# Simple calculus - interpretation Euler-Lagrange equation

1. Sep 2, 2006

### oldstudent

This is not a homework question but one that is part of the course material and I can't really move on until I understand the basic calculus.

I have a problem interpreting "d by dx of partial dF by dy' equals partial d by dy' of dF by dx" in the following question, which I set out and then highlight my problem.

For F = (x^2 + y'^2)^1/2, find;

partial dF/dx = x(x^2 + y'^2)^-1/2

partial dF/dy = 0

partial dF/dy' = y'(x^2 + y'^2)^-1/2

dF/dx = p.dF/dx + p.dF/dy.y' + p.dF/dy'.y''

Show that: d/dx(p.dF/dy') = p.d/dy'(dF/dx)

Taking the RHS I believe is:

p.d/dy'(dF/dx) = p.d/dy'[x(x^2 + y'^2)^-1/2 + p.dF/dy.y' +

y'(x^2 + y'^2)^-1/2 y'']

ditto = - xy'(x^2 + y'^2)^3/2 + p.d/dy' pdF/dy.y'

- y'^2 y''(x^2 + y'^2)^3/2

My problem is that I have forgotten how to interprete p.d/dy' pdF/dy.y' I know that the answer is y''(x^2 + y'^2)^-1/2 but I don't understand how one gets this answer when partial dF/dy = 0

and I don't know how to interpret the LHS - d/dx(partial dF/dy') at all although I appreciate that it gives the same answer for this function as the RHS.
I would appreciate some help on the fundamentals. Many thanks.

2. Sep 2, 2006

### StatusX

In this case $\partial F/\partial y =0$, so that term drops out. In general you would write out that term like the others and do just what you did with them.

The following:

$$\frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right)$$

is computed the same way as $dF/dx$. Remember that $\partial F/\partial y'$ is a function you know, just like F. Just write it out like your fifth line above.

3. Sep 7, 2006

### oldstudent

Thank you StatusX. I've got it at last.