Calculus of Variations: interesting substitution

[member 428835]

Homework Statement

Find the externals of the functional
$$\int\sqrt{x^2+y^2}\sqrt{1+y'^2}\,dx$$
Hint: use polar coordinates.

Homework Equations

$x=r\cos\theta$
$y=r\sin\theta$

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?

 

[haruspex]

Homework Statement

Find the externals of the functional
$$\int\sqrt{x^2+y^2}\sqrt{1+y'^2}\,dx$$
Hint: use polar coordinates.

Homework Equations

$x=r\cos\theta$
$y=r\sin\theta$

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?

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

 

[member 428835]

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

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

 

[member 428835]

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 $F = r\sqrt{1+\left(r'(\theta)\sin\theta+r\cos\theta\right)^2}$.

 

[haruspex]

For completeness, here is the rest (I think)

$$\int r\sqrt{1+\left(r'(\theta)\sin\theta+r\cos\theta\right)^2}\,d\theta$$

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

 

[member 428835]

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

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).

 

[member 428835]

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?

 

[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?

 

[member 428835]

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?

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?

 

[haruspex]

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?

You seem to have dropped a sqrt operation.

 

[member 428835]

You seem to have dropped a sqrt operation.

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?