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njiokm Group Title

how to find the vector valued function forming the boundaries by the closed curve defined by connected points (0,0), (2,1), and (1,4)

  • 2 years ago
  • 2 years ago

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  1. dape Group Title
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    You can define the function piecewise, and just let the vector vary along each of the straight line segments connecting the points.

    • 2 years ago
  2. dape Group Title
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    So for the first segment \[ \vec{f_1}(t)=(2t,t) \]

    • 2 years ago
  3. njiokm Group Title
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    so the second is (t,4t) right?

    • 2 years ago
  4. dape Group Title
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    Not really, at t=1 we are at the point (2,1), so you need to trace the function from there. You think correctly, but you need a term to align the end-points of the segments. With your definition we would jump from (2,1) to (1,4) instantaneously at t=1.

    • 2 years ago
  5. dape Group Title
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    So you basically want \[ \vec{f_1}(1)=\vec{f_2}(1) \\ \vec{f_2}(2)=\vec{f_3}(2) \\ \vec{f_3}(3)=(0,0) \] If each segment should take 1 t-unit to trace.

    • 2 years ago
  6. dape Group Title
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    Then the full function is just \[ \vec{f}(t)= \begin{cases} \vec{f_1}(t) & \text{if } 0≤t≤1 \\ \vec{f_2}(t) & \text{if } 1<t≤2 \\ \vec{f_3}(t) & \text{if } 2<t≤3 \end{cases} \]

    • 2 years ago
  7. njiokm Group Title
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    ar.....sorry dude I don't really get it can u give me a final answer i will figure it out myself

    • 2 years ago
  8. dape Group Title
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    The second function could be \[ \vec{f_2}(t)=(-t+3,\,3t-2) \] and the third \[ \vec{f_3}(t)=(-t+3,\,-4t+12)\]

    • 2 years ago
  9. njiokm Group Title
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    thanks i will try to figure it out

    • 2 years ago
  10. dape Group Title
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    Here is a picture I think will help a lot, if you think visually |dw:1348317929963:dw|

    • 2 years ago
  11. dape Group Title
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    I got the lines equations by finding t in terms of x or y in the parametrizations

    • 2 years ago
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