*x*

^{2}+

*y*

^{2}+

*z*

^{2}= 16 and the cylinder is (

*x*− 2)

^{2}+

*y*

^{2}= 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**=

*x*

**i**+

*y*

**j**+

*z*

**k**, 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

*r*

^{2}+

*z*

^{2}= 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**+

*z*

**k**.

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**+

*z*

**k**.

What can we do about that

*z*? According to the sphere, we must have

*z*= (16 −

*r*

^{2})

^{1/2}. Substituting

*r*= 4 cos

*θ*into the equation for

*z*, we get

*z*= (16 − (4 cos

*θ*)

^{2})

^{1/2}= (16 − 16 cos

^{2}

*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) = 4

**i**and

**r**(

*π*) = 4

**i**, 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( cos

^{2}

*θ*− sin

^{2}

*θ*)

**j**+ (4 cos

*θ*)

**k**.

Plug in

*θ*= 0 and

*θ*=

*π*to find the pertinent tangent vectors:

**r′**(0) = 4

**j**+ 4

**k**

**r′**(

*π*) = 4

**j**− 4

**k**.

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