Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Why is this true? If $\alpha$ is a time dependent curve, $T$ is the unit tangent and $N$ is a normal field along $\alpha$, then $$\langle \partial_s N, T \rangle = -\langle N, \partial_s T \rangle$$ where $s$ is the arc length parametrisation. I assume this inner product thing is a dot product but I don't understand why this holds.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Differentiate $\langle N, T\rangle \equiv 0 $ with respect to $s$.

(yes, $\langle, \rangle$ is presumably a scalar product, and in the setting I assume you are working in an equality $$\frac{d}{ds}\langle X,Y\rangle = \langle\partial_s X,Y\rangle + \langle X, \partial_s Y \rangle$$ almost certainly holds true. Since you did not mention the source of this question I can only guess this. Check your source for this kind of formulae.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.