Parametrize the Curve of Intersection

[peroAlex]

Hi everyone!
I'm a student of electrical engineering. At my math class, we were given a problem to solve at home. Now, from what I've managed to gather, this is a trick question, but I would like to get someone else's opinion on the task. It's also worth mentioning that parametrization is a relatively new concept as it was only yesterday we were dealing with it.
1. The problem statement, all variables, and given/known data
Parametrize the curve of intersection of $x^2 + y^2 + z^2 = 1$ and $x - y = 0$.

Homework Equations

Pardon me, but I was unable to collect "relevant equations" in this section.

The Attempt at a Solution

$x^2 + y^2 + z^2 =1$ represents a sphere with radius 1, while $y = x$ represents a line parallel to x-axis. If I equate both sides and express for $z$ I acquire $$ z=-\sqrt{x^2-x+y^2+y-1} $$
I've plotted this function using Mathematica and it shows a plane. From this, I concluded that the intersection isn't a curve but a plane and cannot be parametrized.

I would like to ask for advice and maybe point me towards a solution. It seems a bit too trivial to hold.

-------------------------

I also tried plugging $x = y$ into sphere equation and simplifying for $y$. This returned a curve, "upside-down" parabola with the equation $$ y = \sqrt{\frac{1-x^2}{2}}$$
Could it be that parametrization follows from this procedure instead of the upper one?

I would like to thank in advance for your time and effort. Hope you're having a fantastic Saturday ;)

 

[andrewkirk]

$x^2 + y^2 + z^2 =1$ represents a sphere with radius 1, while $y = x$ represents a line parallel to x-axis.

The first statement is correct but not the second. The set of points satisfying the equation $y=x$ is a plane perpendicular to the x-y plane (or, if this is easier to visualise, a plane whose normal vector lies in the x-y plane) and passing through the line of points in the x-y plane that satisfy $y=x$.

It's not a trick question. It is unambiguous and the answer is quite simple when one finds it.

If I equate both sides and express for $z$ I acquire $$ z=-\sqrt{x^2-x+y^2+y-1} $$

That equation is not quite right. The signs have gone wrong somewhere. But rather than try to fix it, instead try to visualise the shape of the curve of intersection between the sphere and the plane. If you get the right image in your head, it'll be easy to parametrise using spherical coordinates.

 

[peroAlex]

The first statement is correct but not the second. The set of points satisfying the equation $y=x$ is a plane perpendicular to the x-y plane (or, if this is easier to visualise, a plane whose normal vector lies in the x-y plane) and passing through the line of points in the x-y plane that satisfy $y=x$.

It's not a trick question. It is unambiguous and the answer is quite simple when one finds it.

That equation is not quite right. The signs have gone wrong somewhere. But rather than try to fix it, instead try to visualise the shape of the curve of intersection between the sphere and the plane. If you get the right image in your head, it'll be easy to parametrise using spherical coordinates.

Now that's actually helpful. So if $y=x$ represents such plane, then the curve I'm looking for must be a circle! Is my assumption correct? Because if it's circle I'm after, then I could use $x = \cos{t}$, $y=\sin{t}$ and $z=\cos{t} + \sin{t}$ for $t \in [0,2\pi]$.

Please correct me if I'm wrong on this one. Also, thank you so much for going through my post and helping me.

 

[Mark44]

I also tried plugging $x = y$ into sphere equation and simplifying for $y$. This returned a curve, "upside-down" parabola with the equation $$ y = \sqrt{\frac{1-x^2}{2}}$$

You have two mistakes here.
1. If you do the substitution you describe, you should get either $2x^2 + z^2 = 1$ or $2y^2 + z^2 = 1$, depending on whether you substitute for x or for y.
2. The second equation can be solved for y to get $y = \pm \sqrt{ \frac{1 - z^2}{2}}$. The graph of this equation is not a parabola.

Now that's actually helpful. So if $y=x$ represents such plane, then the curve I'm looking for must be a circle! Is my assumption correct? Because if it's circle I'm after, then I could use $x = \cos{t}$, $y=\sin{t}$ and $z=\cos{t} + \sin{t}$ for $t \in [0,2\pi]$.

You didn't show how you got this, but I don't think it's right. The curve of intersection between the sphere and the plane is a circle, but it's oriented so that its axis is parallel to the x-y plane, and in the same direction as the line y = -x in the x-y plane. If you project this circle onto either the x-z plane or y-z plane, what you get are ellipses.

 

[andrewkirk]

if $y=x$ represents such plane, then the curve I'm looking for must be a circle! Is my assumption correct? Because if it's circle I'm after, then I could use $x = \cos{t}$, $y=\sin{t}$ and $z=\cos{t} + \sin{t}$ for $t \in [0,2\pi]$.

As Mark pointed out, it is indeed a circle, but not the one given by those coordinates.

For me the easiest way to do it is to first parametrise the circle using a new set of planar coordinates in the plane in which it lies. You can use the existing z coordinate for the vertical axis, plus another coordinate, say u, for the horizontal axis, and t for your parameter variable. That will give you formulas of z and u in terms of t. Then all you need to do is express x and y in terms of u and you'll have a parametrisation that expresses x, y and z in terms of s.