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#1 2008-09-18 17:38:54

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Is this rubbish?

here is something taken directly from my teachers lecture notes in my Vector Calc class. It didn't make any sense to me, and i won't get a chance to ask him about it until next week, so I was wondering if someone can verify that this is wrong:

The circle


on the yz plane (a,r are positive real numbers) is revolved around the z-axis, forming a torus T. Find the equation of this torus.

Solution: Let (x,y,z) be a point on T. If this point were on the yz plane, it would be on the circle, and the of the distance (notice the odd typo, i think it was supposed to read 'and the distance') to the axis of rotation would be


where sign(t) is the sign function. Anywhere else, the distance from x,y,z to the z axis is the distance of this point to the point

(that sentence doesn't seem syntactically correct)We must have:

STOP! the proof continues by applying more algebra, but that statement seems utterly senseless. How the devil can we assume that? if the point is no longer on the yz plane, how can we equate the distance from the z axis (the axis of rotation) to the point (x,y,z) with the distance from the z axis of the point (0,y,z), using an equation which we derived by assuming (0,y,z) was on that circle, (and thus a-r <= y <= a + r) ??? 

further, if I'm not mistaken, since the torus cuts through the points (0, a+r, 0) and (0,a-r,0) on the yz plane, and its being revolved around the z axis, then it should also cut through the points (a+r, 0,0) and (a-r,0,0) on the xz plane. 

So now let a = 5, and r = 1, consider the point (a+r,0,0) = (6,0,0)

inserting these values into the equation above:


which reduces to
36 = 16.

Furthermore, the proof continues with the expression


by multiplying out on the right side, solving for sign(y-a) on side, and squaring again to obtain sign(y-a)^2 on one side. The book then claims:
"sign(y-a)^2 = 1, it could not be 0 (why?)"

...why indeed? is not the point  (0, a,+-r) a part of this torus? At those points, sign(y - a) = 0, certainly.

Either this is absolute rubbish, or I'm missing something crucial here!

Last edited by mikau (2008-09-18 18:05:42)


A logarithm is just a misspelled algorithm.

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#2 2008-09-18 18:47:13

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Is this rubbish?

If a point P has coordinates (x,y,z), the distance of P to the z-axis is

. Does this make sense?

If P lies on the torus, let P′ be the corresponding point on the torus in the yz-plane. That is, P′ is the point with coordinates (0,y,z). Then P and P′ have the same distance from the z-axis. Does this also make sense?

[EDIT: Sorry, the above is wrong.]

Last edited by JaneFairfax (2008-09-19 02:18:42)

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#3 2008-09-18 23:51:16

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Is this rubbish?

If P lies on the torus, let P′ be the corresponding point on the torus in the yz-plane. That is, P′ is the point with coordinates (0,y,z). Then P and P′ have the same distance from the z-axis. Does this also make sense?

no it doesn't. I might be too tired right now, so forgive me if I'm way off, but that seems to be stating that


for any point on the torus, which clearly implies x = 0, and thus the entire torus is on the yz plane. what

Lets see... let a = 3 and r = 1, you would agree that the point (0,a+r,0) =  (0,4,0) is a point on the torus that lies on the yz axis, right? It even satisfies that equation.

But this is a torus, revolving around the z axis, therefore (4,0,0) is a point on the torus, no?

Plug x = 4, y = 0, z = 0, r = 1, a = 3 into the equation, and you get:

4^2 + 0^2 = (3 + sign(0-3)sqrt(1^2-0^2))^2
~ 16 = 4

(4,0,0) is a point on the torus, thus it should satisfy the equation, no?
What am I missing here? dunno

Last edited by mikau (2008-09-19 01:34:29)


A logarithm is just a misspelled algorithm.

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#4 2008-09-19 00:14:04

LQ
Real Member
Registered: 2006-12-04
Posts: 1,285

Re: Is this rubbish?

Sounds important, what are you doing with this problems solution?


I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...

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#5 2008-09-19 01:47:21

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Is this rubbish?

what am I doing with the solution? Well the point of the problem is just to find a Cartesian equation of the torus, thats all. Nothing that follows is contingent on this, its just an example problem.


A logarithm is just a misspelled algorithm.

