Question

Help understanding quadric surface equations. Tricks?

Answer 1

For example, how can we distinguish that

\[\large \frac{x^2}{2} - \frac{y^2}{4} + \frac{z^2}{9} = 1\]
Is a hyperboloid of 1 sheet, but
\[\large \frac{x^2}{2} - \frac{y^2}{4} + \frac{z^2}{9} = -1\]
is a hyperboloid of 2 sheets. when the only thing that has changed is the fact they equal 1 and -1 respectively

Answer 2

sketching the curves in xy, yz , zx planes is a good start point

Answer 3

one way to make sense of two sheets is by noticing that when y=0, you get x^2/2 + z^2/9 which can never equal a negative number

Answer 4

Alright so by the plane ideas, so the first equation

xy plane \(\large \frac{x^2}{2} - \frac{y^2}{4} = 1\)
xz plane \(\large \frac{x^2}{2} + \frac{z^2}{9} = 1\)
yz plane \(\large -\frac{y^2}{4} + \frac{z^2}{9} = 1\)

and right I see that now...when y=0 then it will never equal a negative number

Answer 5

however based on the graph from wolfram alpha i dont see that

http://www.wolframalpha.com/input/?i=x%5E2%2F2+-+y%5E2%2F4+%2B+z%5E2%2F9+%3D+1

Answer 6

*I'm talking about the 1 sheet whereas that comment you made was about the 2 sheets...noted*

Answer 7

So indeed it looks like there are circles in terms of every plane...so it would make sense that it would be a hyperboloid.

but lets see for the 2 sheet

xy plane \(\large \frac{x^2}{2} - \frac{y^2}{4} = -1\)
xz plane \(\large \frac{x^2}{2} + \frac{z^2}{9} = -1\)
yz plane \(\large -\frac{y^2}{4} + \frac{z^2}{9} = -1\)

so lets see...you said when y = 0 so in the xz plane...

in the xz plane I see that we cannot ever equal -1....so what does that mean? it means we never have anything in the xz plane?

Sorry, just not quite getting this yet, I'm sure I will with time

Answer 8

sorry was on phone.. .yes exactly, the surface stays left and right of xz plane without touching it

Answer 9

ahh right right, duh >.<

Answer 10

Mind helping me understand a few more?

Answer 11

wil try.. ellipsoid and paraboloid should be easy to *see* compared to hyperboloid

Answer 12

I think I have the cone thing down

Since the basic equation of a cone looks something like

\[\large x^2 + y^2 = z^2\]

^that meaning it would be a cone opening on the z-axis starting at the origin right?

Answer 13

So lets see

\[\large z^2 = \frac{x^2}{4} + \frac{y^2}{9}\] would be a cone that opens along the z-axis and starts at the point (4,9,0) right?

Answer 14

not quite, it should pass throught he origin (0, 0, 0) and notice that thats the only point at which it cuts the xy plane because 0 = x^2+y^2 is a degenerated ellipse

Answer 15

ohhh wait that's right....that second equation I have passes through the origin but it would be an elliptical cone!

Answer 16

it might be easy to think of it as :

\[\large k^2 = \dfrac{x^2}{4} + \dfrac{y^2}{9}\]

Answer 17

think of \(\large k\) as parameter : for each value of k you get an ellipse whose center of axis is z axis

Answer 18

^^^^^ very good way of putting that! that is so much better than what I was thinking!

Answer 19

So hmm, sphere looks pretty straight-forward

\[\large x^2 + y^2 + z^2 = 9\] would be a sphere radius 3 correct?

Answer 20

yes collection of all the points that are 3 units away from origin

Answer 21

Great!

hmm so lets see

it looks to me that paraboloids are equations that have 1 variable that is not a square...makes sense because if I did that "plane breakdown" again I would be left with z = y^2 - 3 for example

so

\[\large y = \frac{z^2}{4} - \frac{x^2}{9}\] would be a paraboloid correct?

Answer 22

wait...but in the xz plane...it would be a hyperbola wouldnt it?

so this would be a hyperbolic paraboloid?

Answer 23

yes its a hyperbolic paraboloid

Answer 24

its easy to see that it meets the zx plane only at (0, 0, 0)

Answer 25

yz and zx planes give you parabolas

Answer 26

Now what about ellipsoids? I dont see any examples in my book lol, terrible

Answer 27

have u sketched that ? looks we get a saddle point

Answer 28

http://erikdemaine.org/hypar/maple_hypar_plot.gif

Answer 29

Hmm no I didnt see that, i thought it just looked like a sheet of parabolas, i didnt notice that it would have a saddle point O.o

Answer 30

A left handed system?