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How to interpret a function in 3 dimentions

  1. Nov 20, 2013 #1
    1. The problem statement, all variables and given/known data
    I have calculus, and a lot of tasks ask for drawing of the functions in 3 dimentions. I have problems thinking of how to do this, often it can be a parabolid or other Objects. Example, one I cant think of how im suppose to Draw, here Im going to solve for Stokes (that is a pretty easy function) x^2 +y^2 +5z =1 , where z≥0.

    Is there any good technique to use for drawing With Three variables?

    2. Relevant equations
    I dont know what to use here


    3. The attempt at a solution
    I think that if I put z=0, you get a circle in the xy-plane With a radius 1, but when z=1, I get X^2 +y^2 =-4, so the New radius is -4 ( or is it √-4?) i dont think im thinking correctly here.
     
  2. jcsd
  3. Nov 20, 2013 #2

    Mark44

    Staff: Mentor

    That's right - you aren't.
    For this surface, you can get a lot of information from cross-sections that are parallel to the x-y plane. To do this, rewrite the equation as x2 + y2 = 1 - 5z.

    Since x2 + y2 ≥ 0 for all real x and y, then it must be that 1 - 5z ≥ 0 as well. If you substitute z = 1, you don't get a circle with an imaginary radius. It means that there are no points on the surface for which z = 1.

    If 1 - 5z = 0, what you get is a degenerate circle - a single point. Try substituting a few values of z for which 1 - 5z > 0. All of these will give you cross-sections at various heights that will help you recognize what the surface looks like.

    Another technique that can be used is to look at the three traces, cross-sections of the curve in the x-y plane (set z = 0), x-z plane (set y = 0), and y-z plane (set x = 0).
     
  4. Nov 20, 2013 #3
    Thanks! But how do you see that 1-5z=0 is a single point?
    And to get 1 - 5z > 0, dosent z have to be negative? to have z=, -1, -2 -3 , but that cant be becous z≥0. and does this gives us the radius of the parabolid?
     
  5. Nov 20, 2013 #4

    haruspex

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    It clearly fixes the value of z. If you substitute that value in the original equation, what do you get? I think you'll find there is only solution for x and y as well.
    No, think that through again.
     
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