Intersecting yz-plane: x^2 + y^2 - 4 Curve

[madachi]

Homework Statement

Does the surface $z = x^2 + y^2 - 4$ * intersects the $yz$-plane? If so, find the equation of the curve and write down the points of intersection.

The Attempt at a Solution

$yz-$plane, so $x=0$

1) * becomes $z = y^2 - 4$ and this is the equation of the curve that intersects the $yz$-plane.
2) So the intersection points are $(0,-2,0)$ and $(0,2,0)$

Are 1) and 2) correct? Thanks in advance!

 

[Je m'appelle]

Shouldn't there be a third intersection point?

You found the intersection points of the curve on the y-axis, what about the z-axis?

 

[madachi]

Is it $(0,0,-4)$ ? But I am slightly confused, doesn't the whole curve intersect yz-plane and so there are infinitely many points of intersection?

Thanks.

 

[lanedance]

the graph of $z = x^2 + y^2 - 4$, is a parabaloid, rotationally symmetric about the z axis

the intersection of the surface with the yz plane (x=0) is the parabola $z = y^2 - 4$ which has, as you say, infinitely many points.

I'm not too sure what the last part is asking for, maybe it is where the curve intersects the y & z axis...

 

[madachi]

the graph of $z = x^2 + y^2 - 4$, is a parabaloid, rotationally symmetric about the z axis

the intersection of the surface with the yz plane (x=0) is the parabola $z = y^2 - 4$ which has, as you say, infinitely many points.

I'm not too sure what the last part is asking for, maybe it is where the curve intersects the y & z axis...

I wanted to do the question myself so I only show a part of the question.
The original question is:
If the surface intersects the $xy, xz$, and $yz$ coordinate planes, find the equation of each of the parabolic boundary curves. Where do the curves intersect with the three coordinate axes?

I thought if the surface intersects the three coordinate planes, then the points of intersection must be $(0,0,0)$?

Thanks.

 

[lanedance]

I wanted to do the question myself so I only show a part of the question.
The original question is:
If the surface intersects the $xy, xz$, and $yz$ coordinate planes, find the equation of each of the parabolic boundary curves. Where do the curves intersect with the three coordinate axes?

I thought if the surface intersects the three coordinate planes, then the points of intersection must be $(0,0,0)$?

Thanks.

nope, that is not a point on the surface, and so none of the curves pass through that point

the easiest way to work out what i going on is to try & draw a picture, so i hope you'e attempted that. This problem is essentially showing you agood way to draw simple 3D surfaces, by drawing the intersection on each coordinate plane.

you should get 2 parabolas & and a circle

if you think about the shape of the surface and curves, the curve constrained by x = 0, will never intersect the x axis. So each point of intersection with axis will have 2 curves passing through it

due to the symmetry (which can save you some time) there will be one point of intersection on the z axis and 2 each for the x & y...

 

[madachi]

nope, that is not a point on the surface, and so none of the curves pass through that point

the easiest way to work out what i going on is to try & draw a picture, so i hope you'e attempted that. This problem is essentially showing you agood way to draw simple 3D surfaces, by drawing the intersection on each coordinate plane.

you should get 2 parabolas & and a circle

if you think about the shape of the surface and curves, the curve constrained by x = 0, will never intersect the x axis. So each point of intersection with axis will have 2 curves passing through it

due to the symmetry (which can save you some time) there will be one point of intersection on the z axis and 2 each for the x & y...

I have drawn the graph, it's a paraboloid. But I don't understand about the 2 parabolas and a circle.

I understand that the level curves are parabola and circle, but why the intersecting ones are only the specific 2 parabolas and 1 circle that you mentioned?

Thanks.

 

[lanedance]

ok, so what is the equation for each curve...

 

[madachi]

ok, so what is the equation for each curve...

$z = x^2 - 4$
$z = y^2 - 4$
$x^2 + y^2 = 4$

Are these correct?

 

[lanedance]

yep so what do the forms of those equations look like?

 

[madachi]

yep so what do the forms of those equations look like?

The first two are parabolas. The last one is circle.
So are the intersection points
1) $(-2,0,0),(2,0,0)$ for the first curve.
2) $(0,-2,0),(0,2,0)$ for the second curve.
3) $(0,0,-4)$ for the third curve mentioned above?

Thanks.

 

[lanedance]

The first two are parabolas. The last one is circle.
So are the intersection points
1) $(-2,0,0),(2,0,0)$ for the first curve.
2) $(0,-2,0),(0,2,0)$ for the second curve.
3) $(0,0,-4)$ for the third curve mentioned above?

Thanks.

the points look ok, but they correspond to an axis, not a specific curve, each will have 2 curves passing through it

 

[madachi]

the points look ok, but they correspond to an axis, not a specific curve, each will have 2 curves passing through it

I am confused. Could you show me the equation for one the curves so I can try to figure what you mean? Didn't I already show the equation of the 3 curves? Are they different from the ones that you just mentioned? Thanks.

 

[lanedance]

you've got it all, so pulling it together from your posts (though check it as i did pretty quick):
curve 1 - yz plane (x=0)
$z = y^2 - 4$
curve 2 - xz plane (y=0)
$z = x^2 - 4$
curve 3 - xy plane (z=0)
$x^2 + y^2 = 4$

axes intersection points
x axis - (-2, 0, 0) and (2,0,0) by curves 2&3,
y axis - (0 ,-2, 0) and (0,2,0) by curves 3&1,
z axis - (0 , 0, 4) by curves 1&2

 

[madachi]

you've got it all, so pulling it together from your posts (though check it as i did pretty quick):
curve 1 - yz plane (x=0)
$z = y^2 - 4$
curve 2 - xz plane (y=0)
$z = x^2 - 4$
curve 3 - xy plane (z=0)
$x^2 + y^2 = 4$

axes intersection points
x axis - (-2, 0, 0) and (2,0,0) by curves 2&3,
y axis - (0 ,-2, 0) and (0,2,0) by curves 3&1,
z axis - (0 , 0, 4) by curves 1&2

I see, thanks!