anonymous
  • anonymous
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)
Mathematics
schrodinger
  • schrodinger
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dape
  • dape
You can define the function piecewise, and just let the vector vary along each of the straight line segments connecting the points.
dape
  • dape
So for the first segment \[ \vec{f_1}(t)=(2t,t) \]
anonymous
  • anonymous
so the second is (t,4t) right?

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dape
  • dape
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.
dape
  • dape
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.
dape
  • dape
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
anonymous
  • anonymous
ar.....sorry dude I don't really get it can u give me a final answer i will figure it out myself
dape
  • dape
The second function could be \[ \vec{f_2}(t)=(-t+3,\,3t-2) \] and the third \[ \vec{f_3}(t)=(-t+3,\,-4t+12)\]
anonymous
  • anonymous
thanks i will try to figure it out
dape
  • dape
Here is a picture I think will help a lot, if you think visually |dw:1348317929963:dw|
dape
  • dape
I got the lines equations by finding t in terms of x or y in the parametrizations

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