Find the intersection of the surface $z=2x^2+y^2$ with the $x-y$ plane I was asked to find the partial derivatives of $z=2x^2+y^2$ and the intersection of this surface with the $x-y$ plane. Is finding it relates to the partial derivatives?
Here are my solutions for the partial derivatives:
$z_x=4x$
$z_y=2y$
$z_{xx}=4$
$z_{xy}=0$
$z_{yy}=2$
$z_{yx}=0$
 A: No need for partial derivates.  The plane defined by the $x$ and $y$ axes has the formula $z=0$.  Combine both formulas:
\begin{align}
z &= 2x^2+y^2 &&\text{your curve} \\
z &= 0 &&\text{the $x$-$y$ plane} \\
\end{align}
The answer is $\ldots$?
A: You write such good posts, and so I will in kind give you some more detail considering that test on Tuesday. Here are three ways to think about this problem. The first will require spotting a property of an equation, the second will be purely algebraic, and the third, my favorite, will be geometric.

1. When presented with the question of intersection of $z=2x^2+y^2$, and the $x-y$ plane, the first thing you want to do is recognize, as your other poster dutifully mentions, that the $x-y$ plane is the place where $z=0$. More on this in a second.
So the problem is equivalent to solving $z=2x^2+y^2$, and $z=0$. You should immediately think of solving this like any other system of equations, and my first inclination is substitution. Plug $0$ in for $z$ in the other equation to get $0=2x^2+y^2$. Now here is the bright thought in Carlos' post. We know that $z=0$, so consider values of $x$ and $y$ so that
$$2x^2+y^2=0.$$
After thought, it might occur to you that $x^2 \geq 0$, and $y^2 \geq 0$, and so the only way that $2x^2+y^2$ can possibly equal zero is if both $x$ and $y$ are zero. Game over. The solution is the single point $(x,y,z)=(0,0,0)$.

2. Suppose you did not see this. Well given any one equation in two variables, there will either be one, none, or many solutions. Translate this as a point, nothing, or a curve. Let us just go with the possibility that the solution is some curve. We solve for $y$ in the usual manner to get 
$$2x^2+y^2=0 \Rightarrow y^2=-2x^2 \Rightarrow y= \pm \sqrt{-2x^2}= \pm  i\sqrt{2} x.$$
So this is presented as a non real (complex) solution. This would mean no solution, but notice how if we let $x=0$, then $y=0$, and so we have again, one real solution, $x=0,y=0,z=0$.

3. When I took cal 3 we were forced to actually draw these things with a pencil. I doubt professors make students do that anymore, and the art is probably vanishing. In any case, we could call the equation $z=2x^2+y^2$ an elliptic paraboloid, and $z=0$ is a flat plane as your question implies. Here is what they look like (custom made just for you in Mathematica :)) ):

The elliptic paraboloid just touches the origin, and so if you know what it looks like, you could just instantly answer this particular question as $(0,0,0)$, although the solutions to your other problems probably do not just work as sweet as these. Note that the graph does not tell us per se that the solution in the origin, but our understanding of what these two surfaces look like (or how they behave) can tell us this.
So which method do I recommend? All of the above, but in real life, even knowing what an equation looks like in your head, you are probably going to need to apply method 2. The interesting problems wind up being some sort of curve, or several curves piecewise defined in terms of one of the variables.
I will conclude by saying that, while learning how to draw these things is not necessary, learning what the various common surfaces look like is very worthwhile. Playing with a grapher along with a book or list of surfaces is one way. I wager there are about 10 or so shapes to know just by looking at the equation such that if you knew them well you would be seriously advantaged. Note that this is actually supposed to be fun, so do the work, but don't forget to enjoy the fun part at the same time. Hey what's not to like about a picture! Best of luck on your test.
