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I know how to find the level surfaces for a $2$ variable functions, $z=(x,y)$, by finding the $3$ planes. How would you find the level surfaces for a $3$ variable function, $w=(x,y,z)$. Would you find $4$ traces? $(wxy, wxz, wyz, xyz)$? Here is a sample question of what I'm referring to incase what I am asking is completely incorrect.

Math the functions with the verbal description of the level surfaces.

$$\begin{align} w&=x+2y+3z&\hbox C\\ w&=\sqrt{x+2y+3z}&\hbox F\\ w&=x^2+2y^2+3z^2&\hbox B\\ w&=\sqrt{x^2+2y^2+3z^2}&\hbox B\\ w&=\sqrt{x^2+y^2+z^2}&\hbox E\\ w&=x^2+y^2+z^2&\hbox A\\ w&=x^2-y^2-z^2&\hbox D \end{align}$$

A. a collection of unequally spaced concentric spheres
B. a collection of concentric ellipsoids
C. a collection of equally spaced parallel planes
D. two cones and two collections of hyperboloids
E. a collection of equally spaced concentric spheres
F. a collection of unequally spaced parallel planes

Most of them make sense, except the one with the hyperboloids and cone. (I know it varies when w differs $(w>0, w<0, w=0)$, but would you just have to plug in a bunch of $w$ to see that?

Also can someone explain the plane problems properly. I have a feeling of how to do them, but I would like a set way to look at them and know if equally spaced, or not.

grg
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Winndie
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1 Answers1

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From a quick look these level surfaces maintain their "general" shape no matter the value of $w$. Except for the one answer that is two different surfaces. If you make $ w $ a constant you should be able to recognize the equation of the closed surface/solid or surface it represents.

dylan7
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  • Thanks for your commend dylan7, yea all of them make sense now except that one where its a hyperboloid of sheet one/two and a cone. How would you go about doing that question? – Winndie Sep 26 '14 at 03:36
  • The cones come from $ w=0$, solve for $ z $ you get +/- sqrt and the two collections of hyperboloids come from $ w<0$ and $ w> 0$. Greater than 0 are circular hyperboloids and less than are parabolic hyperboloids. If you plug in some sample values for the three case you should see that equation changes in each case resulting in three different "types" of surfaces. – dylan7 Sep 26 '14 at 03:43
  • Thanks for your reply again dylan, so if I am given a similar question later, I would have to just plug in a bunch of w's on both side of $w=0$ and $w=0$ itself? Also for number 1 and 2, the plane functions, all thats changing is the $d$ in $ax+by+cz=d$, how can you determine if the surfaces are equally spaced, or unequally spaced from just the change in $d$ . – Winndie Sep 26 '14 at 03:49
  • For the unequally spaced planes, if that is what your talking about. Take the square of both sides plug in $ w $= 1,2,3... and realize you get 1,4,9. Thus unequal spacing. The other plane one gave you 1,2,3 for $ w $=1,2,3, equal spacing. You don't have to plug in a bunch. You can try to think generally. Like it was good you realized the hyperboloid cone one differed for the three case and all you had to do was plug $ w $=-1,0,1. Then it's all about recognizing the surface. To recognize the surface you can play around with x, y, z one you've set $ w $ to something. – dylan7 Sep 26 '14 at 03:50
  • And I meant solve for $ x $ not $ z $ in my previous comment about the cone. – dylan7 Sep 26 '14 at 03:56
  • And the hyperboloids I said are wrong greater than 1 is a two sheet, less than 1 is a one sheet. If that matters. – dylan7 Sep 26 '14 at 04:05
  • Yes, it was the unequally spaced planes that I was referring to. Since in the 2nd one the $d$ in $ax+by+cz-d=0$ is changing from 1 to 4, to 9 for $w=1,2,3$, then that means d is the distance from the plane to the origin? I always thought it was just a unit less constant for that specific function. Thanks alot for ur help :) – Winndie Sep 26 '14 at 04:06
  • It's not the distance from the origin. It helps you determine the intercepts though. Like plug in 0 for two variable solve for the other. If plane $ w=1$ has intercepts $(0,0,1), (0,1,0), (1,0,0) $ and plane $ w=2$ has intercepts $(2,0,0), (0,2,0), (0,0,2) $. Well then their equally space. You're moving it over 1 in all directions. But if plane $ w=1$ has the same as before, but $ w=2$ has intercepts $(0,4,0), (0,0,4), (4,0,0) $ well then plane 2 has intercepts 3 more than plane 1. I would put a couple planes from $ w $ and $ w^2$ in a 3D grapher and take a look. – dylan7 Sep 26 '14 at 04:13
  • If you go to plane $ w=3$ for $ w $ you get intercepts with 3 in them. You went up by 1 each time but for $ w^2$ you get 9 for plane three. For $ w^2$ you get intercepts with 1 in them then 4 then 9. Not equally spaced – dylan7 Sep 26 '14 at 04:17
  • Wow ok thanks, that made so much sense. Thanks so much for everything – Winndie Sep 26 '14 at 04:38
  • No problem. Glad to have helped. – dylan7 Sep 26 '14 at 04:40