I know how to prove that the shortest distance between two points is a straight line by applying E-L equations to

$$ L = \int_a^b ds = \int_a^b \sqrt{dx^2+dy^2} = \int_a^b \sqrt{1+(y')^2} \ dx$$

But this is considering $y$ as a function of $x$. If, instead, I look for both $x$ and $y$ as functions of a parameter $t$, I can't get the expected parametric solution.

Applying the E-L equations to

$$ \int_a^b \sqrt{dx^2+dy^2} = \int_a^b \sqrt{(x')^2+(y')^2} \ dt $$

Leads to

$$ \frac{\partial \sqrt{(x')^2+(y')^2}}{\partial x'} = C_x\ \ \ \ \ \frac{\partial \sqrt{(x')^2+(y')^2}}{\partial y'} = C_y $$

$$ \frac{x'}{\sqrt{(x')^2+(y')^2}} = C_x\ \ \ \ \ \ \frac{y'}{\sqrt{(x')^2+(y')^2}} = C_y $$

Which simplifies to

$$ x' = A y' $$ $$ y' = B x' $$

From here I can conclude that $x$ and $y$ trace a straight line, since any equation can be solved to give $$\frac{\Delta x}{\Delta y} = Const.$$

But I was hoping I'd get a parametric solution for a straight line

$$ x' = Const. $$ $$ y' = Const. $$

And I just can't see how I could get this or why this is not the straightforward solution.


1 Answer 1


Why should $dx/dt$ and $dy/dt$ be constant? A point can move along a line in any fashion. It can even move non-differentiably or non-continuously.

For example, the point $(\sin t, \sin t)$ oscillates sinusoidally along the line $y=x$ back and forth between the points $(-1,-1)$ and $(1,1)$. Neither the horizontal nor the vertical speed is constant.

A more extreme example is $(\lfloor t\rfloor,\lfloor t\rfloor)$ which jumps along discrete points, dwelling at each $(n,n)$ for one unit of time ($n\in\mathbb Z$). Its speed is zero at almost every point in time, and the speed is undefined at the moment of each jump (or if you like, it is $+\infty$).

Of course, I realize part of the hypothesis of your question is that $x(t)$ and $y(t)$ are at least differentiable (and so continuous), but there are other modes of movement.

  • $\begingroup$ You are right! I was thinking of a physical particle moving with constant speed. But there is no reason to think that way. The shortest distance between two points is a relationship between x and y, regardless of how one parameterize x and y. $\endgroup$
    – Chegon
    Sep 9, 2020 at 15:07

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