1. Not finding help here? Sign up for a free 30min 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!

Path for a Level curve

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the path ##\vec\gamma(t)## which represents the level curve ##f(x,y) = \displaystyle\frac{xy + 1}{x^2 + y^2}## corresponding to ##c=1##.
    Similarly, find the path for the curve ##x^{2/3} + y^{2/3} = 1##

    2. Relevant equations
    None.


    3. The attempt at a solution
    Since the level set corresponds to ##c=1##, ##xy + 1 = x^2 + y^2##.

    At this point, I know that ##x^2 + y^2 - xy = 1## is an ellipse, but I cannot put it into a form that is similar to how the path for the general ellipse ##\displaystyle\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1## is ##\vec\gamma(t) = (acos(t), bsin(t))## because of the ##xy## term.

    Same thing for ##x^{2/3} + y^{2/3} = 1##. It almost looks like the general form, but not quite. Any suggestions?
     
    Last edited: Feb 5, 2013
  2. jcsd
  3. Feb 5, 2013 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    This [itex]x^2- xy+ y^2= 1[/itex] is an ellipse but its axes are rotated and not parallel to the x and y axes can be written in that form in either of two ways.

    The first is very elementary but tedious: Let [itex]x= x' cos(\theta)+ y' sin(\theta)[/itex], [itex]y= -x'sin(\theta)+ y'cos(\theta)[/itex]. Putting those into the equation, [itex](x' cos(\theta)+ y'sin(\theta))^2- (x' cos(\theta)+ y' sin(\theta)(-x' sin(\theta)+ y'cos(\theta))+ (-x sin(\theta)+ y' cos(\theta))^2= -1[/itex].

    Multiply those out, combine like powers and set the coefficient of x'y' to 0 and solve for [itex]\theta[/itex] that will make the coefficient 0. Use that [itex]\theta[/itex] to find the coefficients of [itex]x'^2[/itex] and [itex]y'^2[/itex].

    More "sophisticated" (and so simpler) is to write this as a matrix equation:
    [tex]\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix}1 & -\frac{1}{2} \\ -\frac{1}{2} & 1\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= 1[/tex]

    That is, necessarily, a symmetric matrix and so has two independent eigenvalues. The eigenvalue equation is [itex]\left|\begin{array}{cc}1- \lambda & -\frac{1}{2} \\ -\frac{1}{2} & 1- \lambda\end{array}\right|= (1- \lambda)^2- \frac{1}{4}= 0[/itex]
    That can be written as [itex](1- \lambda)^2= \frac{1}{4}[/itex] so that [itex]1- \lambda= \pm\frac{1}{2}[/itex]. The two eigenvalues are [itex]\lambda= \frac{3}{2}[/itex] and [itex]\lambda= \frac{1}{2}[/itex].

    That tells us that the matrix can be diagonalized in the form
    [tex]\begin{bmatrix}x'& y'\end{bmatrix}\begin{bmatrix}\frac{3}{2} & 0 \\ 0 & \frac{1}{2}\end{bmatrix}\begin{bmatrix}x' \\ y' \end{bmatrix}= \frac{3x^2}{2}+ \frac{x^2}{2}= 1[/tex]

    x' and y' are axes in the direction the eigenvectors of the original matrix. They are given by
    [tex]\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix}1 & -\frac{1}{2} \\ -\frac{1}{2} & 1\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \frac{3}{2}\begin{bmatrix}x \\ y\end{bmatrix}[/tex]
    and
    [tex]\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix}1 & -\frac{1}{2} \\ -\frac{1}{2} & 1\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \frac{1}{2}\begin{bmatrix}x \\ y\end{bmatrix}[/tex]

    That first gives x- y/2= 3x/2 and -x/2+ y= 3y/2 or x= -y and the second gives x- y/2= x/2 and -x/2+ y= y/2 or y= x. (So the [itex]\theta[/itex] above is [itex]\pi/4[/itex].) That tells us that [itex]x^2- xy+ y^2= 1[/itex] can be written as [itex](3/2)(x- y)^2+ (1/2)(x+ y)^2= 1[/itex]. That is, you can use parametric equations [itex]x- y= \sqrt{2/3} sin(t)[/itex] and [itex]x+ y= \sqrt{2}cos(t)[/itex]. You can easily solve those for x and y separately as functions of the parameter, t.

    [itex]x^{2/3}+ y^{2/3}= 1[/itex] is NOT an ellipse or any conic section. You can, however, take [itex]x= sin^3(t)[/itex], [itex]y= cos^3(t)[/itex] so that [itex]x^{2/3}+ y^{2/3}= (sin^3(t))^{2/3}+ (cos^3(t))^{2/3}= sin^2(t)+ cos^2(t)= 1[/itex]
     
    Last edited: Feb 5, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Path for a Level curve
  1. Level curves (Replies: 7)

  2. Level curves (Replies: 2)

  3. Level Curve (Replies: 1)

  4. Level curves? (Replies: 2)

  5. Level curves (Replies: 5)

Loading...