partial differentiation in a geodesic problem

Question

I'm trying to understand the example (which is given in a textbook):

Determine $$y\in\{f\in C^2[0,1]:\ f(0)=0,\ f(1)=1\}$$

such that $J(y)=\int_0^1\sqrt{1+y'(x)^2}dx$ is minimum.

From a theorem, we know $y$ has to satisfy the following differential equation:

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

where $f(x,y(x),y'(x))=\sqrt{1+y'(x)^2}$.

But it's unclear to me how to find $\partial f/\partial y'$ and $\partial f/\partial y$. I'm not really sure what these partial derivatives even mean. The work in the textbook gives

$$\frac{\partial f}{\partial y'}=\frac{y'}{\sqrt{1+(y')^2}}$$

$$\frac{\partial f}{\partial y}=0$$

and that "sorta" makes sense if we treat $y$ and $y'$ as real variables... but they're not? Maybe this is super basic stuff from multivariable calculus, but I haven't done that in years.

Answer 1

They are variables when you differentiate with respect to them. Maybe if you write $$f(x,y,y') = \sqrt{1+(y')^2}$$ then it is more clear that ${\displaystyle \frac{\partial f}{\partial y} = 0}$ and by the chain rule ${\displaystyle \frac{\partial f}{\partial y'} = \frac{y'}{\sqrt{1+(y')^2}}}.$

If you wanted to differentiate with respect to $x$, instead, then you should consider that $y'$ depends on $x$.

Answer 2

Suppose you have instead

$$ \sqrt{\alpha y^2+\beta y y'+\gamma y'^2}\equiv\sqrt{\alpha u^2+\beta u v+\gamma v^2} = f(u,v) $$

Regarding the partial derivative $u$ and $v$ are considered as independent so

$$ \frac{\partial f}{\partial u} = \frac{1}{2}\frac{2\alpha u+\beta v}{\sqrt{\alpha u^2+\beta u v+\gamma v^2}}\\ \frac{\partial f}{\partial v} = \frac{1}{2}\frac{2\gamma v+\beta u}{\sqrt{\alpha u^2+\beta u v+\gamma v^2}} $$

and finally

$$ \frac{d}{dt}\frac{\partial f}{\partial v} - \frac{\partial f}{\partial u} = \frac{d}{dt}\left(\frac{1}{2}\frac{2\gamma y'+\beta y}{\sqrt{\alpha y^2+\beta y y'+\gamma y'^2}}\right)-\frac{1}{2}\frac{2\alpha y+\beta y'}{\sqrt{\alpha y^2+\beta y y'+\gamma y'^2}} $$