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anonymous
 3 years ago
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)
anonymous
 3 years ago
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)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You can define the function piecewise, and just let the vector vary along each of the straight line segments connecting the points.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So for the first segment \[ \vec{f_1}(t)=(2t,t) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the second is (t,4t) right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Not 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 endpoints of the segments. With your definition we would jump from (2,1) to (1,4) instantaneously at t=1.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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 tunit to trace.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then the full function is just \[ \vec{f}(t)= \begin{cases} \vec{f_1}(t) & \text{if } 0≤t≤1 \\ \vec{f_2}(t) & \text{if } 1<t≤2 \\ \vec{f_3}(t) & \text{if } 2<t≤3 \end{cases} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ar.....sorry dude I don't really get it can u give me a final answer i will figure it out myself

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The second function could be \[ \vec{f_2}(t)=(t+3,\,3t2) \] and the third \[ \vec{f_3}(t)=(t+3,\,4t+12)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks i will try to figure it out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here is a picture I think will help a lot, if you think visually dw:1348317929963:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got the lines equations by finding t in terms of x or y in the parametrizations
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