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Math Help - how to find this area on the plane?

  1. #1
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    how to find this area on the plane?

    you can replace x y instead of u v
    i cant think of a way to find the are described here
    ?
    Last edited by transgalactic; February 6th 2010 at 11:38 AM.
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    you can replace x y instead of u v
    i cant think of a way to find the are described here
    ?
    If I didn't made a mistake the area in question are 2 rectangles. See attachment.
    Attached Thumbnails Attached Thumbnails how to find this area on the plane?-flaeche_unglg.png  
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  3. #3
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    you made a mistake its a rectangle wit a hole inside

    but i cant see how?
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    you made a mistake its a rectangle wit a hole inside

    but i cant see how?
    As I've presumed I've made a mistake. You are right. It looks like that:
    Attached Thumbnails Attached Thumbnails how to find this area on the plane?-flaeche_unglg2.png  
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  5. #5
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    Yes but i cant take a computer to my test

    how i figure out that there is a hole inside the rectangle?
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    Yes but i cant take a computer to my test

    how i figure out that there is a hole inside the rectangle?
    1. To be honest: The only thing my computer did was drawing the coordinate system.

    2. Consider D=\{(u,v) | max(|u|, v) \le 2~\wedge~v\ge 0\}

    Then you'll get a rectangle from u = -2 to u = 2 and from v = 0 to v = 2.

    3. If you take the additional condition max(|u|, v) \ge 1 into account you have to exclude all those points whose both coordinates u and v start with 0.

    To give you an example:

    P(-1.2 , 0.3) belongs to the area because |u| \ge 1

    Q(0.7 , 0.2) does not belong to the area because both coordinate values are smaller than 1.

    4. All points R(|0.###| , 0.***) are situated in a rectangle from u = -1 to u = 1 and from v = 0 to v = 1
    Last edited by earboth; February 8th 2010 at 12:30 AM. Reason: typo
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  7. #7
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    As earboth says, that hole is because of the condition max(|u|,v)\ge 1.

    If, in the u,v plane, you draw the lines v= 1, u= 1, u= -1, notice that any point inside those lines has v< 1 and -1< u< 1 or |u|< 1. Saying " max(|u|, v)\ge 1" means neither of those can happen and so we are outside or on those lines. Similarly, the lines v= 2, |u|= 2 (or u= -2 and u= 2) form the outer "box", from max(|u|,v)\le 2. The lower boundary is at v= 0, of course, because of the condition v\ge 0.

    If the fact that these are u and v, rather than x and y is bothering you, just go ahead and replace with u with x and v with y- it is exactly the same thing in the xy-plane. If fact, I notice that earboth, in his last post, has accidently replace the condition v\ge 0 with y\ge 0!
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