# Parameterising a certain circle

• Jan 24th 2011, 11:24 AM
FGT12
Parameterising a certain circle
Parameterise the circle that results from intersecting the unit sphere in $\displaystyle \mathbb{R}^3$ centered at the origin with the plane $\displaystyle [(x,y,z) \in : x = z ]$

I tried $\displaystyle x^2 + y^2 + z^2 = 1 \Rightarrow 2x^2 + y^2 = 1$
however the last equation is an ellipse????
• Jan 24th 2011, 11:43 AM
ahaok
Quote:

Originally Posted by FGT12
Parameterise the circle that results from intersecting the unit sphere in $\displaystyle \mathbb{R}^3$ centered at the origin with the plane $\displaystyle [(x,y,z) \in : x = z ]$

I tried $\displaystyle x^2 + y^2 + z^2 = 1 \Rightarrow 2x^2 + y^2 = 1$
however the last equation is an ellipse????

the intersection is an ellipse and no circle!(Wondering)
• Jan 24th 2011, 06:23 PM
xxp9
You result is only the projection of the circle to the xy plane, which is really an ellipse.
• Jan 25th 2011, 06:36 AM
FGT12
so how would you parameterise the circle in $\displaystyle \mathbb{R}^3$
• Jan 25th 2011, 08:57 AM
Mazerakham
It's definitely going to be a circle, because really: how could a plane intersect a sphere and make an ellipse! You're right that it's absurd.

Have you tried drawing a sketch? Try visualizing the picture with the y axis pointing directly out of the paper, and that may give you some insight. I solved the problem by first coming up with an equation based on the sketch, then proving that it was right. So, like I said. Start with the picture.
• Jan 25th 2011, 10:06 AM
HallsofIvy
ahaok is wrong, a sphere cut by a plane gives a circle, not an ellipse. xxp9 is right- a circle, projected onto a plane at an angle to a line perpendicular to the circle gives an ellipse. That is why you get the equation of an ellipse- you are projecting the circle onto the xy-plane, not parameterizing it.

Here, your sphere is the unit sphere, $\displaystyle x^2+ y^2+ z^2= 1$ and the plane is given by x= z. The "standard" parameterization of the unit sphere can be derived from spherical coordinates:
$\displaystyle x= \rho cos(\theta)sin(\phi)$
$\displaystyle y= \rho sin(\theta)sin(\phi)$
$\displaystyle z= \rho cos(\phi)$

by taking the radius variable, $\displaystyle \rho$ equal to 1:
$\displaystyle x= cos(\theta)sin(\phi)$
$\displaystyle y= sin(\theta)sin(\phi)$
$\displaystyle z= cos(\phi)$

The fact that z= x means that we must have $\displaystyle cos(\theta)sin(\phi)= cos(\phi)$ so that $\displaystyle cos(\theta)= \frac{cos(\phi)}{sin(\phi)}= cot(\phi)$.

So one way of getting the parameterization of the circle is to replace $\displaystyle \theta$ by $\displaystyle cos^{-1}(cot(\phi))$, reducing from two parameters to one:
$\displaystyle x= cos(cos^{-1}(cot(\phi)))sin(\phi)$
$\displaystyle y= sin(cos^{-1}(cot(\phi))sin(\phi)$
$\displaystyle z= cos(\phi)$

Of course, $\displaystyle cos(cos^{-1}(cot(\phi))= cot(\phi)$. And, we can use $\displaystyle sin(\theta)= \sqrt{1- cos^2(\theta)}$ to say that
$\displaystyle sin(cos^{-1}(cot(\phi))= \sqrt{1- (cos(cos^{-1}(cot(\phi)}= \sqrt{1- cot^2(\phi)}$ and so have

$\displaystyle x= cot(\phi)sin(\phi)= cos(\phi)$
$\displaystyle y= \sqrt{1- cot^2(\phi)}sin(phi)$
$\displaystyle z= cos(\phi)$.
• Jan 25th 2011, 10:34 AM
TheEmptySet
Quote:

Originally Posted by FGT12
Parameterise the circle that results from intersecting the unit sphere in $\displaystyle \mathbb{R}^3$ centered at the origin with the plane $\displaystyle [(x,y,z) \in : x = z ]$

I tried $\displaystyle x^2 + y^2 + z^2 = 1 \Rightarrow 2x^2 + y^2 = 1$
however the last equation is an ellipse????

The vector form for the plane is

$\displaystyle \vec{r}(s,t)=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}t+\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}s$

This gives that $\displaystyle x=z=t$ Using the equation of the sphere gives

$\displaystyle t^2+y^2+t^2=1 \implies y=\pm\sqrt{1-2t^2}$

So a parametric representation of the circle is
$\displaystyle x=t,\quad y=\pm\sqrt{1-2t^2},\quad z=t$

Since the plus or minus is really messy it can be eliminated using trigonometry.

If we define $\displaystyle \sin(\alpha)=\sqrt{1-2t^2}$ this gives
$\displaystyle \cos(\alpha)=\frac{t\sqrt{2}}{1} \implies t=\frac{1}{\sqrt{2}}\cos(\alpha)$

Using this we get

$\displaystyle \displaystyle \vec{r}(\alpha)=\left( \frac{1}{\sqrt{2}}\cos(\alpha)\right)\vec{i}+\sin( \alpha)\vec{j}+\left( \frac{1}{\sqrt{2}}\cos(\alpha)\right)\vec{k}$

Attachment 20599
http://www.mathhelpforum.com/math-he...2&d=1295983968

Edit: too slow