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sc0tt
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Background info for technical question:
I just started a summer Calculus III course and the introductory topic is graphing in 3-space (just a warning I tend to over think everything). Our assignment was to visualize the graph of
My math teacher and everyone else in the class drew a paraboloid, meaning the top was a flat circle not pointed. The teacher argued that there is a point where all the Z values would be the same and that's what gives the circular shape. I argued that keeping to scale, the shape has corners and that choosing an arbitrary Z-value is creating the picture you want, not the actual graph.
Technicalities:
Does this function have a circular top? I believe if you were to expand the shape, say to -4, 4 for x and y, a section with a delta Z would be added to the previous graph. If we were to approach infinity we would be adding more and more pieces that were shaped like before creating the corners. At the same time I realize this is adding sort of a 4th dimension, because all the points, whether you go to infinity or not, exist simultaneously and so the thought of this delta-Z shape wouldn't exist. Right?!
Also, the thought of this delta Z shape led me to think of what the 3d cross section of this shape would be. Would it be sort of like a frustum from a cone, or would it be the before mentioned shape i.e. the difference between a graph with x and y bounds of -3 and 3 vs. -4 and 4?
My Conclusion:
My conclusion is that everything is arbitrary, it doesn't matter whether you think of it as a circular top or corners because the shape goes to infinity so you could end it however you want.
For the 3d cross sections, its also arbitrary. I would think sticking to the essence of the shape it would be a weird shaped 3d cross section with the delta Z, not a frustum-like shape, it also depends on what you view this 3d cross section as.
Hopefully some doctor out there has some insight on my problem and that everything isn't arbitrary.
Thanks if you read this looonnng post,
-Scott
I just started a summer Calculus III course and the introductory topic is graphing in 3-space (just a warning I tend to over think everything). Our assignment was to visualize the graph of
Z = X[tex]^{2}[/tex] + y[tex]^{2}[/tex]
I plotted points in a table from -3, 3 for both x and y. I visualized this as a topographic table with the Z-values reaching out of my paper. This gives a shape that is similar to a bowl, but the corners stretch upwards (at 18 units) and the center at 0. That's what I drew.My math teacher and everyone else in the class drew a paraboloid, meaning the top was a flat circle not pointed. The teacher argued that there is a point where all the Z values would be the same and that's what gives the circular shape. I argued that keeping to scale, the shape has corners and that choosing an arbitrary Z-value is creating the picture you want, not the actual graph.
Technicalities:
Does this function have a circular top? I believe if you were to expand the shape, say to -4, 4 for x and y, a section with a delta Z would be added to the previous graph. If we were to approach infinity we would be adding more and more pieces that were shaped like before creating the corners. At the same time I realize this is adding sort of a 4th dimension, because all the points, whether you go to infinity or not, exist simultaneously and so the thought of this delta-Z shape wouldn't exist. Right?!
Also, the thought of this delta Z shape led me to think of what the 3d cross section of this shape would be. Would it be sort of like a frustum from a cone, or would it be the before mentioned shape i.e. the difference between a graph with x and y bounds of -3 and 3 vs. -4 and 4?
My Conclusion:
My conclusion is that everything is arbitrary, it doesn't matter whether you think of it as a circular top or corners because the shape goes to infinity so you could end it however you want.
For the 3d cross sections, its also arbitrary. I would think sticking to the essence of the shape it would be a weird shaped 3d cross section with the delta Z, not a frustum-like shape, it also depends on what you view this 3d cross section as.
Hopefully some doctor out there has some insight on my problem and that everything isn't arbitrary.
Thanks if you read this looonnng post,
-Scott