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I want to show that the straight line segment joining two points $p_1$ and $p_2$ in a plane is the shortest path between $p_1$ and $p_2$.

I have tried the following:

The straight line segment joining two points $p_1$ and $p_2$ is $l(t)=p_1+t(p_2-p_1)$. The length of the straight line segment is $\int_0^t \sqrt{l'(t)^2}=|p_2-p_1|$. Suppose that the shortest path is given by $f(t)$ that does not represent a straight line segment. We have $\int_0^t \sqrt{g'(t)^2} dt < \int_0^t \sqrt{l'(t)^2}dt \Rightarrow g'(t)<l'(t) \Rightarrow g(t)<l(t)+c$ since $l,g$ are positive functions. So $g$ could be either a polynomial of first order or a constant. If it is a polynomial of first order, it will be a straight line segment, contradiction. So $g$ has to be a constant. But a constant can't join two different points. Thus, the straight line segment joining two points $p_1$ and $p_2$ in a plane is the shortest path between $p_1$ and $p_2$.

Is it right? Or have I done something wrong?

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    $\begingroup$ $g'(t)<l'(t) \not\Rightarrow g(t)<l(t)+c$ in general. But more importantly: $l'(t)$ is a vector, so what is $l'(t)^2$? $\endgroup$ – John B Jan 20 '16 at 21:33
  • $\begingroup$ @Jonas If we replace $l'(t)^2$ and $g'(t)^2$ by their norm ? $\endgroup$ – Evinda Jan 21 '16 at 10:35
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You want a function $y(x)$ such that, given two points $A=(a,y(a))$ and $B=(b,y(b))$, the length of the path from the two points is minimized.

This length is given by the integral: $$ l=\int _a^b\sqrt{1+y'^2} dx $$ and your problem is classical problem in Calculus of variation:

Find the function $y(x)$ such that the functional $$ \int _a^bL(x,y,y')dx=\int _a^b\sqrt{1+y'^2} dx $$ is minimized.

For the Lagrangian $L=\sqrt{1+y'^2}$ we have: $$ \frac{\partial L}{\partial y}=0 \qquad \frac{\partial L}{\partial y'}=\frac{y'}{\sqrt{1+y'^2}} $$ so the Euler-Lagrange equation is: $$ \frac{d}{dx}\frac{y'}{\sqrt{1+y'^2}}=0 $$ that gives: $$ \frac{y'}{\sqrt{1+y'^2}}=C \iff y'^2=C^2(1+y'^2) \iff y'=\frac{C}{\sqrt{1-C}}=m $$ so : $y=mx+q$ that is a straight line and the boundary conditions for the points $A$ and $B$ give the line between these points.

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  • $\begingroup$ I see... Thank you!!! Is my approach completely wrong? $\endgroup$ – Evinda Jan 21 '16 at 10:43
  • $\begingroup$ Your element line $\sqrt{y'^2} dx$ is wrong and the key point in this kind of problems is that the minimizing process depend on a variation of the function that cannot be siply reduced to a variation of the independent variable. $\endgroup$ – Emilio Novati Jan 21 '16 at 11:39

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