- #1
RyanSchw
- 36
- 0
I've been using various methods of finding curvature, and using the forumla
[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
[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