Shortest distance between two non-intersecting differentiable curves is along their common normal

Question

I want to formally prove the following property:

The shortest distance between two differentiable non-intersecting curves is along their common normal.

I looked at the discussion on this on Quora, (Link: https://www.quora.com/How-can-I-show-that-the-shortest-distance-between-2-non-intersecting-curves-always-lies-along-their-common-normal), but the proofs were not formal there.

My approach: Let the curves be denoted by f and g respectively. And let AB be a line segment with A lying on f and B lying on g. WLOG, take A as the origin of the coordinate axis, and the tangent to f at A to be the x-axis. Let $B \equiv (b,g(b))$, $A = (0,0)$. Let the angle AB makes with the x-axis be $\theta \in (0,\pi)$. Let 2 rays from A making an angle of $\theta + \epsilon$ and $\theta - \epsilon$ ($\epsilon > 0$) with the x-axis intersect $y = g(x)$ at $B_1$ and $B_2$ respectively.

I wish to show that $\theta \neq \frac{\pi}{2} \Leftrightarrow |AB| > \min\{|AB_1|, |AB_2| \}$.

Suppose $\theta \neq \frac{\pi}{2} $. Then let $B_1 \equiv (b_1,g(b_1))$ and $B \equiv (b_2,g(b_2))$.Then \begin{align} g(b) &= b\tan(\theta)\\ g(b_1) &= b_1\tan(\theta + \epsilon)\\ g(b_2) &= b_2\tan(\theta - \epsilon) \\ \Rightarrow |AB| &= |b \sec(\theta) | \\ |AB_1| &= |b_1 \sec(\theta + \epsilon) | \\ |AB_2| &= |b_2 \sec(\theta - \epsilon) | \end{align} I wanted to relate $b_1$ and $b_2$ to $b$ using the continuity of $g$. However I am not able to do so. How do I proceed from here? - or is there an easier formal approach?

Answer 1

Suppose $a(s),b(t)$ are two curves in $\mathbb R^2$ with parameter interval $(0,1).$ Assume $a'(s),b'(t)$ never vanish, otherwise normal vectors make no sense. Define

$$f(s,t)= |a(s)-b(t)|^2.$$

Suppose $f(s_0,t_0)>0$ is the minimum value of $f$ (hence $\sqrt {f(s_0,t_0)}$ is the distance between the two curves). Then $(s_0,t_0)$ is a critical point of $f.$ Thus

$$\frac{\partial f}{\partial s}(s_0,t_0) = 2a'(s_0)\cdot(a(s_0)-b(t_0)) = 0$$

and

$$\frac{\partial f}{\partial t}(s_0,t_0) = 2b'(t_0)\cdot(b(t_0)-a(s_0)) = 0.$$

(Here $\cdot$ denotes the dot product.) Thus both $a'(s_0),b'(t_0)$ are perpendicular to the vector $b(t_0)-a(s_0).$ Thus the vector $b(t_0)-a(s_0)$ is perpendicular to $a$ at $a(s_0),$ and is perpendicular to $b$ at $b(t_0).$ This is the desired conclusion.

Answer 2

The distance between two differentiable curves is a differentiable function. If each of the curves is represented by a parametric function: $[x_1, y_1] = c_1(t_1), [x_2, y_2] = c_2(t_2)$, then this distance is a function of $t_1$ and $t_2$. For our purpose, we use the square of the distance, which has the same minimum and is given by: $D=(x_1(t_1)-x_2(t_2))^2+(y_1(t_1)-y_2(t_2))^2$.

A local minimum of a differentiable function can only occur at a point where the derivative of that function is $0$. Since the distance is a function of two variables, each partial derivative must be $0$:

$$\frac{\partial D}{\partial t_1}=2(x_1(t_1)-x_2(t_2))\frac{\partial x_1}{\partial t_1}+2(y_1(t_1)-y_2(t_2))\frac{\partial y_1}{\partial t_1}=0$$ $$\frac{\partial D}{\partial t_2}=-2(x_1(t_1)-x_2(t_2))\frac{\partial x_2}{\partial t_2}-2(y_1(t_1)-y_2(t_2))\frac{\partial y_2}{\partial t_2}=0$$

This is equivalent to :

$$0=(x_1(t_1)-x_2(t_2))\frac{\partial x_1}{\partial t_1}+(y_1(t_1)-y_2(t_2))\frac{\partial y_1}{\partial t_1}$$ $$0=(x_1(t_1)-x_2(t_2))\frac{\partial x_2}{\partial t_2}+(y_1(t_1)-y_2(t_2))\frac{\partial y_2}{\partial t_2}$$

Note that each of these equations is just the dot product of the vector between the two points and the derivative of that curve:

$$0 = (c_1(t_1)-c_2(t_2))\cdot\begin{bmatrix}\frac{\partial x_1}{\partial t_1}\\\frac{\partial y_1}{\partial t_1}\end{bmatrix}$$ $$0 = (c_1(t_1)-c_2(t_2))\cdot\begin{bmatrix}\frac{\partial x_2}{\partial t_2}\\\frac{\partial y_2}{\partial t_2}\end{bmatrix}$$

The dot product of two non-zero vectors is $0$ if and only if those vectors are perpendicular. We know that $c_1(t_1)-c_2(t_2)$ cannot be zero because then the curves would intersect, and the at least one partial derivative of a differentiable curve must always be nonzero. So if some value of $[t_1, t_2]$ yields a local minimum of the distance, then the curves $c_1$ and $c_2$ must be normal to to the line between $c_1(t_1)$ and $c_2(t_2)$ at those points.

This is only true of local minimums, so if the curves have endpoints then the absolute minimum might contain an endpoint on one or both of the curves. It is also true that not all common normals are even local minima, they could be local maxima, or they could even be saddle points where changing the position of the point on one curve increases the distance and changing it for the other curve decreases the distance, so the condition of a common normal is necessary but not sufficient for points on nonintersecting differentiable curves to be a local minimum of the distance.