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.
Best Answer
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.