Normal of curve at extremum with respect to some coordinates

Question

Suppose we have some curve given by parameterized coordinates $\gamma(t) = \left[ x_1(t),x_2 (t) , x_3(t) \right]$ in rectangular coordinate's, then prove that when the curve is extremized with respect to one of the coordinates, the normal vector the curve at that point is parallel to the coordinate basis of coordinate which it is extremized with respect to. Eg: if $ \frac{dx_2}{dt}=0$, then the curve is extremized w.r.t $x_2$ , this should suggest that the normal should be $\hat{x_2}$ direction according to previous claim

My try:

The normal is defined as:

$$ \vec{N} = \frac{\dot{T} }{|\dot{ T} |} = \frac{ \left[\ddot{x_1}, \ddot{x_2}, \ddot{x_3} \right]}{ \sqrt{ \sum_i (\ddot{x_i})^2}}$$

Now I need to prove that $ \vec{N} \times \hat{x_i}=0$ if $ \frac{dx_i}{dt}=0$.. but I am not sure how to introduce proceed further since the assumptions are with first derivative and I have above the second derivative.

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Motivation

Answer 1

Part I: Solution in 2-D

If we assume that $\gamma$ is contained in a 2D plane, then we can proceed as follows:

$\gamma$ extremizes in direction $\hat{u}$ at $s_0$ and let $P_u$ be orthogonal projection onto the subspace $\mathbb{R}\hat{u}$, which we identify with $\mathbb{R}$.

Then $s_0$ is an extremum of $P_u\gamma(s)$, and so $$0=\left.\frac{d}{ds}(P_u\gamma(s))\right|_{s=s_0}=P_u\vec{T}(s_0)$$ Thus $\vec{T}(s_0)$ is perpendicular to $\hat{u}$. Since $\vec{N}(s_0)$ is perpendicular to $\vec{T}(s_0)$, and there are only 2 dimensions, $\vec{N}(s_0)\in\mathbb{R}u$ — that is, parallel (or antiparallel) to $\hat{u}$. (This slight emendation of your question is necessary: maxima are extrema too.)

Part II: Your claim is not true in 3-D

Take any curve contained in a 2D plane $\Pi$; say a circle in the $xy$-plane. Extremize it with respect to some direction $\hat{u}$ that is neither contained in that plane nor perpendicular to it (say, $(0,1,1)$). Then the normal at the extremal point is contained in $\Pi$, and thus not parallel to $\hat{u}$.

(It's worth noting that if the entire curve is contained in $\Pi$, then we can project $\hat{u}$ into $\Pi$ and recover Part I. But we only need $\gamma$ locally contained in $\Pi$ around the extremum for this counterexample.)

Part III: Technique

Although my solution elides it, you mention in your question another difficulty: we only know about $\vec{T}$. How can we address its derivative?

The key is arclength parameterization that $\vec{T}$ is a unit vector. By the definition of an orthogonal projection, for any $s$, $$1=|\vec{T}(s)|^2=|P_u\vec{T}(s)|^2+|(I-P_u)\vec{T}(s)|^2$$ In particular, since $P_u\vec{T}(s_0)=0$ $$1=|(I-P_u)\vec{T}(s_0)|^2$$ This is the largest possible value $|(I-P_u)\vec{T}|^2$ can attain (remember, we're projecting a unit vector). So $s_0$ is an extremum of $|(I-P_u)\vec{T}|^2$, and its derivative is $0$.

It turns out that this extra property isn't really useful for this problem,* but it's a useful trick to have in your back pocket.

* If $I-P_u$ projects onto a 1-D subspace, then we can drop the norm and square before differentiating. In that case, we do get an alternate solution to Part I.