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Given a two surfaces say: $z=1-y$ and $ x^2+y^2+z^2=1$, we find that they intersect at: $$x^2-2yz=0$$

How is the above a space curve? Is it not just another surface?

And why do we need to introduce a parameter, say $y=t$ , for it to be a space curve? I really don't understand the purpose of parametrization.

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  • $\begingroup$ In particular the first surface already gives $z = 1-y$, $z$ parameterized by $y$. The "intersection" you describe gives $y$ parameterized by $x$. So the context supplies the kind of parameterization you seek. $\endgroup$
    – hardmath
    Jul 3, 2015 at 2:32
  • $\begingroup$ en.wikipedia.org/wiki/Implicit_function_theorem $\endgroup$ Jul 3, 2015 at 2:33
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    $\begingroup$ Civil, you typed in something incorrect for the intersection. What you ought to get is $x^2 - 2 y z = 0,$ which is a circular cone $\endgroup$
    – Will Jagy
    Jul 3, 2015 at 2:42
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    $\begingroup$ There's all sorts of nonsense being written here. The intersection of these two lines is a point, not another line. When you combine two equations to get another equation, that is not the intersection. $\endgroup$ Jul 3, 2015 at 3:23
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    $\begingroup$ No, this is not right at all. In general, it will indeed be a curve, not necessarily in any plane. At any rate, you gave the equation of another surface when you said "they intersect at ... ." All you can say is that their curve of intersection lies in that surface. Make sure you understand. $\endgroup$ Jul 3, 2015 at 9:50

3 Answers 3

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You are right. $$x^2-2yz=0$$ is not a space curve. Is it not just another surface but one that shares a commonalty. It is new a cone, a surface similar to the plane and sphere given, with which it shares the same curved intersection. It is one among a set of possible infinitely many surfaces with a common track or locus.

$ z=1-y $ and $ x^2+y^2+z^2=1$, do not intersect at $$x^2-2yz=0.$$

In fact no two surfaces intersect along another surface!

Just as three curved lines can be concurrent, three surfaces can have a ( I don't know if a proper word exists .."concur-line" :) for lack of one).

If any two of the surfaces is given, the third and indefinitely many more can be set up or found.

$$ (x,y,z) = (\pm \sqrt {2 t ( 1-t)} , t , (1-t) ) \tag{1}$$ is an easy parameterization of the common curved intersection. It has fixed or unique position in 3-space. Here also we can have several possibilities of changing the velocity vector. It is like a race-track. The track is fixed but so many speed / gradient variations are possible prescribed at any point on common line. The purpose of such parameterization is to find embedding in 3-space for this common line of cutting.

This line gives (x,y,z) coordinates as a function of a needed single parameter.

A simpler example. You can imagine all spheres passing through $ x^2 + y^2 = 1$ with their centers on z-axis and varying radii.

EDIT1:

2 errors corrected. $ x^2- 2 y z = 0$ is a cone, not a hyperboloid. But my remarks do not suffer. Other is typo about $\sqrt (..) $ in parameterization.

The red circle is the "concur-line" of the cone, plane and sphere of parameterization (1).

Concurline

EDIT2:

Picturization of all possible coniciods passing through this line (including the above three) is indicated in code given (please make the correction):

https://mathematica.stackexchange.com/questions/87639/manipulated-surfaces-to-include-parametric3d-plot-of-concurrent-line

The animation shows successive changes, as morphing parameter $lambda$ ( zero for circular cone) is varied:

Double plane $\rightarrow$ oblate ellipsoid $\rightarrow$ hyperboloid of 1 sheet $\rightarrow$ cone $\rightarrow$ sphere $\rightarrow$ hyperboloid of 2 sheets $\rightarrow$ prolate ellipsoid $\rightarrow$ Double plane

This I hope completes an answer to a question " How is the parameterization of all surfaces inducing a common single space curve of intersection "?

