Using traces
y=2x^2+z^2

Question

Using traces
y=2x^2+z^2

babynini · Answer

x=0, y=z^2 parabola
y=0, 0=2x^2+z^2 is..a circle?
z=0, y=2x^2 which i thought is a parabola

I think this one is all wrong heh

jim_thompson5910 · Answer

The first and third statements are correct, but 0=2x^2+z^2 isn't a circle

babynini · Answer

I wasn't sure about that one. What is it?
z=sqrt(2x^2) might be a better form..?

jim_thompson5910 · Answer

btw the graph looks like this http://i.stack.imgur.com/FVvJx.jpg source page http://math.stackexchange.com/questions/531357/find-the-intersection-of-the-surface-z-2x2y2-with-the-x-y-plane

babynini · Answer

but turned in a different way since they have z and y^2 and I have z^2 and just y

jim_thompson5910 · Answer

`but turned in a different way since they have z and y^2 and I have z^2 and just y`
yes correct, swap y and z

babynini · Answer

hmhm excellent.
I am still unsure of the steps to get the xz plane

jim_thompson5910 · Answer

Plug in y = 0 and solve for 'z' $$\Large y = 2x^2+z^2$$ $$\Large 0 = 2x^2+z^2$$ $$\Large z^2 = -2x^2$$ $$\Large z = \pm\sqrt{-2x^2}$$ $$\Large z = \pm i*x \sqrt{2}$$ So 'z' is not a real number for any nonzero x. However, if x were zero, then z would be zero as well. No other real number for x will make z a real number. So what this means is that the trace on the xz plane is simply a point. The point is at the origin (0,0). There is nothing else shown on this plane. Think of the very tip of the paraboloid touching the xz plane. I'm referring to something like this where the red point is touching the brown plane https://www.math.rutgers.edu/~greenfie/mill_courses/math251/gifstuff/min.gif

babynini · Answer

ahhh tricksy! Thanks so much for explaining that :D

jim_thompson5910 · Answer

no problem