Loser66
  • Loser66
parameterize segment. (1-i) to (1+i) . Please help
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Loser66
  • Loser66
If I work as usual, I get 1-i + 2it , t from 0 to 1 but my friend makes it easy by 1 + it , t from -1 to 1 I would like to know how he gets it forward, not backward. I meant if we know the parametric equation, then we can check whether it is right or wrong easily. Assume that we don't know it, like I am now, how can we get it from the given information?
Loser66
  • Loser66
|dw:1446433146889:dw|
freckles
  • freckles
is this not right: v=(1,-1)-(1,1)=(0,-2) so answer is (1,1)+t(0,-2) or (1,1-2t) or 1+(1-2t)i

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freckles
  • freckles
http://mathinsight.org/line_parametrization_examples I used this as example
Loser66
  • Loser66
I don't get what you mean but for the first line segment from 1-i to 1+i, the parameterized equation is not that
Loser66
  • Loser66
if t from 0 to 1, that is my solution. As I said above, my friend get from -1 to 1 with different equation which is faaaaaaaaaaaaar easier than my way on next step. But I don't know how he gets that way (Note: his equation is perfect right)
freckles
  • freckles
could you show how he gets backwards?
Loser66
  • Loser66
|dw:1446434405668:dw|
Loser66
  • Loser66
you see, for his solutions line a \(1+it~~~-1\leq t \leq 1\) Check: if t = -1, then 1 + i(-1) = 1-i , the first point if t =1, then 1 +i(1) = 1+ i , the last point perfect.
Loser66
  • Loser66
line b) \(t +i ~~-1\leq t\leq 1\) if t = -1, then -1 + i (the last point) if t =1, then 1+i (the first point) nothing is wrong.
Loser66
  • Loser66
Same as line c and d But I do be mad because I do not know the logic how to get it from him!!! :(
freckles
  • freckles
\[\text{ for } d \\ y=-1 , x \in [-1,1] \\ \text{ set } x(t)=t \text{ and } y(t)=-1 , t \in [-1,1] \\ (x,y)=(t,-1)=t-i\]
freckles
  • freckles
\[\text{ for } b \\ y=1 , x \in [-1,1] \\ \text{ set } x(t)=t \text{ and } y(t)=1 , t \in[-1,1] \\ (x,y)=(t,1)=t+i\]
freckles
  • freckles
|dw:1446435977770:dw|
freckles
  • freckles
\[\text{ for } a \\ x=1 , y \in [-1,1] \\ \text{ set } y(t)=t \text{ and } x(t)=1 , t \in [-1,1] \\ (x,y)=(1,t)=1+ti\]
freckles
  • freckles
|dw:1446436071639:dw| the last one we have a vertical line x=-1
freckles
  • freckles
\[\text{ for } c \\ x=-1, y \in [-1,1] \\ \text{ set } y(t)=t \text{ and } x(t)=-1 , y \in [-1,1] \\ (x,y)=(-1,t)=-1+ti\]
Loser66
  • Loser66
Thank you so much. I got it. :)
freckles
  • freckles
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0CCoQFjACahUKEwiNiOqO6fDIAhXKNj4KHVP5DSQ&url=http%3A%2F%2Fwww.math.uh.edu%2F~jiwenhe%2FMath1432%2Flectures%2Flecture14_handout.pdf&usg=AFQjCNHse8F_z_EXocM7UKXS43_fFIXwPw&sig2=XZn3Y1tXYC_9Xc4wphEm9g example 1.2 is what gave me the idea
freckles
  • freckles
we basically just need to notice the equation of the line
freckles
  • freckles
|dw:1446436552457:dw| well we have y=1 where x is between -2 and 1. if we choose t to be x then t is between -2 and 1 no matter y will just be 1 so this would be (x,y)=(t,1) where t is in [-2,1] now say we wanted t+1 to be x then t+1 is between -2 and 1 and so t is between -3 and 0 no matter y will just be 1 so this would be (x,y)=(t+1,1) where t is in [-3,0]
freckles
  • freckles
this makes sense to me too now :p I feel so dumb
freckles
  • freckles
I didn't understand what was going on at first
freckles
  • freckles
that is not to do with your problem by the way
Loser66
  • Loser66
Thanks for the link also. Because of this, we can parameterize the curve on any interval we like.
freckles
  • freckles
yep yep
Loser66
  • Loser66
Again, thank you so much. I learn a lot.
freckles
  • freckles
me too :)
freckles
  • freckles
I'm so glad I live in the age where there is internet

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