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Vector calculus problem

  1. May 13, 2006 #1
    Does anyone know how do #1? I thought I just needed to find the binormal and it will give me the equation of the tangent plane, however the calculations are too insane for that to be the way to solve this. I can't think of any other way, since we're dealing with a vector function and not the equation, ie the are no defined points on the 3D function.

  2. jcsd
  3. May 13, 2006 #2
    For part 1, the normal to the plane is given by the grad vector. Then, using point A, the equation of the tangent plane can be found. I think that would work
    [tex] n = \frac{\partial{z}}{\partial{x}}i + \frac{\partial{z}}{\partial{y}}j + \frac{\partial{z}}{\partial{{z}}k [/tex]
    [tex] n.(r-a) = 0 [/tex] where a is the position vector of A, and [tex] r = xi + yj + zk [/tex]

    Sorry about the maths bits, it was my attempt to do partial derivatives.
    Last edited by a moderator: May 13, 2006
  4. May 13, 2006 #3


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    Try writing the equation for the surface in the form f(x,y,z)=0, ie, treat it as a constant surface of some function in R^3. Then the gradient of this function at a point represents the vector normal to the surface at the point, and from this it should be easy to find the tangent plane.
  5. May 13, 2006 #4
    But how do I know what point will A be on the surface?
  6. May 13, 2006 #5
    I think I have it: the point A is just (1/25,1/10,3/25). Oh my God how could I be so stupid to miss that!!! OMG!!!

    Now its super easy to find the solution to Q1.
  7. May 13, 2006 #6
    I'm pretty sure you got it. But just to clarify what [tex] \vec{0A} [/tex] means.

    A vector is an object that has direction and length. Given [tex] \vec{0A} = \left(\frac{1}{25}, \frac{1}{10}, \frac{3}{25}\right)[/tex] this means that you have a vector that starts at position 0, which is [tex] \vec 0 = (0,0,0) [/tex] and then moves to the right out in the x direction by 1/25 then in the y direction by 1/10 and then in the z direction by 3/25. So in x-y-z space, this vector points to something. What does it point to?

    Well you know your surface is defined by:
    [tex] z=\left (\frac{1}{2}-y \right)^2 - x [/tex]

    And you have a vector [tex] \vec{0A} = \left(\frac{1}{25}, \frac{1}{10}, \frac{3}{25}\right)[/tex].

    Which in your coordinate system is:
    [tex] x = \frac{1}{25} [/tex]
    [tex] y = \frac{1}{10} [/tex]
    [tex] z = \frac{3}{25} [/tex]

    So if you want to see if the "point" A is actually on your surface. Then the vector must point to an actual element of the set that makes up the surface.

    So take your surface:
    [tex] z=\left (\frac{1}{2}-y \right)^2 - x [/tex]

    then take your vector:
    [tex] x = \frac{1}{25} [/tex]
    [tex] y = \frac{1}{10} [/tex]
    [tex] z = \frac{3}{25} [/tex]

    and you get:
    [tex] z= \left(\frac{1}{2}-\frac{1}{10}\right) ^2 - \frac{1}{25}[/tex]
    [tex] z=\left( \frac{2}{5} \right)^2 - \frac{1}{25} [/tex]
    [tex] z=\frac{4}{25}-\frac{1}{25}=\frac{3}{25} [/tex]

    So you know the point WILL be on the surface.
    Last edited: May 13, 2006
  8. May 13, 2006 #7
    Now for question #4 I found the volume of the bottle to be 0.012414 cubic meters, anyone else got a similar answer? What integral did you set up?
  9. May 13, 2006 #8
    By the way, for #6 am I correct in doing grad(f) at A dot product with r'(1/20)?
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