Here's the question you clicked on:
njiokm
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)
You can define the function piecewise, and just let the vector vary along each of the straight line segments connecting the points.
So for the first segment \[ \vec{f_1}(t)=(2t,t) \]
so the second is (t,4t) right?
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.
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.
Then the full function is just \[ \vec{f}(t)= \begin{cases} \vec{f_1}(t) & \text{if } 0≤t≤1 \\ \vec{f_2}(t) & \text{if } 1<t≤2 \\ \vec{f_3}(t) & \text{if } 2<t≤3 \end{cases} \]
ar.....sorry dude I don't really get it can u give me a final answer i will figure it out myself
The second function could be \[ \vec{f_2}(t)=(-t+3,\,3t-2) \] and the third \[ \vec{f_3}(t)=(-t+3,\,-4t+12)\]
thanks i will try to figure it out
Here is a picture I think will help a lot, if you think visually |dw:1348317929963:dw|
I got the lines equations by finding t in terms of x or y in the parametrizations