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

In summary: 1 & 2y axis - (0,-2,0) and (0,2,0) by curves... 2 & 1z axis - (0,0,-4) by curve... 3hope that clears it up a bityou'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 pointsx axis - (-
  • #1
madachi
29
0

Homework Statement



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

The Attempt at a Solution


[itex]yz-[/itex]plane, so [itex] x=0 [/itex]

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

Are 1) and 2) correct? Thanks in advance!
 
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  • #2
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?
 
  • #3
Is it [itex] (0,0,-4) [/itex] ? But I am slightly confused, doesn't the whole curve intersect yz-plane and so there are infinitely many points of intersection?

Thanks.
 
  • #4
the graph of [itex] z = x^2 + y^2 - 4 [/itex], is a parabaloid, rotationally symmetric about the z axis

the intersection of the surface with the yz plane (x=0) is the parabola [itex] z = y^2 - 4 [/itex] 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...
 
  • #5
lanedance said:
the graph of [itex] z = x^2 + y^2 - 4 [/itex], is a parabaloid, rotationally symmetric about the z axis

the intersection of the surface with the yz plane (x=0) is the parabola [itex] z = y^2 - 4 [/itex] 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 [itex]xy, xz [/itex], and [itex] yz [/itex] 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 [itex] (0,0,0) [/itex]?

Thanks.
 
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  • #6
madachi said:
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 [itex]xy, xz [/itex], and [itex] yz [/itex] 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 [itex] (0,0,0) [/itex]?

Thanks.
nope, that is not a point on the surface, and so none of teh 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...
 
  • #7
lanedance said:
nope, that is not a point on the surface, and so none of teh 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.
 
  • #8
ok, so what is the equation for each curve...
 
  • #9
lanedance said:
ok, so what is the equation for each curve...

[itex] z = x^2 - 4 [/itex]
[itex] z = y^2 - 4 [/itex]
[itex] x^2 + y^2 = 4 [/itex]

Are these correct?
 
  • #10
yep so what do the forms of those equations look like?
 
  • #11
lanedance said:
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) [itex] (-2,0,0),(2,0,0) [/itex] for the first curve.
2) [itex] (0,-2,0),(0,2,0) [/itex] for the second curve.
3) [itex](0,0,-4) [/itex] for the third curve mentioned above?

Thanks.
 
  • #12
madachi said:
The first two are parabolas. The last one is circle.
So are the intersection points
1) [itex] (-2,0,0),(2,0,0) [/itex] for the first curve.
2) [itex] (0,-2,0),(0,2,0) [/itex] for the second curve.
3) [itex](0,0,-4) [/itex] for the third curve mentioned above?

Thanks.

teh points look ok, but they correspond to an axis, not a specific curve, each will have 2 curves passing through it
 
  • #13
lanedance said:
teh 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.
 
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  • #14
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)
[itex] z = y^2 - 4 [/itex]
curve 2 - xz plane (y=0)
[itex] z = x^2 - 4 [/itex]
curve 3 - xy plane (z=0)
[itex] x^2 + y^2 = 4 [/itex]

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
 
  • #15
lanedance said:
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)
[itex] z = y^2 - 4 [/itex]
curve 2 - xz plane (y=0)
[itex] z = x^2 - 4 [/itex]
curve 3 - xy plane (z=0)
[itex] x^2 + y^2 = 4 [/itex]

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!
 

1. What is the equation for the intersecting yz-plane of the curve x^2 + y^2 - 4?

The equation for the intersecting yz-plane of the curve x^2 + y^2 - 4 is simply y^2 - 4 = 0. This is because the x^2 term drops out when the yz-plane is considered.

2. What is the shape of the curve x^2 + y^2 - 4?

The curve x^2 + y^2 - 4 is a circle with center at the origin and a radius of 2 units. This can be seen by simplifying the equation to (x-0)^2 + (y-0)^2 = 2^2.

3. At what points does the intersecting yz-plane intersect the curve x^2 + y^2 - 4?

The intersecting yz-plane will intersect the curve x^2 + y^2 - 4 at two points: (0,2) and (0,-2). This can be found by solving the equation y^2 - 4 = 0 for y.

4. What is the significance of the intersecting yz-plane in relation to the curve x^2 + y^2 - 4?

The intersecting yz-plane can be thought of as a cross-section of the curve x^2 + y^2 - 4. It shows the points where the curve intersects the yz-plane, and can be used to visualize the shape of the curve in three-dimensional space.

5. How can the intersecting yz-plane of x^2 + y^2 - 4 be graphed?

The intersecting yz-plane of x^2 + y^2 - 4 can be graphed by first plotting the curve on the xy-plane. Then, for each point on the curve, plot a corresponding point on the yz-plane at the same x-value. This will result in a circle on the yz-plane with center at the origin and a radius of 2 units.

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