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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!

Lauren Bathers
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  • +1 for your effort, pretty rare these days on Math.SE, unfortunately. – Alex M. Jan 09 '17 at 17:41
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    There are lots of similar questions on MSE. The problem comes needing to do chain rule corrections when you have a non-arclength-parametrized curves. Because most textbooks don't give enough examples, I refer you to my (free) text, which may be of general use to you. – Ted Shifrin Jan 09 '17 at 19:33
  • @TedShifrin Sorry about not recognizing the question as an obvious repetitive theme. This is what happens when I step outside my regular tags :-/ If you can also point at a suitable question, where the OP's error is spelled out, we can close this one as an "abstract" duplicate. – Jyrki Lahtonen Jan 10 '17 at 06:52
  • @JyrkiLahtonen: Not a reason to delete your own answer (which seems fine) once you've already posted it, though. – Alex M. Jan 10 '17 at 15:34

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