Vector Function of Cone & Plane Intersection Curve

[Litcyb]

Homework Statement

Find a vector function that represents the curve of intersection of the two surfaces:
The cone z = sqrt( x^2 + y^2) and the plane z = 1+y.

Homework Equations

z = sqrt( x^2 + y^2) and the plane z = 1+y.

The Attempt at a Solution

This problem can be solved as following using x as the parameter.
x^2+y^2 = z^2 = (1+y)^2 = 1+2y+y^2. => x^2 = 1 + 2y.

x=t; y = (t^2-1)/2; z = 1+(t^2-1)/2 = (t^2+1)/2

My question is, what if we use y as the parameter,

i get ,

y=t, x=(2t+1)^(1/2) z=t+1,

is this answer also correct?

 

[Dick]

Homework Statement

Find a vector function that represents the curve of intersection of the two surfaces:
The cone z = sqrt( x^2 + y^2) and the plane z = 1+y.

Homework Equations

z = sqrt( x^2 + y^2) and the plane z = 1+y.

The Attempt at a Solution

This problem can be solved as following using x as the parameter.
x^2+y^2 = z^2 = (1+y)^2 = 1+2y+y^2. => x^2 = 1 + 2y.

x=t; y = (t^2-1)/2; z = 1+(t^2-1)/2 = (t^2+1)/2

My question is, what if we use y as the parameter,

i get ,

y=t, x=(2t+1)^(1/2) z=t+1,

is this answer also correct?

Sure, with the proviso that it only represents part of the curve. For example, points with negative values of x won't appear in the second parametrization (where you also should specify t>=(-1/2)). You'd need the x=(-(2t+1)^(1/2)) solution as well to get them all.