From the way I'm looking at it, there should be two ways to find the principal normal vector of a plane curve $C$ given by vector equation $$ \pmb{r}(t) = x(t)\pmb{i} + y(t)\pmb{j} + 0\pmb{k}. $$ That is, the way as laid out in TNB Frames: $$ \pmb{N}(t) = \frac{T'(t)}{\|T'(t)\|} \quad (1)$$ (where $T$ is the unit tangent vector) and by simply rotating the unit tangent vector $T$ by $\frac{\pi}{2}$ counterclockwise : $$ \pmb{N}(t) = -\frac{y'(t)}{\|\pmb{r}'(t)\|}\pmb{i} + \frac{x'(t)}{\|\pmb{r}'(t)\|}\pmb{j} \;\quad (2) $$ (this is similar to how the normal vector to the tangent is found in the vector form of Green's Theorem). But when I solve out equation $(1)$ to get $$ \pmb{N}(t) = \frac{T'(t)}{\|T'(t)\|} = \dfrac{\left\langle \frac{x''(t)}{\|\pmb{r}'(t)\|}, \frac{y''(t)}{\|\pmb{r}'(t)\|} \right\rangle}{\sqrt{\frac{(x''(t))^2 + (y''(t))^2}{\|r'(t)\|^2}}} = \dfrac{\langle x''(t), y''(t)\rangle}{\|r''(t)\|} $$ and compare components between $(2)$ to get $$ -\frac{y'(t)}{\|\pmb{r}'(t)\|} = \frac{x''(t)}{\|\pmb{r}''(t)\|} \;\text{ and }\; \frac{x'(t)}{\|\pmb{r}'(t)\|} = \frac{y''(t)}{\|\pmb{r}''(t)\|},\;\;\; $$ things don't seem to work out. For example, the above equalities do not hold for curve $$ \pmb{r}(t) = \langle t^2, t^3 \rangle, $$ and I'm not sure why. Am I misunderstanding how the principal normal vector is supposed to work, or is my math just wrong? Any insight would be appreciated.

Edit: For clarification, if I had a closed curve oriented counterclockwise, I would expect both methods to yield an orthogonal vector to $T$ of unit length pointing towards the region bounded by the curve. I would expect both derivations to produce the exact same vector in this case. That's essentially the crux of my problem.

  • $\begingroup$ When calculating (1), make sure to use the quotient rule for calculating $T'(t)$. $\endgroup$ – user197427 Jul 21 '15 at 5:14
  • $\begingroup$ Thank you a bunch- I can't believe I missed that. I'll go through the calculations later, but I'm fairly confident that was the problem $\endgroup$ – Joshua Potter Jul 23 '15 at 15:06

So, as user197427 pointed out, I did miss the quotient rule when solving for the normal. The math got really ugly really quickly once I correctly applied the derivative, to the point that solving it out would be a real tedious chore.

Nonetheless I wanted to close this question with the answer of "my math was wrong."


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