Principal Unit Normal and Curvature

**cana** · May 17th 2011, 08:28 AM

Just wanted to check these... my notes don't have many (read any!) examples, so I'm not sure if what I'm doing is correct.

Given the arclength parameterisation r(t) of C, find the curvature and principal unit normal of C

$r(t) = (\frac{1}{4}\cos(2t), \frac{1}{2}t - \frac{1}{4}\sin{2t}, \sin{t})$

$\kappa(t) = ||r''(t)|| = ||(-\cos{2t}, \sin{2t}, -\sin{t}|| = \sin{t}$

and the principal unit normal $N(t) = \frac{r''(t)}{||r''(t)||}$

$N(t) =\frac{1}{\sin{t}} (-\cos{2t}, \sin{2t}, -\sin{t})$

**Opalg** · May 17th 2011, 11:39 AM

Originally Posted by cana

Just wanted to check these... my notes don't have many (read any!) examples, so I'm not sure if what I'm doing is correct.

Given the arclength parameterisation r(t) of C, find the curvature and principal unit normal of C

$r(t) = (\frac{1}{4}\cos(2t), \frac{1}{2}t - \frac{1}{4}\sin{2t}, \sin{t})$

$\kappa(t) = ||r''(t)|| = ||(-\cos{2t}, \sin{2t}, -\sin{t})|| = \sin{t}$
Should be $\color{red}\sqrt{\cos^22t + \sin^22t + \sin^2t} = \sqrt{1+\sin^2 t}.$

and the principal unit normal $N(t) = \frac{r''(t)}{||r''(t)||}$

$N(t) =\frac{1}{\sin{t}} (-\cos{2t}, \sin{2t}, -\sin{t})$

Other than that, it's fine.

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