I'm studying the basic example of Euler's equation that is to obtain the shortest path between two points. $$ f = \sqrt {1 + y' ^ 2} $$

$$ \frac{\partial f}{\partial y} - \frac{d}{dx}(\frac{\partial f}{\partial y'}) = 0 $$

My textbook said that I could replace the first term $ \frac{\partial f}{\partial y} $ to $ 0 $, because there is no $ y$ in the definition of $ f $.

I got no intuition to do so, so I tried to calculate $ \frac{\partial f}{\partial y} $.

$$ \frac{\partial f}{\partial x} \frac{\partial x}{\partial y} = \frac{\partial f}{\partial x} \frac{1}{y'} = {(y'^2 + 1)}^{-1/2} = \frac{1}{2}{(y'^2 + 1) ^ {-1/2}}{2y'y''}\frac{1}{y'} = {(y'^2+1)}^{-1/2}y'' $$ And I don't think $ {(y'^2+1)}^{-1/2}y'' = 0 $. This derivation doesn't seem to give the same result of the textbook.

What did I do wrong?

  • $\begingroup$ $f$ is a function of $y'$ only. So you should get $\frac{df}{dy'}=C\implies \frac{y'}{\sqrt{1+y'^2}}=C$ for some constant $C$. Now show that $y=ax+b$ for constants $a$ and $b$. $\endgroup$
    – Mark Viola
    Sep 15, 2020 at 14:34
  • 1
    $\begingroup$ In this case you age considering $y$ and $y’$ to be different variables. This is common notation which I think this is slightly unfortunate. Here the functional to be minimized is $\int^b_a f(x,y,y’)dt$ under the constraint that $y(a)$ and $y(b)$ are the fixed points. Note that $f$ is a function of three variables, and the notation $\frac {df}{dy}$ denotes the derivative of $f$ w.r.t. the second variable. In your case, $f$ doesn’t actually depend on its second variable so the derivative is zero. $\endgroup$
    – User8128
    Sep 15, 2020 at 14:36

1 Answer 1


What you did wrong is computing the partial derivatives, for instance in your case $\frac{\partial f}{\partial x}=0$ (see this).

For example if $f(x,u)=x^{2}+u(x)$, then $\frac{\partial f}{\partial x}=2x,\frac{\partial f}{\partial u}=1$ and $\frac{df}{dx}=2x+\frac{d}{dx}u.$

Continuing from your example, since $\frac{\partial f}{\partial y}=0$, then we have $- \frac{d}{dx}(\frac{\partial f}{\partial y'}) = 0$ or $\frac{d}{dx}(\frac{\partial f}{\partial y'}) =0.$

So $$\frac{d}{d x}(\frac{\partial f}{\partial y'})=\frac{d}{dx}(\frac{\partial }{\partial y'}\big((1+y'^{2})^{\frac{1}{2}}\big)$$ $$=\frac{d }{dx}(\frac{y'}{\sqrt{1+y'^{2}}})=0$$

(You may think of $\frac{\partial}{\partial y'}\big((1+y'^{2})^{\frac{1}{2}}\big)$ the same as computing $\frac{d}{dx}\sqrt{1+x^2}$).

Then you have $$\frac{y'}{\sqrt{1+y'^{2}}}=c_{1}$$ and since the denominator is never $0$ we have $y'^{2}=c_{1}^2(1+y'^{2})$ so re-arranging gives


(Can $c_{1}=\pm 1$?) so you have $$y'=\sqrt{\frac{c_1^{2}}{1-c_{1}^2}}=\text{constant}=a$$

Thus $y(x)=ax+b.$


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