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How to find the equations of the lines that make up the ruled surface z= 2xy? There are a lot of lines and solving this seems non-obvious. I imagine the first thing to know is the equation of a line in 3d. What is the equation for a 3D line?

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  • $\begingroup$ How could $2xy$ be a surface? It's a monomial in two variables. $\endgroup$ – Bernard Apr 1 '18 at 18:14
  • $\begingroup$ @Bernard fixed. $\endgroup$ – User3910 Apr 1 '18 at 18:18
  • $\begingroup$ @Dale Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Apr 3 '18 at 14:40
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The points $\;M=\Bigl(x,\dfrac 2x, 4\Bigr)$ and $\;N=\Bigl(x,-\dfrac 2x, -4\Bigr)$ are on the quadric. One easily check that any point $P=tM+(1-t)N$ also lies on the quadric. Indeed the coordinates of such a point are $$\Bigl(tx+(1-t)x, \dfrac{2t}x-\dfrac{(1-t)2}x, 4t-4(1-t)\Bigr)=\Bigl(x, \dfrac{2(2t-1)}x, 4(2t-1)\Bigr).$$

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  • $\begingroup$ I am not really good with parametric equations. Could you explain what t stands for here: specifically what varying t changes what? Also, how did you arrive at the equations you found? $\endgroup$ – User3910 Apr 1 '18 at 21:44
  • $\begingroup$ It describes the whole line through $M$ and $N$ (any point on this line is a barycentre of $M$ and $N$). $\endgroup$ – Bernard Apr 1 '18 at 21:47
  • $\begingroup$ how did you come up with the points M and N? $\endgroup$ – User3910 Apr 1 '18 at 21:57
  • $\begingroup$ I learnt how to obtain it for the hyperbolic paraboloid when I was a student (b.t.w., it is a doubly ruled surface, i.e. there are two families of lines which generate the surface). $\endgroup$ – Bernard Apr 1 '18 at 22:01
  • $\begingroup$ could you elaborate on how to obtain it from the hyperbolic paraboloid? $\endgroup$ – User3910 Apr 1 '18 at 22:02
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The lines are given by the planes intersection

  • $z=xy$
  • $x=k$

and

  • $z=xy$
  • $y=k$

In parametric form for

  • $z=xy$
  • $x=k$

we have that a generic point is

$$P=(k,t,tk)=(k,0,0)+t(0,1,k)$$

that is the parametric equation for this family of lines and similarly we can find that for the other family of lines.

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  • $\begingroup$ If i wanted to write these equations in math.stackexchange.com/questions/404440/… form how would i do that? Could i just say z=ky=kx ? This seems to be the equation of a surface that is not what we started with. $\endgroup$ – User3910 Apr 1 '18 at 18:43
  • $\begingroup$ in parametric form? $\endgroup$ – gimusi Apr 1 '18 at 18:48
  • $\begingroup$ Before you substituted variables did it look like: $P=(x,y,z)=(x_0,y_0,z_0)+t(a,b,c)$? If so, how did you solve for a, b, and c? $\endgroup$ – User3910 Apr 1 '18 at 21:48
  • $\begingroup$ @Dale for each $k$ we have a line in that form. For example for k=1 we obtain the line $(1,0,0)+t(0,1,1)$ $\endgroup$ – gimusi Apr 1 '18 at 21:51
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A systematic way to find these lines uses the following facts:

  1. If the surface is given by the implicit equation $f(\mathbf x)=0$, then the line $\mathbf p+t\mathbf v$ is a generator iff $f(\mathbf p+t\mathbf v)=0$ for all $t$. For an $n$th-degree surface, this equation is an $n$th-degree polynomial in $t$ and the coordinates of $\mathbf v$.

  2. A generator of the surface lies in a tangent plane to the surface. Therefore, if $\mathbf p$ lies on the surface, then $\nabla f(\mathbf p)\cdot(\mathbf p+t\mathbf v)=0$. If $\nabla f(\mathbf p)\ne0$, this is a linear equation in the coordinates of $\mathbf v$ that can be used to eliminate one of the variables.

  3. The length of $\mathbf v$ is unimportant, but it must be non-zero, so one of its coordinates can be set to $1$. You might have to consider several cases here.

Facts 2 and 3 eliminate two of the variables, leaving you with an equation in $t$ and the remaining variable. Viewed as a polynomial in $t$, all of the coefficients must vanish, which gives you a system of single-variable polynomial equations to solve. You will end up with one or two solutions, depending on whether the surface is singly or doubly ruled.

So, starting with $f(x,y,z)=2xy-z$ we have $$2(p_x+t v_x)(p_y+t v_y)-p_z+t v_z = 0. \tag 1$$ The orthogonality condition of fact 2 gives us $2(p_x v_y+p_yv_x)-v_z=0$, which we can solve for $v_z$ and substitute into (1) to obtain, after simplifying, $$2p_x p_y-p_z+2t^2 v_x v_y = 0. \tag 2$$ Setting $v_x=1$ results in $v_y=0$ and vice-versa, therefore there are two generators through each nonsingular point $\mathbf p$ on the surface: $\mathbf p+t(1,0,2p_y)$ and $\mathbf p+t(0,1,2p_x)$. ${\partial f\over\partial z}=-1$ everywhere, so the gradient never vanishes and there are no singular points to consider.

If you need a directrix curve for this surface, find a plane that’s not parallel to any of these generators and intersect it with the surface. For this one, the plane $x=y$ works, producing a curve that can be parameterized as $(s,s,2s^2)$, which then generates the parameterization $(s,s+t,2s(s+t))$ of the surface.

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