## Monday, November 22, 2010

### The intersecting sphere and cylinder

The bonus task I gave you last week involved looking at the curve created by the intersection of a sphere and cylinder. For purposes of this demonstration, I will assume that the sphere has radius 4 and the cylinder correspondingly has diameter 4. Thus the sphere is x2 + y2 + z2 = 16 and the cylinder is (x − 2)2 + y2 = 4.

We will now show that the self-intersection curve crosses itself at right angles at the point (4, 0, 0).

Here is one way to do it:

We parameterize the path by r = xi + yj + zk, where we need to find appropriate functions for x, y, and z. Let's rewrite everything in cylindrical coordinates, because that form is more useful in this situation. The sphere's equation is r2 + z2 = 16, the cylinder is r = 4 cos θ, and the intersection curve we are looking for is r = (r cos θ)i + (r sin θ)j + zk.

Since the cylinder requires that r = 4 cos θ, we can substitute for r in the expression for the curve r. We obtain r = (4 cos θ cos θ)i + (4 cos θ sin θ)j + zk.

What can we do about that z? According to the sphere, we must have z = (16 − r2)1/2. Substituting r = 4 cos θ into the equation for z, we get z = (16 − (4 cos θ)2)1/2 = (16 − 16 cos2 θ)1/2 = 4 sin θ. We can now write our curve with θ as our parameter:

r(θ) = (4 cos θ cos θ)i + (4 cos θ sin θ)j + (4 sin θ)k.

Observe that r(0) = 4i and r(π) = 4i, indicating that the curve intersects itself for θ = 0 and θ = π. If we find the tangent vectors at these two points, the angle between them is supposed to be a right angle. Of course, the easiest way to find tangent vectors is to use r′(θ), the velocity vector:

r′(θ) = (−8 cos θ sin θ)i + 4( cos2 θ − sin2 θ)j + (4 cos θ)k.

Plug in θ = 0 and θ = π to find the pertinent tangent vectors:

r′(0) = 4j + 4k
r′(π) = 4j − 4k.

If we dot these vectors we get zero. The vectors are perpendicular, so the intersection curve crosses itself at ninety degrees. QED.