Finding the Frenet frame field of a curve $\vec{C}(t)$ = $(\frac{t^3}{3}, 2t-1,t^2+2)$ Working out the unit tangent $T$
$\vec{r} = \frac{t^3}{3}i + (2t-1)j + (t^2+2)k$
$\frac{dr}{dt} = t^2 +2j+2tk$
$\frac{ds}{dt} = |\frac{dr}{dt}| = \sqrt{t^4+4+2t^2}  = t^2+2$
$T = \frac{dr}{ds} = \frac{dr}{dt}/\frac{ds}{dt} = \frac{t^2i+2j+2tk}{t^2+2}$ = $\frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k$
Now working out the Principal Normal N,
$\frac{dT}{dt} = \frac{4t}{t^4+4}i-\frac{4t}{t^4+4}j-\frac{2t^2+4}{t^4+4}$ By the Quotient Rule
$\frac{dT}{ds} = \frac{dT}{dt}/\frac{ds}{dt} = \frac{4t}{t^6+8}i-\frac{4t}{t^6+8}j-\frac{2t^2+4}{t^6+8}k$
Now $N$ is $\frac{1}{k}$$\frac{dT}{ds}$ so in order to find $N$, I must first find $k$.
$k$ = $|\frac{dT}{ds}$| and this is where is becomes problematic. Since $|\frac{dT}{ds}$| = $\sqrt{(\frac{4t}{t^6+8})^2i-(\frac{4t}{t^6+8})^2j-(\frac{2t^2+4}{t^6+8})^2k}$
=$\sqrt{(\frac{4t^2}{t^12+64})i-(\frac{4t^2}{t^12+64})j-(\frac{2t^4+16}{t^12+64})k}$
So $N$ = $\frac{1}{$\sqrt{(\frac{4t^2}{t^12+64})i-(\frac{4t^2}{t^12+64})j-(\frac{2t^4+16}{t^12+64})k}} \times \frac{4t}{t^6+8}i-\frac{4t}{t^6+8}j-\frac{2t^2+4}{t^6+8}k$
Which I don't know what it equals to.
I wonder how I would be able to work out the Binormal B, knowing that $B = T \times N$.
Maybe I'm making a mistake... Any tips?
 A: You've missed an $i$ in your calculations for $\frac{d\vec{r}}{dt}$.
$$\frac{d\vec{r}}{dt} = t^2i +2j+2tk$$
Even though the final result for $\frac{ds}{dt}$ is good, the rightmost term in the radical sign is incorrect.
$$\frac{ds}{dt} = \left| \frac{dr}{dt} \right| = \sqrt{t^4+4+(2t)^2}  = t^2+2$$
Luckily, you put back the $i$ in the expression for $\vec{T}$.
However, you've incorrectly squared the denominator $t^2 +2$ as $t^4 + 4$.
The correct $\frac{d\vec{T}}{dt}$ should be
$$\frac{d\vec{T}}{dt} = \frac{4t}{(t^2+2)^2}i-\frac{4t}{(t^2+2)^2}j-\frac{2(t^2-2)}{(t^2+2)^2}k$$
because $$\frac{d}{dt} \frac{2t}{t^2+2} = \frac{2(t^2+2)-2t(2t)}{(t^2+2)^2} = \frac{4-2t^2}{(t^2+2)^2}.$$
You've got the denominator wrong in the same way.
$$\frac{d\vec{T}}{ds} = \frac{d\vec{T}}{dt}/\frac{ds}{dt} = \frac{4t}{(t^2+2)^3}i-\frac{4t}{(t^2+2)^3}j-\frac{2(t^2-2)}{(t^2+2)^3}k$$
\begin{aligned}
\kappa &= \left| \frac{d\vec{T}}{ds} \right| \\
&= \frac{\sqrt{(4t)^2+(4t)^2 + 4(t^2-2)^2}}{(t^2+2)^3} \\
&= \frac{\sqrt{4(4t^2+4t^2+t^4-4t^2+4)}}{(t^2+2)^3} \\
&= \frac{\sqrt{4(t^4+4t^2+4)}}{(t^2+2)^3} \\
&= \frac{2}{(t^2+2)^2}
\end{aligned}
So
\begin{aligned}
\vec{N} &= \frac1\kappa \frac{d\vec{T}}{ds} \\
&= \frac{(t^2+2)^2}{2} \left[ \frac{4t}{(t^2+2)^3}i-\frac{4t}{(t^2+2)^3}j-\frac{2(t^2-2)}{(t^2+2)^3}k \right] \\
&= \frac{1}{2(t^2+2)} [4ti-4tj-2(t^2-2)k] \\
&= \frac{2ti-2tj-(t^2-2)k}{t^2+2}
\end{aligned}
Finish the exercise with $\vec{B} = \vec{T} \times \vec{N}$.
\begin{aligned}
\vec{B} &= \vec{T} \times \vec{N} \\
&= \left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2-2)k}{t^2+2} \\
&= \frac{1}{(t^2+2)^2} [(-2(t^2-2)-(-2t)(2t))i - ((t^2)(-(t^2-2))-(2t)(2t))j + ((t^2)(-2t)-(2t)(2))k] \\
&= \frac{2(t^2+2)i+t^2(t^2+2)j-2t(t^2+2)k}{(t^2+2)^2} \\
&= \frac{2i+t^2j-2tk}{t^2+2}
\end{aligned}
