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Math Help - surfaces and contour maps

  1. #1
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    surfaces and contour maps

    hi im new to the forum . i needed help on the following question. sorry if i posted this in the wrong area.

    1. draw a level curve diagram (with atleast 3 contours) for the surface: z^2 = 9 x^2 + 4 y^2

    2. sketch a graph of the surface: z^2 = 9 x^2 + 4 y^2

    any help much appreciated
    Last edited by mr fantastic; August 15th 2009 at 02:24 AM. Reason: No edit - just flagging the post as having been moved to Calculus subforum (I think this subforum is best for it)
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  2. #2
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    Contour maps

    Imagine the xy-plane as being the ground, and the z-axis represents height. So, z=0 is "ground" level, z=1 is one unit above the ground, z=2 is two units up, etcetera. What we are doing is taking horizontal cross-sections of the 3D graph.

    First, rewrite the function as 1=\left(\frac{x}{z/3}\right)^2+\left(\frac{y}{z/2}\right)^2

    So each slice of the graph is a vertical ellipse with axes z/2 tall and z/3 wide.

    z=0 \to 0=9x^2+4y^2 The only point (x,y) satisfying this is the origin, (0,0)
    z=\pm 6 \to 1=\left(\frac{x}{2}\right)^2+\left(\frac{y}{3}\rig  ht)^2
    z=\pm 12 \to 1=\left(\frac{x}{4}\right)^2+\left(\frac{y}{6}\rig  ht)^2

    See if you can visualize it from here.
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  3. #3
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    Hi im doing a similar question but confused on how you rewrote the equation into the ellipse standard form. Can you please help?
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  4. #4
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    Basic Algebra

    z^2=9x^2+4y^2
    z^2=(3x)^2+(2y)^2
    z^2=\left(\frac{x}{1/3}\right)^2+\left(\frac{y}{1/2}\right)^2
    1=\left(\frac{x}{z/3}\right)^2+\left(\frac{y}{z/2}\right)^2

    Since an ellipse has the form: 1=\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\rig  ht)^2, then the above equation represents an ellipse with major axis z/2 and minor axis z/3.
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  5. #5
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    hyperbola contours

    Thanks that helps, but im doing it and iv realised its a hyperbola. Can i ask you if what iv done so far is correct.

    z^2=4+9x^2-y^2
    z^2=c
    c-4=9x^2-y^2

    for which c>0 and -infinity< z < infinity

    (3x)^2 - (y)^2 = c-4

    (x/(1/3))^2 - (y)^2 = c-4

    (x/(sqrt(c-4)/3)^2 - (y/sqrt(c-4))^2 = 1

    So then i need values of c to draw the initial rectangles, which i can then draw the the curves. The problem is on the question it says for
    c>=4 and c<4. But since its a square root which i got c cant be less than 4. Its infinity but thats not the solution. Theres a hyperbola for c less than 4. If you can, any help would be really appreciated.

    Thanks so much.
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