What is the shape of the curve z = x^2 + 2y^2 and how can it be sketched?

[tensor0910]

Homework Statement

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Sketch the curve z = x² + 2y²

Homework Equations

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The Attempt at a Solution

: [/B]The easiest thing to do is to sketch the traces at x, y and z = 0. I'm 95% sure its an elliptic paraboloid, but the 2 in front of the Y is really throwing me off. Should we divide everything by 2 and have an ellipse with a radius of √2 and 1? If that's the case then when we set Y to 0 ( for example ) we get z = 2x² and were right back where we started...Im kinda lost here.

 

[Charles Link]

For a given $z=z_o$ in a plane parallel to the x-y plane we have $z_o=x^2+2y^2$ which, can be put in the standard form on ellipse with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. What is $a$ and $b$ for this case? $\\$ The case where you looked at the plane $y=0$ is correct, in that going upward in $z$, a cross section of this thing is a parabola. The cross section through any plane $Ax+By=0$ , will give a parabola of the form $z=C (x')^2$ where $x'$ is the coordinate along this plane perpendicular to $z$.

 

[Mark44]

I'm 95% sure its an elliptic paraboloid, but the 2 in front of the Y is really throwing me off. Should we divide everything by 2 and have an ellipse with a radius of √2 and 1?

The coefficient of 2 on the y² term is why this surface is an elliptic paraboloid. If that coefficient had been 1, all the horizontal sections would have been circular. All of the horizontal cross-sections (i.e., in planes parallel to the x-y plane) are ellipses, except for when both x and y are zero.

Sketching traces is a good start, but you should also include cross sections or level curves for various values of z.

For example, when z = 0, you get a single point. When z= 1, the cross section is the ellipse $x^2 + 2y^2 = 1$. The same thinkiing applies for other values of $z \ge 0$.

Edit: Cross-sections are especially helpful when you have the sum of squares of two of the variables (cross sections are circles) or expressions like $ax^2 + by^2$ (elliptical cross sections). This advice applies not just for expressions involving x and y, but any two of the variables.

 

[tensor0910]

I'm starting to get it.

I was doubting my answers b/c I was having trouble visualizing the sketch ( a huge weakness of mine ). I thought that an elliptic paraboloid would make
perfect parabolas on the yz and xz plane, but if that were the case the sketch would turn into a circle. I'm back on track now. Thanks for the help!

 

[Charles Link]

They make "perfect" parabolas on the yz and xz plane, but they are not identical parabolas. The parabola on the yz plane is $z=2y^2$, and the parabola on the xz plane is $z=x^2$.