- #1
diffusion
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Homework Statement
Not sure how to approach this one, does anyone have an idea?
BTW, in the top line it should read xz-plane, not xy.
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- Thread starter diffusion
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In summary: At z0? Sorry, not sure if I really understand what you're asking. Are you asking for a point, the equation of the particular circle generated, or something else?I'm asking for the x,y,z equation(s) of the circle generated by the initial point (x0,0,z0). :smile:
Not sure how to approach this one, does anyone have an idea?
BTW, in the top line it should read xz-plane, not xy.
- #2
tiny-tim
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Hi diffusion! 
(try using the X2 tag just above the Reply box)
Forget all the stuff about the cylinder …
the question is simply to prove that if you rotate z = e-x2 about the z-axis, you get z = e-x2-y2 …
that's easy, isn't it?
(try using the X2 tag just above the Reply box
)Forget all the stuff about the cylinder …
the question is simply to prove that if you rotate z = e-x2 about the z-axis, you get z = e-x2-y2 …
that's easy, isn't it?
- #3
diffusion
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tiny-tim said:Hi diffusion!
(try using the X2 tag just above the Reply box
Forget all the stuff about the cylinder …
the question is simply to prove that if you rotate z = e-x2 about the z-axis, you get z = e-x2-y2 …
that's easy, isn't it?
I'm sure it's easy once I've been set in the right direction. I can see that z = e-(x2-y2) by graphing it, I just don't know how to show mathematically that it does.
- #4
tiny-tim
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ok … what does "rotate z = f(x) about the z-axis" mean? … how would you express the instruction as an equation?
- #5
diffusion
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tiny-tim said:ok … what does "rotate z = f(x) about the z-axis" mean? … how would you express the instruction as an equation?
Well you will rotate the function around the z-axis, so that z = f(x) becomes z = f(x,y). Not sure how to express it as an equation, besides telling you that the equation is a function of both x and y.
- #6
tiny-tim
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Try putting it in words first …
what happens to each individual point on the original curve?
- #7
diffusion
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tiny-tim said:
Produces a circle when rotated.
- #8
tiny-tim
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diffusion said:
Yup! And the equation of a circle is … ?
- #9
diffusion
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tiny-tim said:Yup! And the equation of a circle is … ?
Yep, I recognized this before I posted the question. Again I just don't know how to show it, or justify it. I suppose I would say something like
"Since each point on the graph z = e-r2 will generate a circle when rotated about the z-axis, and the equation for a circle is r2 = x2 + y2, we can make this substitution into our equation for r2, giving us z = e-(x2+y2)."
I don't know, somehow this justification seems vague and inadequate.
- #10
tiny-tim
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diffusion said:
ok, let's formalise that …
the point (x0,0,z0) generates the circle (x,y,z), where … ?
- #11
diffusion
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tiny-tim said:ok, let's formalise that …
the point (x0,0,z0) generates the circle (x,y,z), where … ?
At z0? Sorry, not sure if I really understand what you're asking. Are you asking for a point, the equation of the particular circle generated, or something else?
- #12
tiny-tim
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I'm asking for the x,y,z equation(s) of the circle generated by the initial point (x0,0,z0).
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