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Here is the problem I am working on: Deduce the equation of the main normal and binormal to the curve: $x=t, y=t^2, z=t^3, t=1.$

I remember from Calc-3 that the binormal is unit tangent $\times$ unit normal, and that unit normal is tangent prime /magnitude of tangent prime. However, my text book has the binormal as unit tangent $\times$ principle normal, with principal normal listed as a very long formula.

Is unit normal different from principal normal? I have worked my way through the unit tangent but am not sure about the normal.

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As Mark points out, unit normal and principal normal is just the same thing.

Now we have:

Let $ v(t) = \frac{1}{\sqrt{1 + 4t^2 + 9t^4}} $

Tangent is $ (1, 2t, 3t^2) v $.

Normal is $ (0, 2, 6t) v(t) + (1, 2t, 3t^2) v'(t) $

$ v'(t) = -(1 + 4t^2 + 9t^4)^{-\frac{3}{2}} (4t + 18t^3) $

$ v(1) = \frac{1}{\sqrt{14}} $

$ v'(1) = -{14}^{-\frac{3}{2}} (22) = -\frac{11}{7\sqrt{14}} $

Unit tangent at $ t = 1 $ is $ T = \frac{1}{\sqrt{14}}(1, 2, 3) $

Unit normal at $ t = 1 $ is $ N = \frac{1}{\sqrt{266}}(-11, -8, 9) $

Unit binormal at $ t = 1 $ is $ T \times N = \frac{1}{\sqrt{19}}(3, -3, 1) $.

Together these form the Frenet frame for the curve.

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  • $\begingroup$ your expressions for T and N aren't perpendicular $\endgroup$
    – janmarqz
    Jan 25 '16 at 13:45
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    $\begingroup$ Thanks, indeed, I will fix my answer. I just started learning differential geometry last month. The problem seems to be the given parametrization of the curve is not an arc length parametrization by I wrongly assumed so. No wonder my binormal is not automatically unit length, thanks for pointing out. $\endgroup$
    – Andrew Au
    Jan 26 '16 at 4:51
  • $\begingroup$ The solution is now fixed, thanks for pointing out. $\endgroup$
    – Andrew Au
    Jan 26 '16 at 16:16
  • $\begingroup$ a precision is... the Serret - Frenet frame at the position with $t=1$ $\endgroup$
    – janmarqz
    Jan 26 '16 at 22:39

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