Finding the Path for Level Curves: Solving for Ellipses and Special Cases

[Karnage1993]

Homework Statement

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##

Homework Equations

None.

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?

 

[HallsofIvy]

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]