I have an assignment question to find an equation of the orthogonal projection onto the XY plane of the curve of intersection of twp particular functions.
If some one knows of a good web page that might explain this to me I would be greatly appreciate it.
If you can write the equation of the curve of intersection as parametric equations, just set z= 0. That is, if the curve is given by x= f(t), y= g(t), z= h(t), its projection onto the xy-plane is x= f(t), y= g(t), z= 0.
Those are NOT parametric equations for the two surfaces. Since they involve only a single parameter they are equations of curves, not surfaces.
The two surfaces are [itex]z= y^2[/itex] and [itex]z= 4- x^2- y^2[/itex]. In this particular case, you don't need parametric equations at all. You can just eliminate z from the two equations to get the projection onto the xy plane.
In your previous post you projected the surfaces first (you should have gotten (x,y,0) for both surfaces btw) and then intersected their projection. That doesn't work. You have to intersect first and then project.
That's exactly what you did in your latest post. Usually you would have to set z = 0 to get the projection. However, since we already have a description of just the x- and y-coordinates of the intersection that does not depend on z, we can skip this step.
By the way, what does the answer look like geometrically?
No, the projection of the curve of intersection of two surfaces into the xy-plane is a curve in the xy-plane, not a surface.
Your first surface, [itex]z= y^2[/itex] is a "parabolic cylinder", your second surface is, [itex]z= 4- x^2- y^2[/itex] is an "elliptic paraboloid" (or, more correctly, a "circular paraboloid". The intersection of the two is given by [itex]y^2= 4- x^2- y^2[/itex] or [itex]x^2+ 2y^2= 4[/itex], an ellipse.
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