Parametric Equations Homework - Find Curve & Position Vector

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The discussion focuses on finding parametric equations for the curve formed by the intersection of the cylinders defined by x^2+y^2=4 and z+x^2=4. A suggested approach is to use the angular coordinate in the x,y plane, setting x=2*cos(t) and y=2*sin(t), which simplifies the parametrization. The correct position vector is derived as r = <2cos(t), 2sin(t), 4-4cos^2(t)>, correcting earlier mistakes in the calculations. The conversation highlights the importance of visualizing the curve and understanding the relationship between the variables. Overall, the participants emphasize the need for clarity in defining the parametric equations.
chrsr34
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Homework Statement


Consider the curve of intersection of the cylinders [x^2+y^2=4] and [z+x^2=4]. Find parametric equations for this curve and use them to write a position vector.

Homework Equations


Thats what I am looking for. What to set t equal to.


The Attempt at a Solution


I set t=x and got a square root for y. So if i set t=x^2, i get rid of the square root for y, but I am not sure if this is correct. i really don't know the rules for parametric equations. Any input is appreciated.

Chris
 
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An easy choice for t is the angular coordinate in the x,y plane. So x=2*cos(t), y=2*sin(t). Given this, can you figure out what z is in terms of t? Unfortunately, I don't think there are any general rules for doing this. You just have to make a picture in your mind of what the curve looks like and then look for a choice for t.
 
I see. Yes i could find z with those coordinates. But is there a reason you chose those? Are those cylindrical coordinates? (im a little rusty).
Is there anything wrong with setting t=x^2?
 
The x^2+y^2=4 cylinder intersects the x,y plane in a circle. It's easy to parametrize that in polar coordinates. If you take t=x^2 then you still have a square root for y (contrary to what you said), so you'd have to define the curve as a union of pieces.
 
All yes your right.
So my position vector should be: r = < 4cos^2(t), 4sin^2(t), 4-4cos^2(t) >
Is this correct?
 
Noooo. x=2*cos(t), not 4cos^2(t)!
 
yea i just realized i did that.
So, r = < 2cos(t), 2sin(t), 4-4cos^2(t) >

This look good?
 

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