Problem:
Just a few comments for myself here:
I keep blocking myself on this question for lacking a good 3D plotting mechanism, but looking backwards, I really should not. All it asks for is a sketch, and I should just move forward by describing it.
Doing all the problems in the book is really not an efficient way of learning. Need to figure out something else.
Solution:
4a) It is the unit sphere.
4b) It is an infinitely long cylinder with radius 1, aligning with the z-axis
4c) It is a single point (-2. 1.5, 0)
4d) Given the hint, the factorized expression is x(z2−y). It is zero only if either (or both) factor are zero, so it is the union of the whole x = 0 plane and the y=z2 parabola curve with any x coordinate.
As per the usual right-handed coordinate convention, the x-axis point towards the screen, the whole screen as the x-coordinate plane is included, and the line extends parallel in and out the screen perpendicularly.
It is like a infinitely long trough with a cut by an infinitely large perpendicular plane.
Using k3dsurf, here is a plot of the affine variety. The tool is really useful.
4e) Having the factoring trick, now I can factor the expressions.
x4−zx=x(x3−z)
x3−yx=x(x2−y)
This is a even more funny shape. The whole x plane is a zero for it. If x is not zero, then z=x3 and y=x2.
It should be a curve that look like this (-2, 4, -8), (-1, 1, -1) (0, 0, 0), (1, 1, 1), (2, 4, 8), ...
A really hard to visualize 3D curve.
4f) The intersection of the two unit sphere centered at (0, 0, 0) and (0, 0, 1). Intuitively, that should be a circle.
Consider a general point (x, y, z) such that both expression is true We have
x2+y2+z2=1 and x2+y2+(z−1)2=1
Subtracting, we get z2−(z−1)2=0, that gives z=0.5.
Subsituting that into the central unit sphere, we get x2+y2=0.75, so that indeed gives a circle!
Dimension-wise, a surface intuitively has dimension 2 as it can be parameterized with two parameters. (4e) Would be an interesting case, as it is a union of a plane with a curve, so it is kind of hard to tell what dimension is it.
Just a few comments for myself here:
I keep blocking myself on this question for lacking a good 3D plotting mechanism, but looking backwards, I really should not. All it asks for is a sketch, and I should just move forward by describing it.
Doing all the problems in the book is really not an efficient way of learning. Need to figure out something else.
Solution:
4a) It is the unit sphere.
4b) It is an infinitely long cylinder with radius 1, aligning with the z-axis
4c) It is a single point (-2. 1.5, 0)
4d) Given the hint, the factorized expression is x(z2−y). It is zero only if either (or both) factor are zero, so it is the union of the whole x = 0 plane and the y=z2 parabola curve with any x coordinate.
As per the usual right-handed coordinate convention, the x-axis point towards the screen, the whole screen as the x-coordinate plane is included, and the line extends parallel in and out the screen perpendicularly.
It is like a infinitely long trough with a cut by an infinitely large perpendicular plane.
Using k3dsurf, here is a plot of the affine variety. The tool is really useful.
4e) Having the factoring trick, now I can factor the expressions.
x4−zx=x(x3−z)
x3−yx=x(x2−y)
This is a even more funny shape. The whole x plane is a zero for it. If x is not zero, then z=x3 and y=x2.
It should be a curve that look like this (-2, 4, -8), (-1, 1, -1) (0, 0, 0), (1, 1, 1), (2, 4, 8), ...
A really hard to visualize 3D curve.
4f) The intersection of the two unit sphere centered at (0, 0, 0) and (0, 0, 1). Intuitively, that should be a circle.
Consider a general point (x, y, z) such that both expression is true We have
x2+y2+z2=1 and x2+y2+(z−1)2=1
Subtracting, we get z2−(z−1)2=0, that gives z=0.5.
Subsituting that into the central unit sphere, we get x2+y2=0.75, so that indeed gives a circle!
Dimension-wise, a surface intuitively has dimension 2 as it can be parameterized with two parameters. (4e) Would be an interesting case, as it is a union of a plane with a curve, so it is kind of hard to tell what dimension is it.
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