TNB Frames (Principal Normal Vector)

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
$$ pmb{r}(t) = x(t)pmb{i} + y(t)pmb{j} + 0pmb{k}. $$
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
$$
pmb{N}(t) = frac{T’(t)}{|T’(t)|}
= dfrac{leftlangle frac{x”(t)}{|pmb{r}’(t)|}, frac{y”(t)}{|pmb{r}’(t)|} rightrangle}{sqrt{frac{(x”(t))^2 + (y”(t))^2}{|r’(t)|^2}}}
= dfrac{langle x”(t), y”(t)rangle}{|r”(t)|}
$$
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.