parametric representation of a curve

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October 8th 2008, 10:03 AM
superfets

parametric representation of a curve

Find a parametric representation for the curve x^2 + y + z = 2 and xy + z = 1.
October 8th 2008, 12:52 PM
Opalg

Quote:

Originally Posted by superfets

Find a parametric representation for the curve x^2 + y + z = 2 and xy + z = 1.

This looks a bit tricky, because the curve splits into two separate pieces. If you subtract the sescond equation from the first then you get $x^2+y-xy=1$ , which factorises as $(x-1)(1+x-y)=0$ .

So the curve consists of the two pieces $x=1,\ y+z=1$ and $y=x+1,\ z=1-x-x^2$ . It's easy enough to parametrise the two pices separately, and that is presumably the best that can be done.
October 8th 2008, 01:21 PM
superfets

thanks for the effort

Don't think that's what i'm looking for. The first eq'n is a surface (parabolic cylinder) and the second eq'n is a plane. The intersection of the surface and the plane in 3 dimensions will give you a curve. It can't be in 2 pieces. I need to describe this curve by the three parametric functions. i.e. C; x = x(t) , y = y(t) , z = z(t). But thanks for trying.
October 8th 2008, 03:26 PM
mr fantastic

Quote:

Originally Posted by superfets

Find a parametric representation for the curve X^2 + y + z = 2 and xy + z = 1.

There's an infinite number of possible answers. One possible answer:

Let x = t.

t is the parameter.

Now solve for y and z simulataneously in terms of t:

y + z = 2 - t^2 .... (1)

ty + z = 1 .... (2)

You can think about the possible values t can have ......
October 9th 2008, 12:01 AM
Opalg

Quote:

Originally Posted by superfets

Don't think that's what i'm looking for. The first eq'n is a surface (parabolic cylinder) and the second eq'n is a plane. The intersection of the surface and the plane in 3 dimensions will give you a curve. It can't be in 2 pieces.

The second equation is certainly not a plane. The equation xy + z = 1 is not linear. It represents a hyperbolic surface which contains lines, including the line parametrised by (x,y,z) = (1,t,1-t) that I indicated in my previous comment.

You can also see this by following mr fantastic's method and paying particular attention to his final paragraph (look carefully at what happens at the particular value t=1 in his parametrisation).