Stuck on this problem

  • Thread starter diffusion
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  • #1
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Homework Statement


problem.jpg


Not sure how to approach this one, does anyone have an idea?

BTW, in the top line it should read xz-plane, not xy.
 

Answers and Replies

  • #2
tiny-tim
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Hi diffusion! :smile:

(try using the X2 tag just above the Reply box :wink:)

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? :wink:
 
  • #3
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Hi diffusion! :smile:

(try using the X2 tag just above the Reply box :wink:)

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? :wink:

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? :smile:
 
  • #5
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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? :smile:

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
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Try putting it in words first …

what happens to each individual point on the original curve?​

Produces a circle when rotated.
 
  • #8
tiny-tim
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Produces a circle when rotated.

Yup! And the equation of a circle is … ? :smile:
 
  • #9
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Yup! And the equation of a circle is … ? :smile:

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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"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)."

ok, let's formalise that …

the point (x0,0,z0) generates the circle (x,y,z), where … ? :smile:
 
  • #11
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ok, let's formalise that …

the point (x0,0,z0) generates the circle (x,y,z), where … ? :smile:

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). :smile:
 

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