Orientation Question: re-calculus of variations

Question

Orientation Question:  re-calculus of variations

irishboy123 · Answer

if we have 
$I = \int_a^b \; dx \sqrt{1 + y'^2}$

what are  $\dfrac{dI}{dy}$ and $\dfrac{dI}{dy'}$ ??

solomonzelman · Answer

$\large\color{#000000 }{ \displaystyle f(x)=\int_{a}^{b}g(t)~dx  }$

$\large\color{#000000 }{ \displaystyle F(x)=G(a)+G(b)  }$     (F'=f)

$\large\color{#000000 }{ \displaystyle f'(x)=0   }$      (i suppose that would be for d$\rm I$/dy)

solomonzelman · Answer

because $I$ is constant when evaluated at constant limits [of integration], and the derivative of a constant, blah blah blah. ....

solomonzelman · Answer

or is it something different? I am becoming afraid

irishboy123 · Answer

soz, Sol

y = y(x)

solomonzelman · Answer

ooh

solomonzelman · Answer

you have y' sitting in there didn't see the  '

solomonzelman · Answer

I haven't learned that yet -:(

solomonzelman · Answer

Hopefully, the time comes though ...

solomonzelman · Answer

(please remove the medals)

solomonzelman · Answer

I will play around and see if it is possible to solve this via some elementary calc that I know..

solomonzelman · Answer

probably not... (I wrote down and I will look at some maths.... perhaps that would be a motivation, but you need someone much more knowledgable than a guy)
my apologizes and good luck...