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

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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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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