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Homework Help: Curves and surfaces, Transformations

  1. Nov 29, 2007 #1
    1) http://www.geocities.com/asdfasdf23135/advcal13.JPG
    Let F1 = x^2 - y^2 + z^2 -1 = 0
    F2 = xy + xz - 2 = 0
    F3 = xyz - x^2 - 6y + 6 = 0
    My thought is to compute the gradients, grad F1 and grad F2. Then by taking their cross product, I can get a tangent vector v for the curve. Now, I can feel that if I can show that (v) dot (grad F3) = 0, then I am done. But I don't quite understand why. Can someone explain why this works?

    Also, provided that I've shown (v) dot (grad F3) = 0, can the curve and surface be simply "parallel" to each other, rather than "tangent" to each other? If yes, what should I do? If not, why not?

    2) http://www.geocities.com/asdfasdf23135/advcal14.JPG
    I am OK with parts (i) and (iii), but I am stuck with part (ii), can someone please guide me through the steps of solving this part?

    [Related concepts: Curves and surfaces, smoothness, implicit function theorem, inverse function theorem, transformations and coordinate systems]

    Any help is greatly appreciated!
    Last edited: Nov 29, 2007
  2. jcsd
  3. Nov 29, 2007 #2


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    1) grad F1 cross grad F2 is not the tangent vector to the curve of intersection of the surfaces F1 and F2 (as this cross product will exist at points where the curves do not intersect). Rather set F1=F2 to find the curve of intersection and proceed from there.

    2) (ii) consider v=0
  4. Nov 29, 2007 #3
    1) I don't get it, can someone please explain a bit more...

    2) So do we just have to remove v=0? How do you know that there are no other?

  5. Nov 30, 2007 #4


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    What benorin is saying is what you alluded to in your first post: "can the curve and surface be simply "parallel" to each other, rather than "tangent" to each other? " -yes, it might happen that the curve and surface do not have that point in common. However, here you do NOT have to set F1= F2 because you are given a specific point: (1, 1, 1). The first thing you need to do is determine if THAT point is on both the curve and the surface. Since the curve is defined as the intersection of two surfaces, it will be on the curve if and only if it is on both surfaces. In other words, put x= y= z= 1 into each of the three formulas and see if it satifies them. If it does not satisfy any one of the equations, it is not a point of intersection of the curve and surface and so they cannot be tangent there. If x=y=z=1 does satisfy all three equations the it is on both curve and surface and from there your logic is correct: the cross product of the normal vectors to the first two surfaces will be tangent to the curve of intersection and that will be tangent to the third surface if and only if it is normal to a normal vector of that surface.

    As for (2), you say that you have proved that f(u,v)= (uv,u2v) is not one-to-one on [1, 2]x[0, 1]. That means you can show that f(u1,v1)= f(u2, v2) where at least one u1, v1 is not equal to the corresponding u2, v2. Okay, in doing that where did you find "duplicate" values? You need to remove one of the two sets so that you do not have duplicate values. For example, in one dimension f(x)= x2 because we get the same value for both positive and negative x. Removing x< 0 from the domain gives a one-to-one function. Similarly, removing x> 0 from the domain gives a one-to-one function.
  6. Nov 30, 2007 #5
    1) Thanks, this is very helpful!

    2) For part (i), I just picked v=0 to show that f is not 1-1 on S.
    But the trouble is that "are there any other points that need to be eliminated?" I am not sure about this...
    And also, how can I prove that f is 1-1 on a restricted region S\S1 ?
  7. Nov 30, 2007 #6


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    v= 0 is not the only reason f is not 1-1! what if u= 0?

    If f is NOT 1-1 then there exist distinct points, (u,v) and (x,y), such that f(x,y)= f(u,v). That is, such that uv= xy and such that [itex]u^2v= x^2y[/itex]. Obviously, if v and y are 0, then for any u, x, f(u,0)= [itex](u(0),u^2(0))= (x(0),x^2(0))[/itex]= f(x,0). Just as obviously, if x and u are 0, f(0,v)= [itex](0(v),0(v))= (0(y),0(y))[/itex]= f(0,y).
    If neither none of u, v, x, y are 0 then uv and xy are not 0 and so we can divide each side of [itex]u^2v= x^2y[/itex] by them: [itex]u^2/uv= x^2y/xy[/itex] so u= x. Once we know that u=x, and they are not 0, we can divide both sides of uv= xy to get u= y. The only points that "duplicate" values are (u, 0) and (0, v).
  8. Nov 30, 2007 #7
    But u can't be 0, right? (since we are considering S= [1,2] x [0,1])
    Now restricted to this rectangle S, I am not sure how to locate ALL the points that make f not 1-1...and then I have to eliminate ALL these in order to make f 1-1.
  9. Dec 1, 2007 #8
    I've thought of a proof, can someone please tell me if this works?

    2(ii) Set uv=u'v' and u^2 v = u'^2 v'
    Divide 2nd equation by 1st equation
    =>u=u' (provided u not=0 and v not=0)
    Substitue this into the first equation
    =>v=v' (provided u not=0)

    but u is never 0 in the given rectangle S, so v=0 is the only trouble. Let S1={(u,v)| v=0 and 1<u<2}, then on S\S1, f must be 1-1.

    Am I right?
  10. Dec 1, 2007 #9


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    Yes, you are correct.
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