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Math Help - calcii: contours

  1. #1
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    calcii: contours

    Hi, I am having trouble with contours, is there a procedure to do these?

    the question I am working on is when z=c

    z=((x^2)+(y^2))/2x

    what i tried to do:

    c2x=(x^2)+(y^2)
    c2x-x^2=y^2
    x(2c-x)=y^2
    2c-x=(y^2)/x
    2c=((y^2)/x)+x

    and that is where I got stuck. I cant visualize what is going on here

    thanks for any replys
    Last edited by jameskw; December 7th 2010 at 04:44 PM.
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  2. #2
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    Clarksville, ARk
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    Quote Originally Posted by jameskw View Post
    Hi, I am having trouble with contours, is there a procedure to do these?

    the question I am working on is when z=c

    z=((x^2)+(y^2))/2x

    what i tried to do:

    c2x=(x^2)+(y^2)
    c2x-x^2=y^2
    x(2c-x)=y^2
    2c-x=(y^2)/x
    2c=((y^2)/x)+x

    and that is where I got stuck. I cant visualize what is going on here

    thanks for any replies

    What you have done is equivalent to:

    \displaystyle c={{(x^2)+(y^2)}\over{2x}}

    \displaystyle 2c={{(x^2)+(y^2)}\over{x}}

    \displaystyle 2c=(x)+{{y^2}\over{x}}

    What you need after your line, c(2) x-x^2=y^2 is:

    0=x^2-2cx+y^2, then add \displaystyle c^2 complete the square.

    c^2=x^2-2cx+c^2+y^2

    Finally you have c^2=(x-c)^2+y^2.

    Do you recognize this?
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  3. #3
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    Well, it looks like its close to a circle because of its form. As c increases radius increases and x gets displaced.
    sounds like 2 cones with a slope, how would i find a slope?

    thanks
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  4. #4
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    Clarksville, ARk
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    Calc II, contours

    Quote Originally Posted by jameskw View Post
    Well, it looks like it's close to a circle because of its form. As c increases radius increases and x gets displaced.
    sounds like 2 cones with a slope, how would i find a slope?

    thanks
    Yes! For any fixed value of c, c^2=(x-c)^2+y^2 is the equation of a circle with radius, c, centered on the x-axis at x=c. This circle passes through the origin.

    When looked at in a more general way, c^2=(x-c)^2+y^2 is a family of such circles.

    Therefore, the equation, \displaystyle z={{(x^2)+(y^2)}\over{2x}}, describes a pair of oblique circular cones, each with a vertex at
    (0, 0, 0) and axis along the line z=x in the x-z plane. One opens upward, the other downward.

    You can also look at this as a pair of right elliptical cones, each with a vertex at (0, 0, 0) and axis along the line z={4\over3}x in the x-z plane. (I haven't figured out the eccentricity of the ellipses.)
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  5. #5
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    c= \frac{x^2+ y^2}{2x} gives 2cx= x^2+ y^2, x^2- 2cx+ y^2= 0
    Now "complete the square": x^2- 2cx+ c^2+ y^2= c^2, (x- c)^2+ y^2= c^2,
    a circle with center at (c, 0) and radius c so it is tangent to the y-axis at (0, 0).
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