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Math Help - area4

  1. #1
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    area4

    the graph area4-codecogseqn-2-.gif divides the planes into regions .the area of bounded region is
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  2. #2
    Newbie Auri's Avatar
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    Re: area4

    The graph of the bounded region is rectangle. It has a length of 40, and width of 10.
    A = L * W
    A = 40 * 10
    A = 400.
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  3. #3
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    Re: area4

    how it is a rectangle it can also be a curve
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  4. #4
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    Re: area4

    If you add and subtract y^2 you get x^2+ 2xy+ y^2- y^2+ 40|x|= (x+y)^2- y^2+ |x|= 400. Let u= x+y and v= y. x= u- y= u- v so the equation becomes u^2- v^2+ 40|u- v|= 400.

    For x= u- v>0, |x|= |u-v|= u- v so that becomes u^2+40u- v^2- 40v= 400. Complete the square by adding 400 and subtracting 400 on the left: u^2+ 40u+ 400- (v^2+ 40v- 400)= (u+20)^2- (v+20)^2= 400. That is a hyperbola, in the uv-plane, with center at (-20, -20), vertices (0, -20) and (-40, -20), and vertical asymptotes at 45 degree angles to the axes. That is, one of the asymptotes is the line u= v and, since we require that u- v> 0 here, we have only the branch of the hyperbola on that side.

    If u- v< 0, then |u- v|= v- u> 0 and the equation becomes u^2- 40u- v^2+ 40v= 400 and, after completing the square, (u- 20)^2- (v- 20)^2= 400, a hyperbola with center at (20, 20), vertices (0, 20) and (40, 20), and u= v and u= -v as asymptotes again. Since u< v now, we have the branch of the hyperbola now on the opposite side. Those don't cross: one lies on one side of v= u and the other on the opposite side, so I don't see how they can create a bounded region!
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  5. #5
    Member sbhatnagar's Avatar
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    Cool Re: area4

    I plotted y^2+2xy+40|x|=400 using mathematica 8. You can also check the graph by clicking here.
    Attached Thumbnails Attached Thumbnails area4-graph00.bmp  
    Last edited by sbhatnagar; October 24th 2011 at 01:26 AM.
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