Homework Help: Parametrization of a torus

1. Jun 1, 2010

Jamin2112

1. The problem statement, all variables and given/known data

Consider the parametrization of torus given by:

x=x(ø,ß)=(3+cos(ø))cos(ß)
y=y(ø,ß)=(3+cos(ø))sin(ß)
z=z(ø,ß)=sin(ø),

for 0≤ø,ß≤2π

What is the radius of the circle that runs through the center of the tube, and what is the radius of the tube, measured from the central circle?

2. Relevant equations

?

3. The attempt at a solution

Hmmmm..... I don't get it. The distance from a point on the torus to the center would be √(x2+y2+z2), I think. But the center of the tube? Not sure how that would work. H

Help is appreciated. And after that, I may ask some more questions. I know you folks are ready, willing, and able.

2. Jun 1, 2010

LCKurtz

$$\langle x,y,z\rangle=\langle 3\cos\beta,3\sin\beta,0\rangle+ \langle\cos\phi\cos\beta,\cos\phi\sin\beta,\sin\phi\rangle$$

and look at the lengths of those vectors.

3. Jun 1, 2010

Jamin2112

But what does the length of those vectors have to do with anything?

4. Jun 1, 2010

HallsofIvy

If $\beta= 0$, $x= 3+ cos(\phi)$, $y= 0$, and $z= sin(\phi)$ so that $(x-3)^2+ z^2= 1$. That's a circle in the xz-plane with center at (3, 0, 0) and radius 1. Similarly, if $\beta= \pi/2$, $x= 0$, $y= 3+ cos(\phi)$, and $z= sin(\phi)$ so that $(y-3)^2+ z^2= 1$. That's a circle in the yz-plane with center at (0, 3, 0) and radius 1. With $\beta= \pi$ and $\beta= 3\pi/2$ we get circles with radii 1 and centers at (-3, 0, 0) and (0, -3, 0), respectively. In general, as $\beta$ ranges from 0 to $2\pi$ we get circles with radii 1 and centers sweeping out a circle with center at (0, 0, 0) and radius 3. That's the "center of the tube".

5. Jun 1, 2010

Jamin2112

Oh, okay. It just seems like if you had donut floating in space that if defined by some parametrization x(∂,ß), y(∂,ß), z(∂,ß), values of ∂,ß would map to a point on the surface. Like I know if you have x(t), y(t), you can regions in the x-y plane into cool things like cartoids, that could not be made with just x, y=f(x). I'm just confused about how exactly the equation for the radius of a sphere in x-y-z space gets turned into the a formula for the distance between the origin and the center of a donut.

6. Jun 1, 2010