looks like a fun q

Question

looks like a fun q

sweetburger · Answer

$r(t)=(\cos(t),\sin(t),t)$ show that the helix is not planar.
My guess is a contradiction where we initially say r(t) is a plane and the normal of said plane must be orthogonal to the tangent vector for all points.
So $$r'(t)=(-\sin(t),\cos(t),1)$$.
My confusion here is the norm of the "plane" that we're looking at here is that purely just the currently parametrized equation of the helix?
If so i think this can be done
$$r(t)\bullet r'(t)=0$$ im saying that if the normal of the plane and the tangent vector are going to be orthogonal then the dot product must be 0 for all points but not sure how to write this as in mathier terms. 
$$r(t) \bullet r'(t)= (\cos(t))(-\sin(t))+(\sin(t))(\cos(t))+t=0$$ (yes i realize its not equal to 0)
terms cancel
$$t \neq 0$$
so since this was the result does this sufficiently show that the helix is not planar?

Ultrilliam · Answer

@Vocaloid

Vocaloid · Answer

@sillybilly123 I don't quite remember how to do this can you show me?

sillybilly123 · Answer

simplifying this just to get on the same song-sheet, if you can imagine that you are on a flat surface (2-D) and tied by a rubber band to a tree. you are at $\mathbf r(t) = $, and free to move in any direction you choose in this 2D space. you can complete prefect circles around the tree, with the tree as the origin and maintaining a constant radius, in which case $\mathbf r \cdot \mathbf{r'} = 0$. because the geometry tells you that $\mathbf r \perp \mathbf r' $. on another extreme, you can instead choose to move directly away from the tree, along the length of the rubber band. in that case, $\mathbf{r'}$ is still the tangent vector but clearly $\mathbf r \cdot \mathbf{r'} \ne 0$ because $\mathbf r \parallel \mathbf r' $. that's all in just 2-D, in a plane. can you establish co-planarity, using maths tricks with $\mathbf r$ and it's derivative, when you are already in a 2-D plane?! imagine now that you are tied to the same tree but wearing and using a jetpack. the $\mathbf r$ vector is till just an arrow that points from the base of the tree to you hovering somewhere above the base of the tree. what is interesting is that the $\mathbf r'$ vector is still the tangent vector. but not because $t$ actually means time or that $\mathbf r'$ has anything to do with velocity. if we used $\mathbf r = \mathbf r(\lambda)$ instead of $\mathbf r = \mathbf r(t)$, this might seem more obvious. there is, here, a pretty direct answer to what I am guessing the question is: https://math.stackexchange.com/questions/939275/a-space-curve-is-planar-if-and-only-if-its-torsion-is-everywhere-0 plus, found this vid, it runs through a load of important mechanics if you wanna get back into the idea of curvature: https://www.youtube.com/watch?v=q8gSiUR_ik4 the mathstack thing I can help translate, if that is needed, but there is more to this than I think maybe was anticipated at question inception. maybe.

sillybilly123 · Answer

know that's waffly

sillybilly123 · Answer

hey @Bob, can you pls use yer **supreme genius** to check my thoughts?!?!

Bob · Answer

sure thing

hmmm... it looks like it was 99.348% pasted off of google or another source.

though that leaves a 0.652% chance of it being your own work, i give you the benefit of the doubt