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More vector calculus

  1. Sep 19, 2009 #1

    rock.freak667

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

    Sketch the cone z2=x2+y2 in 3D space.

    Let (x0,y0,z0)≠(0,0,0) be a point on the given cone. By expressing the fiven equation of the cone in the form f(x,y,z)=a, find a normal vector tot he cone at point (x0,y0,z0)

    Find the equation of the tangent plane to the cone at point(x0,y0,z0)

    Show that every plane that is a tangent to the cone passes through the origin

    2. Relevant equations

    [tex]\hat{n} =grad(f) \ at \ (x_0,y_0,z_0) [/tex]

    3. The attempt at a solution

    I was able to do part 2 and 3.

    [tex]\hat{n}= 2x_0 \hat{i} +2y_0 \hat{j} -2z_0 \hat{k}[/tex]

    and then tangent plane is

    2x0(x-x0)+2y0(y-y0)-2z0(z-z0)=0

    so if it passes through the origin x=0,y=0 and z=0, which leaves me with

    [tex]x_0 ^{2}+y_0 ^{2}-z_0 ^{2}=0[/tex]

    But since (x0,y0,z0)≠(0,0,0), then doesn't this mean that it does not pass through the origin?


    Also, how do I sketch a curve in 3D? Do I just randomly plug in values of (x,y,z) and plot or is there some method in doing it?
     
  2. jcsd
  3. Sep 19, 2009 #2

    gabbagabbahey

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    Remember, the point [itex](x_0,y_0,z_0)[/itex] lies on the cone, so [itex]z_0^2=x_0^2+y_0^2[/itex].


    I would plug a few different values of [itex]z[/itex] into the equation of the cone, and sketch the resulting curve at that height (value of [itex]z[/itex]), it should become clear pretty quickly what the cone looks like.
     
  4. Sep 19, 2009 #3

    rock.freak667

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    thanks



    But I don't get how I would know what it looks like :confused:

    on the yz plane z2=y2 which means y=±z. How does the curve have two straight line portions ?


    EDIT: yeah this is the first time I've done these things and the person teaching this to me is a bit...hard to talk to, so I can't ask.
     
  5. Sep 19, 2009 #4

    gabbagabbahey

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    Well, let's see...

    For [itex]z=0[/itex] we have [itex]x^2+y^2=0[/itex], which is a circle of radius zero, i.e. just a point at the origin.

    For [itex]z=1[/itex] we have [itex]x^2+y^2=1[/itex], which is a circle of radius 1 at height [itex]z=1[/itex].

    For [itex]z=-1[/itex] we have [itex]x^2+y^2=1[/itex], which is again a circle of radius 1, this time at a height [itex]z=-1[/itex].

    For [itex]z=2[/itex] we have [itex]x^2+y^2=4[/itex], which is a circle of radius 2 at height [itex]z=2[/itex].

    For [itex]z=-2[/itex] we have [itex]x^2+y^2=4[/itex], which is again a circle of radius 2, this time at a height [itex]z=-2[/itex].

    As you get further and further away from z=0, in either direction, the circles get larger and larger...

    http://img268.imageshack.us/img268/1341/cone2t.th.jpg [Broken]

    You should see clearly now why there are two lines (in the shape of an X) in the yz plane.
     
    Last edited by a moderator: May 4, 2017
  6. Sep 19, 2009 #5

    rock.freak667

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    oh I see now, so the best thing to do is to try to vary one variable and see what shape is formed by the x and y ( a circle in this case with increasing radius)?

    Sketching on paper is gonna be a bit odd

    thanks!
     
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