Sketching the curve

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tensor0910
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Homework Statement


[/B]
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 thats 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.
 
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Charles Link
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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##.
 
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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.
 
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tensor0910
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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!
 
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Charles Link
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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 ##.
 

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