## 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) one year ago one year ago

1. dape Group Title

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

2. dape Group Title

So for the first segment $\vec{f_1}(t)=(2t,t)$

3. njiokm Group Title

so the second is (t,4t) right?

4. dape Group Title

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.

5. dape Group Title

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.

6. dape Group Title

Then the full function is just $\vec{f}(t)= \begin{cases} \vec{f_1}(t) & \text{if } 0≤t≤1 \\ \vec{f_2}(t) & \text{if } 1<t≤2 \\ \vec{f_3}(t) & \text{if } 2<t≤3 \end{cases}$

7. njiokm Group Title

ar.....sorry dude I don't really get it can u give me a final answer i will figure it out myself

8. dape Group Title

The second function could be $\vec{f_2}(t)=(-t+3,\,3t-2)$ and the third $\vec{f_3}(t)=(-t+3,\,-4t+12)$

9. njiokm Group Title

thanks i will try to figure it out

10. dape Group Title

Here is a picture I think will help a lot, if you think visually |dw:1348317929963:dw|

11. dape Group Title

I got the lines equations by finding t in terms of x or y in the parametrizations