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This is a very simple question. I'm stuck on one line in the Wiki proof for proving the shortest distance between two points is a line using the Euler-Lagrange equation. The proof can be found hereunder the "Example" heading.

Basically, I follow everything where they go through setting up the integral and directly calculating the different pieces of the E-L equation. The part where I get lost is when they do the following:

$\frac{y'(x)}{\sqrt{1+y'(x)^2}} = C$ (C is some constant)

$\rightarrow y'(x) = \frac{C}{\sqrt{1-C^2}}=A$

I'm guessing we can say it equals A because the C expression in the middle of the last line is just some constant. Then getting the result that y(x) is a line is straight forward. Where I'm confused is on getting to that C expression in the middle of the second line. How did we get the square root in the denominator? And how did we end up with the minus inside the square root? This may be a silly question but any explanation on that one specific piece is greatly appreciated. I'm just curious about it. Thanks!

k12345
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1 Answers1

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So it did end up being a very silly question. The answer is as follows:

$(y') ^2 = C^2 (1+ (y')^2)$

$\rightarrow (y')^2 - C^2 (y')^2 = C^2$

$\rightarrow (y')^2 (1-C^2) = C^2$

Then take the square root of both sides and get $y'$ alone.

k12345
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    Should note that when you take the square root of both sides that we already knew $C$ and $y'(x)$ both have the same sign. Otherwise your result is just the weaker $|y'| = \frac {|C|}{\sqrt{1-C^2}}$ and $y' = \pm \frac {C}{\sqrt{1-C^2}}$. But because $\frac{y'}{\sqrt{1+y'2}} = C$ and $\sqrt{1 + y'2} > 1 > 0$ we know that $y'$ and $C$ are both the same sign and therefore we can know that $y' = \frac c{\sqrt{1-C^2}}$. – fleablood Jul 05 '22 at 17:14
  • @fleablood great point. Thanks! – k12345 Jul 05 '22 at 17:29