Results 1 to 3 of 3

Math Help - Tricky problem involving 2nd order ODE

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    60

    Tricky problem involving 2nd order ODE

    Hi, my prof gave us this problem in my ODE course. I have no idea how to solve it so help would be very much appreciated. The question is as follows:

    A curved mirror of equation y=y(x) (with respect to some cartesian coordinate frame) has the property that whenever a ray of light emanates from the origin it reflects parallel to the x-axis. Find the equation of the mirror.

    Thanks a lot!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42
    Here are some ideas. Pick a point on your curve (x,y(x)). You can find the slope of the line that passes through the origin and your point. You also know that the slope of the horizontal line is zero. At the point you picked on your curve you can find the slope of the tangent line. There is an angle between your first curve and your tangent line (angle of incidence) and an angle between your tangent line and your horizontal line (angle of reflection) and these are equal. By relating your angles to your slopes you'll find a differential eqution for y(x).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    If You observe the figure...



    ... on the basis of geometrical considerations You can extablish that, defining BFC= \alpha , is FCA=FDC= \frac{\alpha}{2}.Now if x and y are the coordinates of C is...

    \displaystyle \sin \alpha = \frac{x}{\sqrt{x^{2}+ y^{2}}}

    \displaystyle \cos \alpha = \frac{y}{\sqrt{x^{2}+ y^{2}}} (1)

    ... so that the DE of the 'confocal mirror' is...

    \displaystyle y^{'} = \tan \frac{\alpha}{2} = \frac{1-\cos \alpha}{\sin \alpha} = \sqrt{1+\frac{y^{2}}{x^{2}}} - \frac{y}{x} (2)

    The first order DE of the type (2) was resolved in the eighteenth century by the Italian mathematician Gabriele Manfredi and is commonly called 'homogeneous'. The standard procedure consists in the change of variables \frac{y}{x}= t \implies dy= t\ dx + x\ dt, that permits to write the (2) as...

    \displaystyle \frac{dt}{\sqrt{1+t^{2}}-2\ t}= \frac{dx}{x} (3)

    The problem now is solving (3) and that will be done in a succesive post...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: September 11th 2011, 10:08 AM
  2. Replies: 2
    Last Post: September 10th 2011, 10:50 PM
  3. Tricky First Order D.E
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: May 19th 2011, 07:58 AM
  4. Tricky question involving exact differentials
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 3rd 2011, 07:33 PM
  5. Tricky train problem involving ratios
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 22nd 2010, 04:32 PM

Search Tags


/mathhelpforum @mathhelpforum