I asked this question yesterday and some one was nice enough to try to help me with it but after further review of the answer the function that was arived at did not seem to work when checking points.
the two functions that i need to find a vector valued function for where they intersect are
$4x^2+4y^2+z^2=16$
and
$x=z^2$
the solution that I was helped to arive at subbed in the x for the $z^2$ but then added it to the $4x^2$ but since their powers don't agree I don't belive that i can do that. I know that the first equation is an elipse and that the second equation is a cylinder that looks like a folded sheet of paper, so i'm thinking that i need to use
$x=\rho sin(\Phi)cos(\Theta)$
$y=\rho sin(\Phi)sin(\Theta)$
$z=\rho cos(\Phi)$
$\rho=x^2+y^2+z^2$
my answer needs to be in the form of a vector valued function. I know that it will be a line that bounds the ellipse where it is intersected but I don't know how to get there. I can't use the equation for $\rho$ since i can't get the 4's out of the left side without putting the z over 4 and I've tried moving the z, dividing by 4 and setting $(16+z^2)/4$ equal to $r^2$ but i just end up with a mess and no way to get rid of the z. I'm willing to try any suggestions.
Best Answer
If you substitute $x=z^2$ into your other equation you get
$$4z^4+4y^2+z^2=16$$
Solving for $y$,
$$y=\pm\frac 12\sqrt{16-z^2-4z^4}$$
The expression inside the square root must be nonnegative, which means that
$$-1.37073\approx-\sqrt{\frac{-1+\sqrt{257}}8}\le z\le\sqrt{\frac{-1+\sqrt{257}}8}\approx 1.37073$$
So here are two vector-valued functions that together describe the intersection of your two surfaces, given that $z$ is in that interval above.
$$f_1(z)=\left(z^2\right)\hat i +\left(\frac 12\sqrt{16-z^2-4z^4}\right)\hat j+z\hat k$$ $$f_2(z)=\left(z^2\right)\hat i -\left(\frac 12\sqrt{16-z^2-4z^4}\right)\hat j+z\hat k$$
Does this meet your needs? (If you look at my previous edits of this answer you will find another solution where $y$ is the parameter: this current answer is somewhat simpler.)