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[diffusion]

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.

 

[tiny-tim]

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?

 

[diffusion]

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.

 

[tiny-tim]

ok … what does "rotate z = f(x) about the z-axis" mean? … how would you express the instruction as an equation?

 

[diffusion]

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.

 

[tiny-tim]

Try putting it in words first …

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

 

[diffusion]

Try putting it in words first …

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

Produces a circle when rotated.

 

[tiny-tim]

Produces a circle when rotated.

Yup! And the equation of a circle is … ?

 

[diffusion]

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.

 

[tiny-tim]

"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 … ?

 

[diffusion]

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?

 

[tiny-tim]

I'm asking for the x,y,z equation(s) of the circle generated by the initial point (x₀,0,z₀).