What's a fast way to find the normal and binormal vectors?

[Eclair_de_XII]

Homework Statement

$r(t)=(t^2)i+(1+\frac{1}{3}t^3)j+(t-\frac{1}{3}t^3)k$
Find the tangential, normal, and binormal vectors for this TNB frame.

Homework Equations

$T(t)=\frac{v(t)}{|v(t)|}$
$N(t)=\frac{T`(t)}{|T`(t)|}$
$B(t)=T(t)×N(t)$

The Attempt at a Solution

The problem isn't that I don't know how to solve this, it's just that I don't know how to solve this fast enough. I had about forty minutes to figure out this problem, to calculate the normal and tangent components of acceleration, and to find the velocity and acceleration of an unrelated polar equation. In the end, I didn't have enough time to finish all of the quiz. Here's what I did...

$v(t)=(2t)i+(t^2)j+(1-t^2)k$
$|v(t)|=\sqrt{4t^2+t^4+1-2t^2+t^4}=\sqrt{2t^4+2t^2+1}$
$T(t)=\frac{2t}{\sqrt{2t^4+2t^2+1}}i+\frac{t^2}{\sqrt{2t^4+2t^2+1}}j+\frac{(1-t^2)}{\sqrt{2t^4+2t^2+1}}k$

And here's the time-consuming part that took up much of the allotted time.

$T_x`(t)=\frac{2(\sqrt{2t^4+2t^2+1})-2t(\frac{1}{2})(8t^3+4t)(2t^4+2t^2+1)^{-\frac{1}{2}}}{2t^4+2t^2+1}$
$T_x`(t)=\frac{4t^4+4t^2+2-8t^4-4t^2}{(2t^4+2t^2+1)^{\frac{3}{2}}}=\frac{2-4t^4}{(2t^4+2t^2+1)^{\frac{3}{2}}}$

Then I had to do this for the y- and z-components, then the magnitude, and it just drained away so much time. I need a faster way to compute N(t), and in turn, B(t), in other words.

 

[Mark44]

I need a faster way to compute N(t), and in turn, B(t), in other words.

I'm not aware of any faster way than the one you employed. These kinds of problems tend to be time-consuming, especially in calculating the N and B vectors.

 

[Eclair_de_XII]

I see. Maybe I should e-mail the teacher about it, that it's too much to be giving in just one quiz. I just had a stroke of hindsight, by the way, for calculating the normal component of the acceleration. I could have used:

$a_N=\sqrt{|a|^2-|a_T|^2}=\sqrt{4+8t^2-2t^4-2t^2-1}=\sqrt{-2t^4+6t^2+3}$

I should have known this, but I suppose that losing points on this question will help me remember it better in the future. Thanks for your input, by the way.

 

[micromass]

You could apply the non-arclength formulas. The advantage of this is that you don't need to differentiate a fraction: https://en.wikipedia.org/wiki/Frenet–Serret_formulas#Other_expressions_of_the_frame

For example:
\mathbf{r}(t) = (t^2, 1 + t^3/3, t - t^3/3)
\mathbf{r}'(t) = (2t, t^2, 1 - t^2)
\mathbf{r}''(t) = (2, 2t, -2t)

So

\mathbf{r}'(t)\times \mathbf{r}''(t) = (-2t^3 - 2t(1-t^2), -(-4t^2 - 2(1-t^2)), 4t^2 - 2t^2) = (-2t, 2t^2 +2, 2t^2)
So
\mathbf{B}(t) = \frac{(-2t, 2t^2 + 2, 2t^2)}{\|(-2t, 2t^2 + 2, 2t^2\|} = \frac{(-2t, 2t^2 + 2, 2t^2)}{\sqrt{8t^4 + 12t^2+4}}

 

[Eclair_de_XII]

Thank you for that link. I think I'll find those formulas useful next time I must calculate TNB or curvature.