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 $x^2- xy+ y^2= 1$ 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 $x= x' cos(\theta)+ y' sin(\theta)$, $y= -x'sin(\theta)+ y'cos(\theta)$. Putting those into the equation, $(x' cos(\theta)+ y'sin(\theta))^2- (x' cos(\theta)+ y' sin(\theta)(-x' sin(\theta)+ y'cos(\theta))+ (-x sin(\theta)+ y' cos(\theta))^2= -1$.

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

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

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

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

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

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

$x^{2/3}+ y^{2/3}= 1$ is NOT an ellipse or any conic section. You can, however, take $x= sin^3(t)$, $y= cos^3(t)$ so that $x^{2/3}+ y^{2/3}= (sin^3(t))^{2/3}+ (cos^3(t))^{2/3}= sin^2(t)+ cos^2(t)= 1$