What is the vector function for the intersection of a cone and a plane?

[Yae Miteo]

Homework Statement

"Find a vector function that represents the curve of intersection of the two surfaces."

Homework Equations

Cone: $$z = \sqrt{x^2 + y^2}$$ Plane: $$z = 1+y$$

The Attempt at a Solution

I began by setting $$x=cos t$$, so that $$y = sin t$$ and $$z = 1+sin t$$. At this point, however, I am stuck. I think I need to set something equal to something else, but I am not sure what.

 

[D_Tr]

[strike]You have 2 equations in 3 variables and you want to get 3 equations in 4 variables. This means you can only introduce one new equation where your new variable will appear. Otherwise you will have 4 equations in 4 variables and you no longer have the one degree of freedom required to get a curve. Here you have introduced two new arbitrary equations. Try just setting x = t and then try to express y and z in terms of just t.[/strike] Sorry, I thought you arbitrarily set y = sint, but it seems that you just made a calculation mistake (because you write "so that y = sint", which is wrong). It will be much easier if you just set x = t and then you solve y and z in terms of x (=t).

 

[HallsofIvy]

Do you have any particular reason for setting $x= cos(t)$ and $y= sin(t)$?
Surely, not from the first equation?

If you set $x= cos(t)$ and $y= sin(t)$ then you are saying that $$z= \sqrt{sin^2(t)+ cos^2(t)}= 1$$ for all t- and that is NOT true.

The obvious thing to do, since $z= \sqrt{x^2+ y^2}$ and $z= 1+ y$, is to set $x+ y= \sqrt{x^2+ y^2}$. Then, after squaring both sides, $1+ 2y+ y^2= x^2+ y^2$ so that $2y= x^2- 1$ and $y= (1/2)x^2- (1/2)$.

Now, let x= t as the parameter.

 

[Yae Miteo]

Awesome! Thank-you.

 

[LCKurtz]

Nevermind my previous post (deleted).