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Cedric

Normal to functions of two variables (warning: homework question)

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Cedric    158
I think the theorem our teacher is asking us to prove is wrong, so I''d like to have other opinions before beating him up until he admits it. Or, more likely, before I make a fool out of myself . The problem is simple: "Show that all the planes tangent to the surface z = x*f(y/x) pass through the point (0,0,0)" Since we have no information at all on f, I think this can''t be true for any f. Possible proof: Let f(x) = 1 / x z = x * x / y = x² / y The normal of the tangent plane at x,y,z is N = (dz / dx, dz / dy , -1) (Partial derivatives) dz / dx = 2x / y dz / dy = -x² / y² If (x,y) = (3,6) then z = 1.5 N = (1, -81 / 36, -1) The normal has to be perpendicular to (3, 6, 1.5) if the point (0,0,0) is part of the plane at point (3, 6, 1.5) with normal (1, -81 / 36, -1), so Dot(N, (3,6,1.5)) = 0 1 * 3 + 6 * -81 / 36 + 1.5 * -1 = 0 3 - 81 / 6 - 1.5 = 0 3 - 13.5 - 1.5 = 0 -12 = 0 I just can''t figure out how this equation could hold for any f, and I can''t prove it true mathematically. Where did I make a mistake? I find it hard to believe that all the other students followed his suggestion for the demonstration without thinking about it conceptually (although it wouldn''t surprise me _that_ much). And sorry for posting a homework question. Cédric

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Zipster    2365
Error in your calculations.

-x2/y2 with (3,6) is -9 / 36, not -81 / 36. You most likely squared 3 twice by mistake. With those new numbers...

Dot(N, (3,6,1.5)) = 0
1 * 3 + 6 * -9 / 36 + 1.5 * -1 = 0
3 - 9 / 6 - 1.5 = 0
3 - 1.5 - 1.5 = 0
0 = 0

If there's one thing I've learned over the years, it's that the Algrebra always comes back to bite us in the ass

[edited by - Zipster on April 28, 2002 8:17:41 PM]

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Cedric    158
Arggghhh. So it must be right. Fine. I'll go find that proof.

Sigh. I still can't figure out how this can work with any function. I know I can find a stupid demonstration, but I really like to feel mathematics rather than follow recipes.

And thanks for the answer,

Cédric

[edited by - cedricl on April 28, 2002 8:23:22 PM]

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Zipster    2365
Ok, now that I've got my 3D graphing utility up, lets graph f(x) = 1 / x , or z = x2 / y:
(Positive X goes inward towards screen)

(Positive Y goes inward towards screen)

(Skew pic)


Now lets say f(x) = x2 . z = y2 / x. This is similar to the first one.

One last one (I like my graphing program!). f(x) = 1 / x2 . z = x3 / y2.


I know this wasn't exactly the most mathamatical way of doing things, but it looks as though it works out. I think it has to do with the fact that no matter what function you plug in, you're always going to get either some asymptotes or a flat plane through the origin. Thus I suppose all tangeant planes would pass through the origin in those cases.

[edited by - Zipster on April 28, 2002 9:26:15 PM]

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sQuid    149
You look like you already have everything you need to know to prove this - and I can verify that it''s true

So z = x*f(y/x). You are right that at any point P0 = (x0, y0,z0) the dot product between the normal N = (dz/dx,dz/dy,-1) and the vector from the origin to P0 must be zero. Just use the product and chain rules to do the differentiation, substitute in the x0, y0, z0 and calculate the dot product. You''ll find it''s 0.

hint: z0 = x0f(y0/x0)

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Zipster    2365
Gah, never thought of just deriving it in its current form

I''m a little rusty, so when we derive f(y/x), we can''t really find anything explicitly from that, can we? I mean, we don''t know the function, so we can''t assume its just y/x.

I''ve never really worked with deriving function I don''t know

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Cedric    158
Thanks for the replies, but I have already completed the demonstration. I was more looking for graphs like Zipster did, to understand it visually, but I have a hard time visualizing them. I can''t understand how a continuous function can have tangent planes passing through the origin at every point, unless the function itself is a plane. What happens if f(x) = log(x)? x * log(y / x) looks too weird to be a simple plane.

In 2D, it''s even clearer to me that I don''t understand how this is possible How can a graph have tangent lines at every point that pass through the origin if the graph itself is not a straight line??

To Zipster,

(f(u))'' = f''(u)*du
(f(x/y))'' = f''(x/y)*(x/y)''

I had a lot of "fun" doing the demonstration because of stupid sign problems, but it''s not really hard.

Cédric

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Zipster    2365
Alright, so you just leave it in a generic form. That's what I wanted to hear

And you just had to give me something else to plug into my program! Actually, f(x) = log(x) looks a lot clearer than any of the other graphs:



It would appear to me that all tangeant planes pass through the origin.

And just incase you want to explore: MathGV
It's the cool graphing program. Once you plot a function, you can rotate and zoom and do all sorts of things. It's a lot easier to see the shape of the graph when it's rotating.

[edited by - Zipster on April 29, 2002 6:52:59 PM]

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Cedric    158
MathGV sure looks like a nice program, but with such simple visuals, I think it yearns to be done in Java and put on the web.

Now that I think of it, isn''t there any Maple-like software available freely in Java? Explaining stuff would be so much easier if we had a simple java applet that could draw 3D graphs like MathGV and, most importantly, that could format equations so that they look even half-decent. It could be embedded into Gamedev.net''s forum system, like the code snippets are.

Sigh...

Cédric

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grhodes_at_work    1385
Cédric,

This was an interesting problem, and clearly you did work on this independently as well as with some assistance from folks here. But in the future please do not post homework questions here. Its nothing personal, just a matter of principle, .

I''m closing the thread now.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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