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Sketching level curves and the surface of a sphere

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data
    (a) Sketch the level curves of z = (x^2 - 2y +6)/(3x^2 + y) at heights z = 0 and z =1.
    (b) Sketch the surface (x−1)^2 + (y+2)^2 + z^2 = 2 in R^3. Write down a point which is on the surface.

    2. Relevant equations
    --

    3. The attempt at a solution
    (a) From the question, I assume that you would draw the equation in 2D on the xy-axis and then mark the curves z = 0 and z =1. However, what I am unsure of how is to how to actually sketch the level curves and then find at what values would z = 0 and z = 1 be drawn.
    (b) From the equation, I recognise that the surface will be a sphere. But is R^3 the xyz-plane? If so, then how would I be able to sketch this as a sphere (on paper)? Would sketching it within a cube be appropriate? Moreover, how would I find a point which is on the surface? Would a point on the surface be an integer solution? (For example, x=0, y=-3, z=0 so, (0,-3,0)).
     
  2. jcsd
  3. Apr 24, 2012 #2

    Mark44

    Staff: Mentor

    No, you have the steps reversed. First, set z = 0, and then graph that equation. That will be the level curve for z = 0.
    Second, set z = 1, and then graph that curve. You will have a different equation for each level curve.
    There is no xyz plane, not is there an xy-axis, as you said above. You can talk about the x-axis or y-axis or z-axis - these axes are straight lines. You can also talk about the xy-plane or xz-plane or yz-plane. These are coordinate planes.
    Can you draw the part of the sphere that is in the first octant?

    Does the point (0, -3, 0) satisfy the equation (x−1)^2 + (y+2)^2 + z^2 = 2? If so, the point is on the surface of this sphere. If not, it isn't. Since you have three variables and only one equation, you can arbitrarily set any two variables and then solve for the third.

    Since you know that this equation is of a sphere, do you also know the other things of interest, such as the location of its center and its radius? You can pretty much pick this information off the equation.
     
  4. Apr 25, 2012 #3
    Okay, thanks. I set z = 0 and then sketched the level curve. I got a positive parabola, intersecting the y-axis at 3 (no roots). Is that correct?
    For z = 1, I got a negative parabola, intersecting the y-axis at 2, with roots -2 and 2. Is that correct? Also, would I draw each level curve on separate axes or on the same one?

    Sorry, I don't understand what you mean by this. How would you be able to figure out the arc of the circle that would be in the first octant? (Sorry, I have never sketched a sphere before.)

    Okay, so (0, -3, 0) does not satisfy the equation then. I set two variables, x = 2, y = -2. With z = 1. I then substituted them into the equation and the equal to 2. So, does that therefore mean (2, -2, 1) is a point on the surface of this sphere then?

    Would the center be (1, -2, 0) and the radius √2?
     
  5. Apr 25, 2012 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes.

    Yes.


    ehild
     
  6. Apr 25, 2012 #5
    Okay, thank you. However, I am confused with actually sketching the sphere. So, I understand that the center must be (1, -2, 0) with radius √2. Yet, how would I arrange the axes to sketch the sphere? Since y=-2, it exists on the negative y axis. & I assume you would arrange the axes (appropriately and correctly) so that it forms a cube for the sphere to be within, right? That, I'm unsure of how to do as would all the axes still meet at the origin?
     
  7. Apr 25, 2012 #6

    ehild

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    Homework Helper
    Gold Member

    The axes meet at the origin, the centre of the sphere is at (1,-2,0). You can use the level lines to make the sphere look 3-dimensional. Perhaps something like in my figure.

    ehild
     

    Attached Files:

  8. Apr 25, 2012 #7
    Okay, thanks.

    Moreover, is this correct for part (a)?
     
  9. Apr 25, 2012 #8

    Mark44

    Staff: Mentor

    "Positive" doesn't have much meaning here. It would be better to say that it opens up or down or to the left of right. In those terms how would you describe the parabola?
    The parabola opens downward, and has a y-intercept at (0, 2), but your x-intercepts are wrong.

    If you graph both level curves on the same axis, label each one as z = 0 or z = 1. For this problem the two level curves don't give much insight as to the shape of the three-dimensional surface in the problem.

    Topographical maps are essentially the same thing as level curves. On such a map high points and low points are shown with closed curves that are at the same height. Hikers can use these maps to determine how steep a trail will be.
     
  10. Apr 25, 2012 #9
    Okay. Then, the parabola opens upward, with the y-intercept at (0,3). It has no x-intercepts but the opening (with how it expands/terms of width) of the curve is within x=-4, x =4.

    Yes, I realised that my x-intercepts are wrong. The x-intercepts are √3 and -√3, right?

    So, should I draw both level curves on the same axis then? Or is it also fine to draw them both on separate axes? With the level curves as well, it would be the parabola on the xy-axis with "z=0" or "z=1" next to it, right? & That would answer the question?
    Also, the question is just asking for the level curves and not for the shape of the surface itself, right (with the shape of the surface more focused in part (b))?
     
  11. Apr 25, 2012 #10

    Mark44

    Staff: Mentor

    No, the parabola is not confined to the interval [-4, 4].
    Right.
    That's probably best. (BTW, it would be the same axis system. I mistakenly wrote "axis" when I should have written "axes" or "axis system". You can't graph a parabola on just a single axis.
    Again, there is no such thing as "xy-axis". I would advise that you draw both level curves on the same axis system, and close to each curve write z = 0 or z = 1 to identify them.

    Right.
     
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