Calculus of Variations: interesting substitution

1. Feb 4, 2017

joshmccraney

1. The problem statement, all variables and given/known data
Find the externals of the functional
$$\int\sqrt{x^2+y^2}\sqrt{1+y'^2}\,dx$$
Hint: use polar coordinates.
2. Relevant equations
$x=r\cos\theta$
$y=r\sin\theta$

3. The attempt at a solution
Transforming the given functional where $r=r(\theta)$ yields
$$\int r\sqrt{1+\left(\frac{r'(\theta)\sin\theta+r\cos\theta}{r'(\theta)\cos\theta-r\sin\theta}\right)^2}(r'(\theta)\cos\theta-r\sin\theta)\,d\theta$$ which doesn't seem to help much. Does anyone see anyhting I did wrong?

2. Feb 4, 2017

haruspex

That simplifies hugely. Bring the trailing factor inside the square root and expand the brackets.

3. Feb 4, 2017

joshmccraney

Wow, I hate that I didn't see this!

4. Feb 4, 2017

joshmccraney

For completeness, here is the rest (I think)

$$\int r\sqrt{1+\left(r'(\theta)\sin\theta+r\cos\theta\right)^2}\,d\theta$$ Then I suppose simply use Euler's formula where $F = r\sqrt{1+\left(r'(\theta)\sin\theta+r\cos\theta\right)^2}$.

5. Feb 4, 2017

haruspex

No, it should be simpler than that. It's easier, perhaps, if you start by writing $\sqrt{1+y'^2}dx$ as $\sqrt{dx^2+dy^2}$.

6. Feb 4, 2017

joshmccraney

Ok, but $dx = (\partial_r x) dr+(\partial_\theta x) d\theta = \cos\theta dr-r\sin\theta d\theta$, right (I figured I'd ask before going through all the math).

7. Feb 4, 2017

joshmccraney

Before I was thinking $$\frac{d y}{dx} = \frac{d_\theta y}{d_\theta x}=\frac{\partial_ry\cdot\partial_\theta r+\partial_\theta y}{\partial_rx\cdot\partial_\theta r+\partial_\theta x}=\frac{\sin\theta r'(\theta)+r\cos\theta}{\cos\theta r'(\theta)-r\sin\theta}$$ Is this correct?

8. Feb 4, 2017

haruspex

I agree with all your equations in posts 1, 6 and 7, but not the one in post 4.
What do you get for dx2+dy2?

9. Feb 5, 2017

joshmccraney

Oops, I forgot to distribute! $dx^2=( \cos\theta dr-r\sin\theta d\theta)^2$ and $dy^2=( \sin\theta dr+r\cos\theta d\theta)^2$. Then $dy^2+dx^2=dr^2+r^2d\theta^2$ so the integral in post 1 becomes $$\int r dr^2+r^3d\theta^2$$ Now are we free to arbitrarily select the independent variable?

10. Feb 5, 2017

haruspex

You seem to have dropped a sqrt operation.

11. Feb 6, 2017

joshmccraney

Shoot! Ok, so we would actually have $$\int r \sqrt{dr^2 + d\theta^2 r^2} = \int r\sqrt{r'^2+r^2}\,d\theta$$ which implies $F=r\sqrt{r'^2+r^2}$. Since $F$ is independent of $\theta$, Euler's equation reduces to $F-r'F_{r'}=C$, and is stated as
$$C=r\sqrt{r^2+r'^2}-\frac{rr'^2}{\sqrt{r^2+r'^2}}\implies\\ \int\,d\theta=\int \frac{dr}{\sqrt{C_1r^6-r^2}}\implies\\ \theta = C-\frac{arccot\sqrt{-1 + C_1 r^4}}{2}\\ \cot(C_2-2\theta)=\sqrt{-1 + C_1 r^4}$$ Have I made a mistake?

Last edited: Feb 6, 2017