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Math Help - Parametric representation

  1. #1
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    Parametric representation/Lagrange

    Let p0, p1, and p2 be points in the plane.

    a) Specify a parametric representation of a planar curve C(t)
    that satisfies C(t0) = p0, C(t1) = p1, and C(t2) = p2.
    The components of C should be quadratic polynomials in t.

    having alot of trouble with this. Im not quite sure where to begin.
    Last edited by p00ndawg; April 20th 2010 at 02:39 PM.
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  2. #2
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    If you set the x coordinate to at^2+bt+c, and substitute the x-coordinate part of each of your three conditions, you get three equations of the form at_i^2+bt_i+c=\text{x coordinate of }p_i. Since the t_i and p_i are known values, you have 3 equations in the three unknowns a, b, and c. Then you do the same thing for the y coordinate.

    You might find this simpler than solving 3 equations in 3 unknowns (twice):
    Lagrange polynomial - Wikipedia, the free encyclopedia

    Post again in this thread if you're still having trouble.

    - Hollywood
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  3. #3
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    Quote Originally Posted by hollywood View Post
    If you set the x coordinate to at^2+bt+c, and substitute the x-coordinate part of each of your three conditions, you get three equations of the form at_i^2+bt_i+c=\text{x coordinate of }p_i. Since the t_i and p_i are known values, you have 3 equations in the three unknowns a, b, and c. Then you do the same thing for the y coordinate.

    You might find this simpler than solving 3 equations in 3 unknowns (twice):
    Lagrange polynomial - Wikipedia, the free encyclopedia

    Post again in this thread if you're still having trouble.

    - Hollywood
    yes im having trouble setting this up.
    i got help for another question in this thread.

    http://www.mathhelpforum.com/math-he...rpolation.html

    is it similar?

    is it:

    x(t) = C0(t)x0 + C1(t)x1 + C2(t)x2
    y(t) = C0(t)y0 + C1(t)y1 + C2(t)y2

    or am i way off?
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  4. #4
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    It is the same as the other thread, except you have both x and y to solve, and you have variables instead of numbers.

    Let p_0=(x_0,y_0)\text{, }p_1=(x_1,y_1)\text{, and }p_2=(x_2,y_2).

    Also, let C(t)=(x(t),y(t)).

    So we are looking for:

    x(t)\text{ satisfying }x(t_0)=x_0\text{, }x(t_1)=x_1\text{, and }x(t_2)=x_2, and

    y(t)\text{ satisfying }y(t_0)=y_0\text{, }y(t_1)=y_1\text{, and }y(t_2)=y_2.

    Each of these is solved by the same process as the other thread. For the x-coordinate:

    x(t)=x_0\ \frac{t-t_1}{t_0-t_1}\ \frac{t-t_2}{t_0-t_2}+x_1\ \frac{t-t_0}{t_1-t_0}\ \frac{t-t_2}{t_1-t_2}+x_2\ \frac{t-t_0}{t_2-t_0}\ \frac{t-t_1}{t_2-t_1}

    and of course the y-coordinate is similar.

    - Hollywood
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  5. #5
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    Quote Originally Posted by hollywood View Post
    It is the same as the other thread, except you have both x and y to solve, and you have variables instead of numbers.

    Let p_0=(x_0,y_0)\text{, }p_1=(x_1,y_1)\text{, and }p_2=(x_2,y_2).

    Also, let C(t)=(x(t),y(t)).

    So we are looking for:

    x(t)\text{ satisfying }x(t_0)=x_0\text{, }x(t_1)=x_1\text{, and }x(t_2)=x_2, and

    y(t)\text{ satisfying }y(t_0)=y_0\text{, }y(t_1)=y_1\text{, and }y(t_2)=y_2.

    Each of these is solved by the same process as the other thread. For the x-coordinate:

    x(t)=x_0\ \frac{t-t_1}{t_0-t_1}\ \frac{t-t_2}{t_0-t_2}+x_1\ \frac{t-t_0}{t_1-t_0}\ \frac{t-t_2}{t_1-t_2}+x_2\ \frac{t-t_0}{t_2-t_0}\ \frac{t-t_1}{t_2-t_1}

    and of course the y-coordinate is similar.

    - Hollywood
    oh ok so by doing it for the x and for the y ill get C(t)=(x(t),y(t))?

    where x(t) and y(t) will be a function?
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  6. #6
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    Yes, that's correct.

    - Hollywood
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