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July 28th, 2017, 06:01 AM   #1
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parameterization of curve

I need to parameterize the curve K, which is intersection between P: z=sqrt(x^2+y^2) z<or equal 2 and x=1.

And also the curve x^2+y^2-z^2=1, x+y+z=0. And I don't have idea where to start.

Last edited by skipjack; July 28th, 2017 at 08:18 AM.
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July 28th, 2017, 07:37 AM   #2
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The intersection of the surface $\displaystyle z= \sqrt{x^2+ y^2}$ and the plane x= 1 is, of course, $\displaystyle z= \sqrt{y^2+ 1}$. I would think that x= 1, y= t, $\displaystyle z= \sqrt{t^2+ 1}$ would be an obvious parameterization.
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July 29th, 2017, 04:54 AM   #3
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Quote:
Originally Posted by Country Boy View Post
The intersection of the surface $\displaystyle z= \sqrt{x^2+ y^2}$ and the plane x= 1 is, of course, $\displaystyle z= \sqrt{y^2+ 1}$. I would think that x= 1, y= t, $\displaystyle z= \sqrt{t^2+ 1}$ would be an obvious parameterization.
I did that too. But I was not sure about it cos I later got some weird derivatives. Thank you.
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July 29th, 2017, 05:28 AM   #4
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I did that too. But I was not sure about it cos I later got some weird derivatives. Thank you.
If you tell us the weird derivatives, we can tell you whether they are correct or not!
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July 29th, 2017, 06:29 AM   #5
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If you tell us the weird derivatives, we can tell you whether they are correct or not!
I used y by mistake it should be t.
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July 29th, 2017, 06:59 AM   #6
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Can you provide clear images or type your work?
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July 30th, 2017, 12:44 AM   #7
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Can you provide clear images or type your work?
I hope this is better.
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July 30th, 2017, 03:37 AM   #8
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You didn't say anything about differentiating in your first post. What, exactly, does the exercise ask you to do?

Last edited by Country Boy; July 30th, 2017 at 03:39 AM.
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July 30th, 2017, 04:48 AM   #9
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You didn't say anything about differentiating in your first post. What, exactly, does the exercise ask you to do?
First parameterization and then to find flexion of the parametrized curve.
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