TITLE: Level Curves in multiple variabes
QUESTION [1 upvotes]: I want to sketch some level curves to the function $f(x,y) = 3x^2 + 4xy  + 3y^2$. I have tried to set the equation to C but I'm not sure how to proceed.
I've gotten hints to substitute $u = x + y$ and $v = x - y$ but it doesn't seem to make sense to me.
Thanks in advance!

REPLY [1 votes]: The curves are clearly elliptical.
To be more precise, notice that
$$
f(x,y) = (x,y)A\binom xy
\quad\text{where}\quad
A = \begin{bmatrix}3 & 2\\ 2& 3\end{bmatrix}
$$
Infinite many such $A$ exist, the one above is the one which is symmetric and hence diagonalizable.
Its eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 5$, with respective eigenversors $v_1 = \frac{1}{\sqrt 2}(1,-1)$ and $v_2 = \frac{1}{\sqrt 2}(1,1)$.
This means that
\begin{align}
f(x,y)
= &
(x,y)
\begin{bmatrix}3 & 2\\ 2& 3\end{bmatrix}
\binom xy
\\
= &
(x,y)
\frac{1}{\sqrt 2}
\begin{bmatrix}1 & 1\\ -1& 1\end{bmatrix}
\begin{bmatrix}1 & 0\\ 0& 5\end{bmatrix}
\sqrt 2
{\begin{bmatrix}1 & 1\\ -1& 1\end{bmatrix}}^{-1}
\binom xy
\\
= &
(x,y)
\begin{bmatrix}\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2}\\ -\frac{1}{\sqrt 2}& \frac{1}{\sqrt 2}\end{bmatrix}
\begin{bmatrix}1 & 0\\ 0& 5\end{bmatrix}
\begin{bmatrix}\frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2}\\ \frac{1}{\sqrt 2}& \frac{1}{\sqrt 2}\end{bmatrix}
\binom xy
\end{align}
Letting
$$\tag{1}
\begin{cases}
\hat x & = \displaystyle\frac{x-y}{\sqrt 2} \\
\hat y & = \displaystyle\frac{x+y}{\sqrt 2} \\
\end{cases}
$$
you have
$$
f(\hat x,\hat y) = \hat x^2 + 5\hat y^2
$$
therefore the level curves of $f$ in $(\hat x,\hat y)$ coordinates are solutions to
$$
\hat x^2 + 5\hat y^2 = C
$$
For $C<0$ you have no solutions, for $C=0$ you have only the origin, while for $C>0$
$$
\frac{\hat x^2}{(\sqrt C)^2} + \frac{\hat y^2}{(\sqrt{C/5})^2} = 1
$$
which are ellipses centered in the origin and with semiaxes $\sqrt C$ and $\sqrt{C/5}$ parallel to $\hat x$ and $\hat y$ respectively.
Basically they are all concentrical ellipses with ratio horizonal/vertical axis equal to $1/\sqrt 5$.
Notice that the change of variable $(1)$ is simply a rotation of $\frac{\pi}{4}$, so that in the original coordinates $(x,y)$ your curves are such ellipses rotated by $-\frac{\pi}{4}$.

REPLY [0 votes]: On substituting $u=x+y$ and $v=x-y$, you get, $$x=(u+v)/2$$ and $$y=(u-v)/2$$
Thus,
$$3x^2 + 4xy + 3y^2 = 3(u+v)^2/4 + (u+v)(u-v) + 3(u-v)^2/4$$
On simplification, it reduces to:
$$f(x,y)=g(u,v)=3(u^2+v^2)/2 + u^2-v^2 =( 5u^2 + v^2)/2$$
which is the equation of an ellipse in the $u-v$ plane.
The transformation is like rotating the $x-y$ axes and scaling it. So the cross section's shape wouldn't change.
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