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  • $\begingroup$ If you could, I would appreciate it! $\endgroup$
    – CivilSigma
    Jul 3, 2015 at 22:47
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When "geometric objects" intersect transversely, or in "general position" ,the dimension of their intersection is the dimension of the ambient space (which I am assuming here is $\mathbb R^3$) minus the sum of their respective codimensions, say "Cod" i.e.,

$$Dim (S_1 \cap S_2)_{\mathbb R^3}=Dim(\mathbb R^3)-Cod(S_1)-Cd(S_2) $$

Where $S_1, S_2$ are the two surfaces , giving $$3-1-1=1 $$

So the dimension of the set of points in the intersecting surfaces is $1$, i.e., the intersection is a curve.

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  • $\begingroup$ I never knew this. This definitely adds more depth to my understanding. Thank you. $\endgroup$
    – CivilSigma
    Jul 3, 2015 at 3:06
  • $\begingroup$ Glad to help; let me know if you want a source beyond this :mathworld.wolfram.com/TransversalIntersection.html . $\endgroup$
    – Gary.
    Jul 3, 2015 at 3:07
  • $\begingroup$ And the transversal intersection of homology classes is a well-defined homology class. $\endgroup$
    – Gary.
    Jul 3, 2015 at 3:11
  • $\begingroup$ So if 3 surfaces intersect, we have dimension of intersection of single points = zero, ok? $\endgroup$
    – Narasimham
    Jul 4, 2015 at 21:03
  • $\begingroup$ @Narasimham: I would think that if the intersection is pairwise transversal yes, but give me some time to think. Maybe we could argue by using pairs of surfaces. $\endgroup$
    – Gary.
    Jul 4, 2015 at 21:04
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The intersection of two surfaces will be a curve. Here, one surface is the sphere defined by

$$x^2+y^2+z^2=1 \tag 1$$

while the second surface is the plane defined by

$$z=1-y \tag 2$$.


Now, the curve of intersection can be defined by a parametric curve

$$\vec r(t)=\hat xx(t)+\hat yy(t)+\hat zz(t)$$

where $t\in [t_1,t_2]$.

And since this curve lies on both the sphere defined by $(1)$ and the plane defined by $(2)$, we have simultaneously

$$z(t)=1-y(t) \tag 3$$

and

$$x^2(t)+y^2(t)+z^2(t)=1 \tag 4$$

Now, $(3)$ and $(4)$ have a common solution

$$x^2(t)+2y^2(t)-2y(t)=0 \tag 5$$

and

$$z(t)=1-y(t) \tag 6$$


We rewrite $(5)$ as

$$x^2(t)+2(y(t)-1/2)^2=1/2 \tag 7$$

from which we can see that a parametric curve describing the intersection of $(1)$ and $(2)$ can be expressed as

$$x(t)=\frac{1}{\sqrt{2}}\cos t$$

$$y(t)=\frac12+\frac12 \sin t$$

$$z(t)=\frac12-\frac12 \sin t$$

for $0\le t\le 2\pi$.


NOTE:

The parametric description is not unique and in fact, we need not use a parametric description to describe the curve of intersection. Here, we could have as easily described the curve in terms of one of the coordinates. For example, we can write

$$x=\pm\sqrt{y(1-y)}$$

$$y=y$$

$$z=1-y$$

for $0\le y\le 1$.

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  • $\begingroup$ note that the OP incorrectly typed the initial conclusion, it should read $x^2 - 2 y z = 0,$ which is a circular cone. $\endgroup$
    – Will Jagy
    Jul 3, 2015 at 2:50
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    $\begingroup$ @WillJagy Yes, I wasn't using the OP's original conclusion. $\endgroup$
    – Mark Viola
    Jul 3, 2015 at 2:56
  • $\begingroup$ Thank you for the answer. One final thing, why do we all of a sudden introduce vector functions when our original functions are scalars ( I assume ) ? $\endgroup$
    – CivilSigma
    Jul 3, 2015 at 3:05
  • $\begingroup$ You're quite welcome!! My pleasure. We don't need to introduce the vector notation. That just makes things a bit more compact to write. $\endgroup$
    – Mark Viola
    Jul 3, 2015 at 3:22

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