Compute the Frenet Frame at each point of the curve $c(t)=(3t-t^3,3t^2,3t+t^3)$
I first found $c'(t)=(3-3t^2,6t,3+3t^2)$ and checked for arc length parameterization but $||c'(t)||=\sqrt{9-6t^4+9t^4+36t^2+9+6t^2+9t^4}=\sqrt{18+36t^2+18t^4}\neq0$
Therefore I tried to follow a similar example to solve. So
$c'(t)=(3−3t2,6t,3+3t2)$
$c''(t)=(-6t,6,6t)$
$c'''(t)=(-6,0,6)$
Then using Gram-Schmidt
$T=\frac{c'(t)}{||c'(t)||}=(\frac{\sqrt2(1-t^2)}{2(t^2+1)}, \frac{\sqrt2t}{(t^2+1)}, \frac{\sqrt2}{2})$
Now using GS, as a Frenet Curve implies linear indepence, I would presume:
$N=\frac{c''(t)}{||c''(t)||}$ however I can't get my answers to match those in the book.
$||c''(t)||=\sqrt{36t^2+36+36t^2}=\sqrt{72t^2+36}=6\sqrt{2t^2+1}$
Therefore I would get:
$\frac{-6t}{6\sqrt{2t^2+1}}=\frac{-t}{\sqrt{2t^2+1}}=\frac{-t\sqrt{2t^2+1}}{2t^2+1}$
$\frac{6}{6\sqrt{2t^2+1}}=\frac{1}{\sqrt{2t^2+1}}=\frac{\sqrt{2t^2+1}}{2t^2+1}$
$\frac{6t}{6\sqrt{2t^2+1}}=\frac{t}{\sqrt{2t^2+1}}=\frac{t\sqrt{2t^2+1}}{2t^2+1}$
So $N=(\frac{-t\sqrt{2t^2+1}}{2t^2+1}, \frac{\sqrt{2t^2+1}}{2t^2+1}, \frac{t\sqrt{2t^2+1}}{2t^2+1})$
However my book gives: $N=(\frac{-2t}{t^2+1}, \frac{1-t^2}{t^2+1}, 0)$
Any help please, preferably solving using the same method I have began to use!