How do I parameterize the intersection of these two surfaces?

[slr77]

Homework Statement

Parameterize the curve of intersection of the two surfaces:

x^2+y^2+z^2=14
z=y^2-x^2

Homework Equations

The Attempt at a Solution

I tried manipulating the equations above but can't seem to get a nice parameterization which I can use to do the rest of the (calculus) problem.

 

[DEvens]

Maybe if you start by understanding the nature of the intersecting surfaces. The first one is what kind of surface? (That's the easy one.) The second one is what kind of surface? So what should the intersection look like? For the second surface you may get someplace by considering a small set of values of z, and seeing what that implies for x and y.

If you are having trouble with understanding those surfaces, try "rounding up the usual suspects." For each surface:
- Is the surface bounded? Or can you have the coordinate values go to infinity?
- Does the surface intersect the origin?
- Does the surface intersect the x=0 plane? The y=0 plane? The z=0 plane? If it does, what does that intersection look like?

 

[slr77]

The first one is a sphere of radius sqrt(14) and and the second is a hyperbolic paraboloid. I even grahed both surfaces before posting this question and am looking at the curve of intersection right now. But I can't figure out what the parameterization should be. I tried look at the intersection on the xz, xy, yz planes but nothing is jumping out at me.

 

[Ray Vickson]

Homework Statement

Parameterize the curve of intersection of the two surfaces:

x^2+y^2+z^2=14
z=y^2-x^2

Homework Equations

The Attempt at a Solution

I tried manipulating the equations above but can't seem to get a nice parameterization which I can use to do the rest of the (calculus) problem.

You could try solving for $x,y$ in terms of $z$. Writing your two equations as
\begin{array}{rcc}x^2 + y^2 &amp;=&amp; 14 - z^2\\<br /> -x^2 + y^2 &amp;= &amp;z \end{array}
allows an easy solution for $x^2$ and/or $y^2$. Those will give two roots each for $x$ and $y$, for a total of four $(x,y)$ pairs, each given in terms of $z$. They will probably correspond to two separate curves ("branches"), with each branch having two formulas: one for positive and one for negative values of $x$ or $y$.

 

[slr77]

You could try solving for $x,y$ in terms of $z$. Writing your two equations as
\begin{array}{rcc}x^2 + y^2 &amp;=&amp; 14 - z^2\\<br /> -x^2 + y^2 &amp;= &amp;z \end{array}
allows an easy solution for $x^2$ and/or $y^2$. Those will give two roots each for $x$ and $y$, for a total of four $(x,y)$ pairs, each given in terms of $z$. They will probably correspond to two separate curves ("branches"), with each branch having two formulas: one for positive and one for negative values of $x$ or $y$.

The full problem requires me to compute r'(t) = <x'(t), y'(t),z'(t)> (well actually I just need a tangent vector to the curve of intersection at a certain point but this is the only way I can think of to compute it) so this parameterization seems too complicated. the closest I was able to get was by solving this:

x^2 + y^2 + z^2 - 14 = y^2 - x^2-z
2*x^2 + (z+1/2)^2 = 57/6

But this also results in a overly complicated parameterization. So it seems that there must be some other way to do this problem without getting a parameterization because these would result in a very hairy problem and I don't think it's meant to be...

 

[slr77]

sorry double post

 

[slr77]

I finally realized how to do it and it was a case of severe tunnel vision as I suspected. I just have to take the gradient of the two level surfaces at the specified point and cross them in order to get the tangent vector. Parameterizing the curve of intersection and computing the tangent vector is not necessary.