# Unit Normal vs Principal Normal

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.

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.

• your expressions for T and N aren't perpendicular Jan 25 '16 at 13:45
• 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. Jan 26 '16 at 4:51
• The solution is now fixed, thanks for pointing out. Jan 26 '16 at 16:16
• a precision is... the Serret - Frenet frame at the position with $t=1$ Jan 26 '16 at 22:39