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Math Help - [SOLVED] Linear substitution

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Linear substitution

    D is the triangle with corners in (0,0), (1,-1) and (3,0).

    If we make the substitution: u=x+y and v=x-2y we end up with a triangle E with corners in (0,0), (0,3) and (3,3).

    The triangle D is formed by the three lines x+y=0, x-2y=3 and y=0, but I'm not quite sure how you're supposed to think when you transform this to uv coordinates (the E triangle).
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  2. #2
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    Quote Originally Posted by Spec View Post
    D is the triangle with corners in (0,0), (1,-1) and (3,0).

    If we make the substitution: u=x+y and v=x-2y we end up with a triangle E with corners in (0,0), (0,3) and (3,3).

    The triangle D is formed by the three lines x+y=0, x-2y=3 and y=0, but I'm not quite sure how you're supposed to think when you transform this to uv coordinates (the E triangle).
    just transform each sides of the triangle of D: the side x+y=0, \ 0 \leq x \leq 1, \ -1 \leq y \leq 0 goes to u=x+y=0, \ v=x-2y=-3y. since -1 \leq y \leq 0, we'll get: 0 \leq v \leq 3.

    so that side of D is transformed to u=0, \ 0 \leq v \leq 3. now let's see what happens to the second side of D, i.e. x-2y=3, \ 1 \leq x \leq 3, \ -1 \leq y \leq 0. we'll have: v=x-2y=3

    and u=x+y=3y+3. since -1 \leq y \leq 0, we'll have 0 \leq 3y+3 \leq 3. so the second side of D goes to 0 \leq u \leq 3, \ v = 3 in uv plane. the third side of D is y=0, \ 0 \leq x \leq 3. thus

    u=x+y=x, \ v=x-2y=x. so u=v=x. hence the image of the third side of D in uv plane is the line segment v=u, \ 0 \leq u \leq 3.
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