parametric equation for curve of intersection of two cylinders

Question

I'm struggling to find a way to express the curve of intersection of two surfaces.

Context: this is for a university assignment about the MuPAD CAS system. It's a correspondence university and I'm not getting very prompt responses from the lecturer. I have asked him, but I'm asking here as well in an attempt to get an answer sooner.

I have the equations of two surfaces:

$$ 9(y+1)^2+9(z+4)^2=16\\ 4(x-1)^2+4(z+4)^2=25 $$

Now I've realized the surfaces are two cylinders and I've plotted them via the parametric equations:

$$ x=u,\;y=\frac{4}{3}\cos(v)-1,\;z=\frac{4}{3}sin(v)-4\\ x=\frac{5}{2}cos(u)+1,\;y=v\;z=\frac{5}{2}sin(u)-4 $$

Now I can do the following:

$$ 9(y+1)^2+9(z+4)^2=16\\ \Rightarrow (y+1)^2+(z+4)^2=\frac{16}{9}\\ \Rightarrow (z+4)^2=\frac{16}{9}-(y+1)^2\\ $$

and

$$ 4(x-1)^2+4(z+4)^2=25\\ \Rightarrow (x-1)^2+(z+4)^2=\frac{25}{4}\\ \Rightarrow (z+4)^2=\frac{25}{4}-(x-1)^2 $$

Now I have two equations with $(z+4)^2$ on one side, so I can equate the other sides:

$$ \frac{25}{4}-(x-1)^2=\frac{16}{9}-(y+1)^2\\ \Rightarrow (y+1)^2=(x-1)^2-\frac{161}{36}\\ \Rightarrow y+1=\pm\sqrt{(x-1)^2-\frac{161}{36}}\\ \Rightarrow y=-1\pm\sqrt{(x-1)^2-\frac{161}{36}}\\ $$

Now I have an expression for y in terms of x, but it has two problems.

1. It doesn't properly describe the curve of intersection. It describes a "prism" surface in the 3d space.
2. It has a $\pm\sqrt{}$ term, which makes it 2 different functions.

So I tried the following:

$$ 9(y+1)^2+9(z+4)^2=16\\ \Rightarrow 9(y+1)^2+9(z+4)^2-16=0\\ \\ 4(x-1)^2+4(z+4)^2=25\\ \Rightarrow 4(x-1)^2+4(z+4)^2-25=0\\ \\ \Rightarrow 9(y+1)^2+9(z+4)^2-16=4(x-1)^2+4(z+4)^2-25\\ \Rightarrow 4(x-1)^2-9(y+1)^2-5(z+4)^2=9 $$

This last form looks somewhat similar to a sphere's equation. I think I might be able to parameterize this one in polar coordinates. How to do that still eludes me, however.

It also still has the issue that it describes a surface as far as I can understand and not a curve in space.

For fun, I can also show that I can solve for $z$:

$$ 9(y+1)^2+9(z+4)^2-16=4(x-1)^2+4(z+4)^2-25\\ \Rightarrow 5(z+4)^2=4(x-1)^2-9(y+1)^2-9\\ \Rightarrow (z+4)^2=\frac{4}{5}(x-1)^2-\frac{9}{5}(y+1)^2-\frac{9}{5}\\ \Rightarrow z+4=\pm\sqrt{\frac{4}{5}(x-1)^2-\frac{9}{5}(y+1)^2-\frac{9}{5}}\\ \Rightarrow z=-4\pm\sqrt{\frac{4}{5}(x-1)^2-\frac{9}{5}(y+1)^2-\frac{9}{5}}\\ $$

My ideas run out at this point. Any help would be appreciated.

Answer 1

The parametric curve of the intersection of the two cylinders can be found as follows:

First, realize that the YZ plane view of the curves of intersection is a circle with radius $r=\frac{4}{3}$ and centre $(y,z)=(-1,-4)$.

To simplify, translate the circle to the origin.

Hence a parametric equation for the $y$ and $z$ coordinates of the curve of interpenetration are given by $y=r\cos\theta=\frac{4}{3}\cos\theta$ and $z=r\sin\theta=\frac{4}{3}\sin\theta$

Now "all" that's left to get a parametric equation in cylindrical coordinates in the YZ plane is finding the $x$ height of that circle on the YZ plane as function of the parameter $\theta$.

The height is the $x$ value of the larger cylinder's outer surface as function of $\theta$.

First translate the circle to the origin. Then it is a circle with radius $R=\frac{5}{2}$ on the XZ plane.

The radius $R$ of the larger circle is the hypotenuse of a right triangle. One perpendicular side is given by $z=r\sin\theta$. The other is given by $\pm\sqrt{R^2-z^2}$.

The curves of interpenetration of the simplified case (with both circles centered on the origin) is therefore given by:

$$ (x,y,z)=\Bigg(\pm\sqrt{\frac{25}{4}-\frac{16}{9}\sin^2\theta}, \frac{4}{3}\cos\theta, \frac{4}{3}\sin\theta\Bigg) $$

Adding back the offsets:

$$ (x,y,z)=\Bigg(1\pm\sqrt{\frac{25}{4}-\frac{16}{9}\sin^2\theta}, \frac{4}{3}\cos\theta-1, \frac{4}{3}\sin\theta-4\Bigg) $$

Hence, we have two curves of interpenetration given by:

$$ (x,y,z)=\Bigg(1+\sqrt{\frac{25}{4}-\frac{16}{9}\sin^2\theta}, \frac{4}{3}\cos\theta-1, \frac{4}{3}\sin\theta-4\Bigg) $$ and $$ (x,y,z)=\Bigg(1-\sqrt{\frac{25}{4}-\frac{16}{9}\sin^2\theta}, \frac{4}{3}\cos\theta-1, \frac{4}{3}\sin\theta-4\Bigg) $$