Why Is the Unit Normal Vector N(s) Defined as r''(s)/||r''(s)||?

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SUMMARY

The unit normal vector N(s) of a curve is defined as r''(s)/||r''(s)|| to ensure it has a unit length, differentiating it from r''(s), which does not necessarily have a length of 1. In arc length parametrization, while the tangent vector T(s) = r'(s) is normalized to a length of 1, the second derivative r''(s) does not share this property. This distinction is crucial for understanding the geometric properties of curves in multivariable differential calculus.

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  • Understanding of arc length parametrization
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  • Knowledge of derivatives and their geometric interpretations
  • Basic proficiency in multivariable differential calculus
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Students and educators in multivariable calculus, particularly those focusing on differential geometry and curve analysis.

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Homework Statement


This question comes from my multivariable differential calculus course, and it pertains to arc length parametrizations the unit normal vector of a curve.

Why does the arc length parametrization of the unit normal vector of a curve, N(s) equal...

_r''(s)__ = N(s) and not just equivalent to r''(s) = N(s)?
||r''(s)||

Homework Equations


T(s) = r'(s)
||r'(s)|| = 1



The Attempt at a Solution


I thought that since ||r'(s)|| = 1, ||r''(s)|| would be equivalent to 1 as well since they are both the normalizations of arc length parametrizations of curves. However, this apparently isn't the case...any help would be appreciated

thanks
 
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Well, take a 'fer instance'. r(s)=(1/2)*(cos(2s),sin(2s)). r'(s) is unit length, r''(s) isn't.
 
I'm just seconding Dick. If a curve is paraemtrised by arclength \vec{r}(s), then it follows from the definition of "arclength", and the chain rule, that the length of \vec{r}' is 1. There is no reason to expect that to be true for \vec{r}'' as well.

Of course, \vec{r}'' is normal to the curve. I just doesn't have length 1.
 

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