Parameterizing A Circle Projected onto a Plane

  • #1

Homework Statement


Find a vector function that parameterizes a curve C which lies in the plane x-y+z=2 and directly above the circle x2 + (y-1)2 = 9

The Attempt at a Solution

So, in order to parameterize the circle, I simply use x=cos(t), y = sin(t) with some adjustments. Namely, I let x=3cos(t) and y=3sin(t)+1. Then, to determine how the z-coordinates of this curve change according to x,y I use the definition of the plane. So I have z=2-x+y, which I convert using my parametric defintions for x,y.

I have, z=3-3cos(t)+3sin(t). Therefore, my parameterized curve is,

<3cos(t),3sin(t)+1,3-3cos(t)+3sin(t)>

Right?
 

Answers and Replies

  • #2
I like Serena
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Looks good.
But... is it always above the circle?
 
  • #3
So I just drew myself a little sketch. Basically I have plane and, beneath it, a cylinder which extends upwards, whose base represents the circle. The intersection of the cylinder and the plane should be the projection of the circle onto the plane, right? In this case, I can't see how the projection could be anywhere else besides over the circle.

As an aside, the vector projection of a onto b is given as follows:

((b [dot] a)/|a|) * a/|a|.

Is it possible to use the idea of projections along with dot products to solve this problem?
 
  • #4
I like Serena
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So I just drew myself a little sketch. Basically I have plane and, beneath it, a cylinder which extends upwards, whose base represents the circle. The intersection of the cylinder and the plane should be the projection of the circle onto the plane, right? In this case, I can't see how the projection could be anywhere else besides over the circle.
Consider t=-pi/4.
Is the corresponding point above or below the circle?


As an aside, the vector projection of a onto b is given as follows:

((b [dot] a)/|a|) * a/|a|.

Is it possible to use the idea of projections along with dot products to solve this problem?
This is the formula for projection onto a line.
It doesn't work for projection onto a plane.

Anyway, you still need a parametrization.
Your method is just fine.
The only catch is that the curve is not entirely above the circle (assuming the problem does make this fine distinction).
 
  • #5
OH! I see what you mean now. So presumably I need only calculate a z-coordinate for above/below orientation:

z=3-3cos(t)+3sin(t), where t=-pi/4
z=3-3sqrt(2).

So I understand that this point is below the xy-plane, but I don't understand how this impacts the parametrization. Does it change how I am allowed to define t? If the intersection of my cylinder with the plane is still the projection, what difference does "above" or "below" make?
 
  • #6
I like Serena
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Well, as I interpret the problem (and I may be wrong), it asks for a parametrization that is "above" the circle.
So I think you're supposed to limit the range of t, such that the curve is indeed "above" the circle.
But perhaps I'm being too nit-picky. You tell me.
Still, it is good practice...
 

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