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Math Help - Parameterize by a continuous map a polygonal curve

  1. #1
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    Parameterize by a continuous map a polygonal curve

    Hi, some help with this problem in intro analysis please:

    Parameterize by a continuous map f:[0,1]-->S in R^3 the polygonal curve L from (0,0,0) to (2,1,1) to (2,3,2) to (3,3,3).

    Thanks a lot.
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  2. #2
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    Quote Originally Posted by AKTilted View Post
    Hi, some help with this problem in intro analysis please:

    Parameterize by a continuous map f:[0,1]-->S in R^3 the polygonal curve L from (0,0,0) to (2,1,1) to (2,3,2) to (3,3,3).

    Thanks a lot.

    The line from (a,b,c) to (x,y,z) in R^3 is given by (a,b,c) + t*(x-a, y-b, z-c), with t in [0,1]. Since you need the parameter in [0,1] for the three poligonal lines you need to "scale" it:

    == (a,b,c)) + 3t*((x-a), (y-b), (z-c)), t in [0,1/3] is from (a,b,c) to (x,y,z) ==> in your case we get:

    (0,0,0) + 3t(2, 1, 1) = {(6t, 3t, 3t) , t in [0,1/3]}

    == (x,y,z) + (3t-1)(p-x, q-y,r-z), with t in [1/3,2/3] is from (x,y,z) to (p,q,r), so in your case is...etc

    Tonio

    ==
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  3. #3
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    Is the last set of parameters:

    == (p,q,r)+(3t-2)(d-p, e-q, f-r) with t in [2/3,1] is from (p,q,r) to (d,e,f)?
    Thanks
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  4. #4
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    Quote Originally Posted by AKTilted View Post
    Is the last set of parameters:

    == (p,q,r)+(3t-2)(d-p, e-q, f-r) with t in [2/3,1] is from (p,q,r) to (d,e,f)?
    Thanks

    Indeed. Very nice. Finally, instead (a,b,c), (x,y,z) etc. use your own vectors.

    Tonio
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