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.

  • $\begingroup$ Your formula for $N(t)$ works only if $|T(t)| = 1$ for all $t$. So find the right formula for $T$ first. $\endgroup$ – Deane Jan 19 at 2:20
  • 1
    $\begingroup$ Normalizing or not normalizing doesn't change the direction of $T(t)$, but as the length of $T(t)$ changes, this effect does change the direction of $T'(t)$. $\endgroup$ – David K Jan 19 at 3:19

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*}

  • $\begingroup$ I'm so used to this assumption for normalization. Thank you! $\endgroup$ – Buddhapus Jan 19 at 21:47

I say this without having seen your computations, but I suspect your issue was subbing the velocity vector $\textbf{v}(t) = \textbf{r}'(t)$ for the unit tangent vector $\textbf{T}(t) = \textbf{v}(t)/\|\textbf{v}(t)\|.$
If I try to compute $$\textbf{N}(t) = \frac{\textbf{v}'(t)}{\|\textbf{v}'(t)\|} = \frac{\textbf{a}(t)}{\|\textbf{a}(t)\|},$$ instead of correctly using $$\textbf{N}(t) = \frac{\textbf{T}'(t)}{\|\textbf{T}'(t)\|},$$ because I mistakenly assume $\textbf{T}(t) = c\textbf{v}(t)$ for some fixed $c$ and all $t$, then of course my resulting $\textbf{N}(t)$ is going to fail to be orthogonal to $\textbf{T}(t)$.
This is because when $\|\textbf{v}(t)\|$ is non-constant (i.e. I'm traversing the curve with changing speed), the acceleration vector $\textbf{a}(t) = \textbf{v}'(t)$ has components both in the direction of $\textbf{v}(t)$ (changing the speed) and in a direction orthogonal to $\textbf{v}(t)$ (changing the direction). You can find this equation in most textbooks on vector calculus:

If the acceleration vector is written as $$\textbf{a} = a_{T} \textbf{T} + a_{N} \textbf{N},$$ then $$a_T = \frac{d}{dt} \| \textbf{v} \| \text{ and } a_N = \kappa \| \textbf{v} \|^2.$$

Because the unit tangent vector's length never changes, $\textbf{T}'(t)$ solely represents the change in direction of $\textbf{T}(t)$, which is the same as the change in direction of $\textbf{v}(t)$. Normalizing then gives us a unit normal vector $\textbf{N}(t)$. Another user has already posted the correct equations for the unit tangent and normal in the specific case of the parabola, as well as verifying that they're orthogonal.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.