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.
