1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculus of Variations: interesting substitution

  1. Feb 4, 2017 #1

    joshmccraney

    User Avatar
    Gold Member

    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. jcsd
  3. Feb 4, 2017 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That simplifies hugely. Bring the trailing factor inside the square root and expand the brackets.
     
  4. Feb 4, 2017 #3

    joshmccraney

    User Avatar
    Gold Member

    Wow, I hate that I didn't see this!:headbang:
     
  5. Feb 4, 2017 #4

    joshmccraney

    User Avatar
    Gold Member

    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}##.
     
  6. Feb 4, 2017 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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}##.
     
  7. Feb 4, 2017 #6

    joshmccraney

    User Avatar
    Gold Member

    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).
     
  8. Feb 4, 2017 #7

    joshmccraney

    User Avatar
    Gold Member

    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?
     
  9. Feb 4, 2017 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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?
     
  10. Feb 5, 2017 #9

    joshmccraney

    User Avatar
    Gold Member

    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?
     
  11. Feb 5, 2017 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You seem to have dropped a sqrt operation.
     
  12. Feb 6, 2017 #11

    joshmccraney

    User Avatar
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Calculus of Variations: interesting substitution
  1. Calculus of Variations (Replies: 3)

  2. Calculus of variations (Replies: 10)

  3. Calculus of variations (Replies: 8)

  4. Calculus by variations (Replies: 2)

  5. Calculus of variations (Replies: 3)

Loading...