I've been using various methods of finding curvature, and using the forumla(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

K=\frac{||\vec{r}'X \vec{r}''||}{||\vec{r}'||^3}

[/tex]

I took the cross product of my two vectors and came up with (just assume this is correct, if my question isn't answered i'll post all related work)

[tex]

128 \pi^3 sin(2 \pi t)^2 - -128 \pi^3 cos(2 \pi t)^2

[/tex]

My question is: Can I factor out the 128[itex] \pi^3[/itex] and then use a trig identity to make sin^2 + cos^2 =1?

I assumed that since these were components of vectors I couldn't do this. However, if I do, I get the correct answer. I left them as is, and then took the magnitude of the resulting cross product.

Which produced

[tex]

128 \pi^3 \sqrt{sin(2 \pi t)^4+cos(2 \pi t)^4}

[/tex]

This would also produce the correct answer if sin(x)^n +cos(x)^n =1, but I'm fairly certain it doesn't.

Any help would be greatly appreciated. Thanks

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# Homework Help: Curvature, More of a cross product question really.

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