On the cylinder that projects a curve on a plane.

Question

Suppose I have in space the curve $C$ given by $(x,y,z) = (t,t^2,t^3)$ and I want to project this curve $C$ on the plane $\alpha$ that has equation $x+y+z=0$.

Then I take every vector orthogonal to the plane that passes through $C$, i. e. $$(x,y,z) = (t,t^2,t^3) + u(1,1,1)$$ and intersect it with the plane $x+y+z=0$, correct?

I wrote that this is a cylinder that projects $C$ on $\alpha$ but I don't understand why this is the case because the equation of a cylinder is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and I don't see it appear here.

Answer 1

Good question. Yes it is an oblique cylinder, or an extruded surface. It is special case of a translational surface. This cylinder or translated surface has a twisted cubic drawn on it.

The image below is a plot of a translated surface of the two parameters you indicated:

{t + u, t^2 + u, t^3 + u }, {t, -2, 3}, {u, -3, 3}.

By rotation of axes ( direction cosines (1,1,1) it is possible to write its equation in terms of only two dependent coordinates.

Answer 2

Each point $p(t)=(t,t^2,t^3)$ of the curve C project into the point $P(t) = (a(t),b(t),c(t))$ such that

• $a(t)+b(t)+c(t) = 0$

• $(a(t)-t,b(t)-t^2,c(t)-t^3) \wedge (1,1,1) = 0$

This yields to

$ a=2/3\,t-1/3\,{t}^{2}-1/3\,{t}^{3},b=-1/3\,t+2/3\,{t}^{2}-1/3 \,{t}^{3},c=-1/3\,t-1/3\,{t}^{2}+2/3\,{t}^{3} $