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#6 2008-09-19 02:01:36

LQ
Real Member
Registered: 2006-12-04
Posts: 1,285

Re: Is this rubbish?

Your current motive is accepted.


I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...

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#7 2008-09-19 02:20:25

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Is this rubbish?

Sorry Mikau, what I did above was wrong. The z-coordinates of P and P′ are the same, but their y-coordinates are not. I was posting in a hurry this morning and didn’t check my equations. sad

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#8 2008-09-19 02:23:00

LQ
Real Member
Registered: 2006-12-04
Posts: 1,285

Re: Is this rubbish?

For me you are allways forgiven for eternity!


I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...

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#9 2008-09-19 05:43:16

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Is this rubbish?

JaneFairfax wrote:

Sorry Mikau, what I did above was wrong. The z-coordinates of P and P′ are the same, but their y-coordinates are not. I was posting in a hurry this morning and didn’t check my equations. sad

Its okay, Jane! smile Thanks for taking the time to help! It appears my teacher made the same mistake you did.

When both you and my teacher said the same thing, I figured it had to be me, but now that you agree, I feel more confident that I'm right.

However, I did run into my teacher at school today, and tried to explain my gripe. I had little success communicating my reasoning, and he still insists he's correct.

I'm pretty sure he's wrong now... but what am i going to do... this might show up on an exam. neutral

Last edited by mikau (2008-09-19 05:44:40)


A logarithm is just a misspelled algorithm.

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#10 2008-09-19 20:54:41

Kurre
Member
Registered: 2006-07-18
Posts: 280

Re: Is this rubbish?

I dont really understand how your teacher did, but i think i can solve this using vectors:
start by choosin a vector w=(x1,y1,0) such that |w|=a. This will be the center of any of the circles that created the torus. We now want all vectors u in the plane spanned by w and  the z axis such that |u|=r, because then adding u and w yields a point on the torus' surface. u can then be written as a linear combination of w and e= (0,0,1), ie u=pw+qe=(px1,py1,q), with:


or equivalently, since

(1)
so we get the vector u+w= ((p+1)x1,(p+1)y1,q)= (x,y,z).
so z=q, x= (p+1)x1, y= (p+1=y1.
now we want to express a relation between x,y and z.
if we multiply the equation
by (p+1)^2 we get



inserting this into (1) yields:

This is a necessary condition, I dont think I proved it to be sufficient though since I guess I used non equivalent transformations, but it may be possible to show that all points satisfying this actually lies on the torus surface.

and mikau, no, I dont think the sign function can be zero. That would be a bit weird since all we want is the sign + or -, if the entry is zero then it doesnt matter which sign we use and probably we can define sgn(0)=1.

Last edited by Kurre (2008-09-19 20:56:51)

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#11 2008-09-20 03:48:05

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Is this rubbish?

The text explicitly states in that solution, that sign(t) := 0 where t = 0, 1 where t > 1 and -1 where t < 0
I left out that remark because I figured we all knew that anyway.

Here is how i think the problem needs to be solved, replace y with sqrt(x^2 + y^2) in the original equation:

observe now that if


then

and so

or just
0 = 1*0
therefore, in the equation,

it is safe to assume sign(t) is not zero, for if it is, we could have replaced sign(t) with 1 in its original location.

for brevity, let

So now, since we can assume sing(t) != 0,


so

his solution technically did the same thing, but only after he masked out the sign(y - a) term by squaring. Before that, the equation remains incorrect.


A logarithm is just a misspelled algorithm.

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#12 2008-09-20 03:49:38

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Is this rubbish?

The text explicitly states in that solution, that sign(t) := 0 where t = 0, 1 where t > 1 and -1 where t < 0
I left out that remark because I figured we all knew that anyway.

Here is how i think the problem needs to be solved, replace y with sqrt(x^2 + y^2) in the original equation:


observe now that if


then

and so

or just
0 = 1*0
therefore, in the equation,

it is safe to assume sign(...) is not zero, for if it is, we could have replaced sign(...) with 1 where it originally appeared.

for brevity, let

So now, since we can assume sing(t) != 0,




so

his solution technically did the same thing, but only after he masked out the sign(y - a) term by squaring. Before that, the equation remains incorrect.

Last edited by mikau (2008-09-20 03:59:54)


A logarithm is just a misspelled algorithm.

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