General equation of cylinder surface

Question

According to the definition "A cylinder is a surface that consists of all lines (called rulings) that are parallel to a given line and pass through a given plane curve. " I'm curious about the general equation of cylinder . I mean finding the equation by using the line and plane curve .

Answer 1

This definition is for a generalized cylinder, so it does not necessarily look like this:

However, to see this standard cylinder as an example of the definition you have, we could take the plane curve to be a circle and the line of parallel to be perpendicular to the plane we have parameterized the circle in.

So for example, $r(t) = (cos(t), sin(t),0)$ for $t \in [0,2\pi)$ is a unit circle in the plane $z = 0$ and the line of parallel is $s(t) = (0,0,t)$ for $t \in \mathbb{R}$. The cylinder determined by this set up, according to your definition, would correspond to a cylinder like the one pictured above.

The flexibility your definition is allowing is that we don't have to use a circle for the plane curve, we don't have to embed the plane curve in the plane $z = 0$, and we don't have to choose the 'rulings' (lines) to be perpendicular to the plane. So you could get something more like this:

My art skills are lacking.

Also, depending on how they defined ‘plane curve’, you don’t even have to use a closed loop for the plane curve - you could have used a straight line, or a parabola, which would be even less similar to a usual cylinder.

Let us see if we can get a general sort of equation for these objects. If we restrict for now to "cylinders" that are going to live in $\mathbb{R}^3$, then if our plane curve is given by some parameterization $r(t)$ and the line of parallel is given by some $s(t) = p_1 + t\cdot p_2$ where $p_1, p_2$ are points in $\mathbb{R}^3$ then according to your definition, to get the cylinder we want to make lines going through the plane curve, parallel to $s(t)$.

Our cylinder parameterization could look something like

$c(u,v) = r(u) + v \cdot p_2$.

Why is that? See if you can figure it out. What happens when the parameter $v = 0$? (it traces out the plane curve). What happens when you fix $u$ and let $v$ vary? (it traces out lines going through a point on the plane curve). Maybe go on 3D graphing software and plug this in for specific $r(t)$ plane curves and lines $s(t)$. Mac computers come with 3D graphing software.

Playing around with parameterizations is fun. When I started writing this answer I actually didn't know what the equation would be, but with practice you learn how to tinker with these things and can usually figure out what you want.

Answer 2

I think one of the approaches you can have is to start from the equation of a special cylinder (like one having a radius of 1 parralel to the z-axis), and apply a rotation, an homothety and a translation on the x, y and z.

The equation of this original cylinder is : (x - a)² + (y - b)² = 1.

Here, a and b are the coordinates of the intersection of the axis of the cylinder with the xy plane. An example of rotation (around the x axis) in the 3-space is given by the matrix :

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos(\theta) & -sin(\theta) \\ 0 & sin(\theta) & cos(\theta) \end{bmatrix}$$

Apply it on the coordinates of the equation and you'll get the equation for a new rotated cylinder, rotated of an angle $\theta$ from the x-axis :

$$(x - a)^2 + (cos(\theta) y - sin(\theta) z - b)^2 = 1$$

After that, you can try another rotation, around the z-axis this time, with matrix :

$$\begin{bmatrix} cos(\phi) & -sin(\phi) & 0 \\ sin(\phi) & cos(\phi) & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

This gives :

$$(cos(\phi) x - sin(\phi) y - a)^2 + (cos(\theta) (cos(\phi) y + sin(\phi) x) - sin(\theta) z - b)^2 = 1$$

Then, to make the cylinder the scale you want, you can use the transformation given by :

$$\begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}$$

This gives :

$$(cos(\phi) \lambda x - sin(\phi) \lambda y - a)^2 + (cos(\theta) (cos(\phi)\lambda y + sin(\phi) \lambda x) - sin(\theta) \lambda z - b)^2 = 1$$

Or, more simply :

$$(cos(\phi) x - sin(\phi) y - a)^2 + (cos(\theta) (cos(\phi) y + sin(\phi) x) - sin(\theta) z - b/\lambda)^2 = 1/\lambda^2$$

A second approach, which I find shorter, is to use the formula for the distance between a line and a point. The distance between the line of vector $\vec{u}$ passing by $B$ and a point $A$ is given by :

$$d^2(A,B,\vec{u}) = \frac{\left\lVert{\vec{AB} \wedge \vec{u}}\right\rVert^2}{\left\lVert{\vec{u}}\right\rVert^2}$$.

It remains to write $d^2(A,B,\vec{u}) = r^2$ and to express it in terms of x, y and z (the coordinates of A), and this will give an algebraic equation for the cylinder.

Answer 3

Suppose the plane curve is given by the equation $f(x,y) = 0$, and the line is parallel to $\vec v = \left<a,b,c\right>$. Given a point $P= (x_1,y_1,z_1)$ in space, we need to:

1. Trace the line through $P$ parallel to $\vec v$;
2. Find the point $Q$ of intersection of that line with the $xy$-plane;
3. Decide if $Q$ is on the curve. If it is, $P$ is on the cylinder.

Let the coordinates of $Q$ be $(x_0,y_0,0)$. The parametric equations for the line are \begin{align*} x &= x_1 + at \\ y &= y_1 + bt \\ z &= z_1 + ct \end{align*} If $x=x_0$, $y=y_0$, and $z=0$, we can solve \begin{align*} x_0 &= x_1 - \frac{a}{c}z_1 \\ y_0 &= y_1 - \frac{b}{c}z_1 \end{align*} [If $c=0$, the vector is parallel to the $xy$-plane and we won't get a cylinder at all.] This point is on the curve if $f(x_0,y_0) = 0$. Replacing the point $(x_1,y_1,z_1)$ with a varying point $(x,y,z)$, the equation for the cylinder is $$ f\left(x-\frac{a}{c}z,y-\frac{b}{c}z\right) = 0 $$