I have a parameterized equation of a path that looks like a parabola, $\textbf{r}(t) = \langle t, t^2 \rangle $. The principal normal vector is $\textbf{N}(t) = \frac{\textbf{T}'(t)}{|\textbf{T}'(t)|}$. Without normalizing, which doesn't change the direction of the vector, we can find the tangent vector $\textbf{T}(t) \propto \langle 1, 2t \rangle$, and the normal vector $\textbf{N}(t) \propto \langle 0, 2 \rangle$. At any time t, the normal vector $\textbf{N}(t)$ and the tangent vector $\textbf{T}(t)$ are not orthogonal to each other. I read somewhere that it has to do with curvature. Is there an easier way to understand why this simple path's tangent and normal vectors are not orthogonal to each other?
The theorem in the Briggs Calculus textbook states:
Let $\textbf{r}$ describe a smooth parameterized curve with unit tangent vector $\textbf{T}$ and principal unit normal vector $\textbf{N}$. Then, $\textbf{T}$ and $\textbf{N}$ are orthogonal at all points of the curve; that is, $\textbf{T} \cdot \textbf{N} = 0$ at all points where $\textbf{N}$ is defined.
I would argue that this path satisfies the conditions to this theorem, so not sure why the two vectors wouldn't be orthogonal.
Best Answer
While it's true that the vector $\langle{1,2t}\rangle$ is in the same direction as the unit vector $\textbf{T}(t)$, the scalar multiple, $c(t)$ say, varies as a function of $t$.
But in your attempt to find a vector in the same direction as $\textbf{T}'(t)$ by taking the derivative of the vector $\langle{1,2t}\rangle$, you are implicitly assuming that $c(t)$ is truly constant (i.e., independent of $t$).
To verify the orthogonality of $\textbf{T}(t)$ and $\textbf{N}(t)$ for the curve $\textbf{r}(t)=\langle{t,t^2}\rangle$ we can proceed as follows . . .$\\[10pt]$ \begin{align*} \textbf{r}'(t)&=\langle{1,2t}\rangle\\[2pt] |\textbf{r}'(t)|&=\sqrt{1+4t^2}\\[16pt] \textbf{T}(t)&=\frac{\textbf{r}'(t)}{|\textbf{r}'(t)|}\\[2pt] &= \left\langle \frac{1}{\sqrt{1+4t^2}} , \frac{2t}{\sqrt{1+4t^2}} \right\rangle \\[16pt] \textbf{T}'(t) &= \left\langle -\frac{4t}{(1+4t^2)^{\frac{3}{2}}} , \frac{2}{(1+4t^2)^{\frac{3}{2}}} \right\rangle \\[2pt] |\textbf{T}'(t)| &= \frac{2}{1+4t^2} \\[16pt] \textbf{N}(t)&=\frac{\textbf{T}'(t)}{|\textbf{T}'(t)|}\\[2pt] &= \left\langle -\frac{2t}{\sqrt{1+4t^2}} , \frac{1}{\sqrt{1+4t^2}} \right\rangle \\[16pt] \textbf{T}(t){\,\cdot\,}\textbf{N}(t) &= \left(\frac{1}{\sqrt{1+4t^2}}\right) \left(-\frac{2t}{\sqrt{1+4t^2}}\right) + \left(\frac{2t}{\sqrt{1+4t^2}}\right) \left(\frac{1}{\sqrt{1+4t^2}}\right) \\[3pt] &= \;0 \\[2pt] \end{align*}