I know that I'm already messing up by putting in a picture instead of mathjax typing. My problem is that it is simply too much for me to type. I tried copying it in, but it didn't copy well.
The given question was:
Find the Binormal vector $B(t)=T(t)\times N(t)$ at $t=0$ and $t=1$. Also, sketch the curve traced out by $r(t)$ and the vectors $T(t),\ N(t),$ and $B(t)$ at these points. $r(t)=<t,2t,t^3>$
This is my work:
My question is mainly on the graph. Aren't $T(t)$ and $N(t)$ unit vectors? Do they look long in my graph because my graph is wrong? Or is my graph simply not made well? Or, is nothing wrong at all? This is really the first time I'm graphing such a thing, which is why I'm feeling uncertain. Thank you!
Edit: @amd wrote in the comments: T'(0)=0, so if you got something other than the zero vector by normalizing T', there’s surely an error in your calculations. Since the curvature vanishes at this point, you’re going to have to do something different to compute the Frenet frame there, anyway.
Logically, this seems true, but when I rechecked my equations, I got the same answer for N(t). As far as I know, $N(t)=\frac{T'(t)}{||T'(t)||}$ - and using this found that $N(0)=<0,0,1)$. Where did I go wrong?
You calculations look fine for the Frenet frame at $t=1$, but at $t=0$ there’s a problem: $T'(0)=0$, so $N$ and $B$ are in fact undefined there. When you simplified the general expression for $T'(t)/\lVert T'(t)\rVert$ that you derived, you didn’t take into account that the denominator vanishes at $t=0$, so that the original expression is in fact undefined there. The curve (which is planar) has an inflection point at $t=0$ and $N$ flips from one side of the curve to the other there, so there’s no unit normal that you could choose to make $N(t)$ continuous at $0$, either.
Your plot of these vectors, on the other hand is obviously wrong—neither unit tangent even looks close to being tangent to the curve. (It also appears that you’re using a left-handed coordinate system.) It looks like you’ve mistakenly plotted the vector from $r(1)$ to $T(1)$ instead of from $r(1)$ to $r(1)+T(1)$, and similarly for the others.