one of the factors. In other words, we know that d ds B · B = 0 = d ds B · T . This means that it has to be in the direction of
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Unformatted text preview: N , and we just deFne the magnitude to be- . In other words, we see that d ds B =- N . Now we approach the derivative of N . We can write N = B T , just by thinking about the right hand rule. Now we dierentiate this, using the previous rules: d N ds = p d B ds P T + B p d T ds P =- N T + B N . Now we just rewrite the cross products using the right hand rule:- N T + B N = B- T . 1...
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