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Calculus 3

  1. Jul 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Describe and sketch the surface.

    (y^2)+(4z^2)=4



    2. Relevant equations
    It appears that the sketch will be an ellipsoid. Because the problem instructs me to describe and sketch the surface, I don't believe there are any useful equations.



    3. The attempt at a solution
    I don't know where to start. I think the graph will wrap around the variable not included in the problem. If you will, please assist me.
     
  2. jcsd
  3. Jul 15, 2009 #2
    Since the equation is independent of x we know that it will be some sort of "cylinder" that has cross-sections in the yz-plane shaped like (y^2)+(4z^2)=4 (which is an ellipse). So not so much an ellipsoid, but an elliptical cylinder (if that's what you call it; if anyone knows a more technically correct term, I would love to learn it).
     
  4. Jul 15, 2009 #3
    That equation describes an ellipse in the yz-plane (look up in your book or online how equations of ellipses work). Sketching the surface in R3, no matter what the variable x is, you will still have the same curve in the yz-plane at the point x. This means that you will end up with an elliptical tube running along the x-axis.

    EDIT: Sorry, I started posting before anyone, but cipher42 finished before me. We said the exact same thing. :)
     
    Last edited: Jul 15, 2009
  5. Jul 15, 2009 #4

    Mark44

    Staff: Mentor

    Elliptical cylinder is the right terms for it.
     
  6. Jul 15, 2009 #5
    I'm most appreciative for your help. I understand that I must have more familiarity, namely, with circles, ellipses, hyperbolas, etc. What was confusing for me is that 2 squared variables equal to 1, I thought, resulted in a circle. And although this might sound stupid, what I'm dealing with now is 2 squared variables equal to 4 resulting in an elliptical cylinder. I'm embarrassed to own my ignorance, but in my ignorance I don't understand the change. Maybe I just need to read the book more. Forgive me for wasting your time.:confused:
     
  7. Jul 15, 2009 #6
    n!kofeyn, i shall do just what you suggested and i appreciate your input for it helped me out even the more. i, again, am most appreciative for everyone's help.

    i am the PotnlMathmtcn.
     
  8. Jul 15, 2009 #7
    PotnlMathmtcn, don't sweat it; these things take time to learn. But you are right, when doing harder problems like this, it definitely helps to be very familiar with the more basic concepts from 2-dimensional geometry.

    Perhaps a little refresher will help clear some stuff up:

    The equation for a circle with radius [itex]r[/itex] centered at the origin is:
    [tex]x^2+y^2=r^2[/tex]
    You can shift the origin to the point [itex](x_0,y_0)[/itex] by subtracting the coordinates from their respective variables to get:
    [tex](x-x_0)^2+(y-y_0)^2=r^2[/tex]
    Next, you can squish/stretch the circle into an ellipse by dividing [itex]x[/itex] and [itex]y[/itex] terms by constants to get
    [tex]\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=r^2[/tex]

    This is what you are looking at in your problem with [itex]x_0=0[/itex], [itex]y_0=0[/itex], [itex]a=1[/itex], and [itex]b=\frac{1}{2}[/itex], [itex]r=2[/itex] (of course, you're in the yz-plane instead of the xy-plane, but that's all just names) so it will be an ellipse with its center at the origin that could have been a been a circle of radius 2, but b=1/2 means that it gets squished by half in the y-direction.
     
  9. Jul 15, 2009 #8
    You haven't wasted our time by any means. We choose to reply to the threads we reply to, and there wouldn't be any point to this forum if questions like yours weren't asked. I myself forget the details of ellipses, but just get to where you recognize the general form of equations and be able to look up or work out the details. cipher42's post should be very helpful.
    Haha. I like this. Is your name "Potential Mathematician"?
     
  10. Jul 15, 2009 #9
    Here's another problem. I'm given the vector (t, cos2t, sin2t) and I'm asked to sketch the curve with the given vector equation. Indicate with an arrow the direction in which t increase.

    What I do know are the parametric equations, x=t, y=cos2t, z=sin2t, but from this point know I not in what direction to pursue this problems solution.

    For all help offered do I appreciate in advance.

    I am the PotnlMathmtcn.
     
  11. Jul 15, 2009 #10

    Dick

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    I don't know how to explain to you how to "sketch a curve" in 3 dimensions. If t=0 then (x,y,z)=(0,1,0), if t=pi/4 then (x,y,z)=(pi/4,0,1) etc, etc. Can't you figure out a way to visualize the curve and draw it? It's a helix.
     
  12. Jul 15, 2009 #11

    fluidistic

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    Is it a helix defined on the surface of a cylinder? If so, what would be the radius of the cylinder? (Take note that I'm not the OP).
     
  13. Jul 15, 2009 #12

    Dick

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    It's certainly on the surface of the cylinder y^2+z^2=1, isn't it, non-OP?
     
  14. Jul 15, 2009 #13

    fluidistic

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    Oh... So the parametrized curve [tex](t, \cos t, \sin t)[/tex] is on the same surface. I doubted a bit and then I thought it was much more complicated to obtain the equation of the cylinder, because [tex]\cos 2t = \cos ^2 t - \sin ^2 t[/tex].
    To the OP : sorry for being a bit curious here, but I'm also learning calculus 3 now. Disregard my question.
    And thanks once again, Dick.
     
  15. Jul 15, 2009 #14

    Dick

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    It's exactly the same curve as (t/2,cos(t),sin(t)), just reparametrize the t. (t,cos(t),sin(t)) is a different curve but on the same cylinder. It just has a different 'pitch'. But I think you know that.
     
  16. Jul 15, 2009 #15

    fluidistic

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    Ah now I fully visualize it in my head. For example if we had the curve [tex](t, \cos (10^9 t), \sin (10^9 t))[/tex], we'd see it as the cylinder (if we are not too close of the cylinder).
    In order to change the radius we'd have to had something like [tex](t, a \cos (t), b \sin (t))[/tex] and if [tex]a \neq b[/tex] then the curve would fit on the surface of an elliptical cylinder.
    Thanks.
     
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