For my Calc III class, I need to find $T(t)$, $N(t)$, and $k(t)$ using Mathematica, given $r(t) = {sqrt(t), (1/2)t^2, t^3 - (5/2)t^2 - 17}$. While the former and latter have been found, attempting to use the same method used to get $T(t)$ in order to find $N(t)$ has had... messy results, as seen in the following image.

(If the image doesn't show, it is basically an eight line mess full of Abs.

The code used is:

N1[t_] = Simplify[T'[t] / Norm[T'[t]], t ∈ Reals]

By hand, it starts getting ridiculously nasty around $||T'(t)||$, and I can't even begin to figure out $N(t)$, so I couldn't tell you if the result given is even remotely correct. What Mathematica gave me for $T(t)$ matched up with what I got by hand, so I know that, at the very least, Mathematica got that right or otherwise as wrong as I have.

I've tried doing a Google search for anything that could help compute $N(t)$, but nothing has proven useful.

For $T(t)$, I have gotten the following:

T(t) = {1 / sqrt(t) sqrt((1/t)(4t^3((3t - 5)^2 + 1) + 1)), 2t / sqrt((1/t)(4t^3((3t - 5)^2 + 1) + 1)), 2t(3t - 5) / sqrt((1/t)(4t^3((3t - 5)^2 + 1) + 1))}

(I apologize if the formatting is nasty; I've never really done this before, so I don't know how to properly clean it up.)

  • 1
    $\begingroup$ Probably better off avoiding Norm[] altogether: Simplify[T'[t]/Sqrt[T'[t].T'[t]]]. You might be interested in looking at FrenetSerretSystem[], too. $\endgroup$ Oct 7, 2016 at 2:11
  • $\begingroup$ It helped... some. Insofar as it reduced it to a two-line mess of fractions full of Abs. $\endgroup$
    – krourou2
    Oct 7, 2016 at 2:17
  • $\begingroup$ I mean, you should use something similar to get your tangent vector, too: Simplify[r'[t]/Sqrt[r'[t].r'[t]]]. Make sure to clear all variables before trying all of these. $\endgroup$ Oct 7, 2016 at 2:19
  • $\begingroup$ ... Wow. Oh wow, that's beautiful. Thank you so much. $\endgroup$
    – krourou2
    Oct 7, 2016 at 2:48
  • $\begingroup$ Is it only closely related Finding unit tangent, normal, and binormal vectors for a given r(t) or simply a duplicate? If not try to explain why? $\endgroup$
    – Artes
    Oct 7, 2016 at 9:04

1 Answer 1


All you have to do is replace the assumption for Simplify with

t > 0

and it will work. The reason is that your function contains a square root of t which precludes evaluation of the Euclidean norm for $t<0$.


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