how (step-by-step wise) do you calculate parametric representations of a function if given two points? ex. line segment (0,0) to (2,0) and (2,0) to (3,2)

- anonymous

- katieb

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- anonymous

do you want a separate parametrization for each line segment?

- anonymous

yes please

- anonymous

\[x(t)=t, y(t)=0, \operatorname
{for} 0\leq t\leq 2\]

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## More answers

- anonymous

how did you get that? I know you use the equation r(t) = (1-t)ro - tr1 but when i try that I get x= 2t and y=0
then for the next one i get x=2+t and y= 2 but those are wrong

- anonymous

your first one is correct
the second should be
\[x(t)=2+t, y(t)=2t, \operatorname{for} 0\leq t\leq 1\]

- anonymous

parameterizations are not unique
as long as the parametric representations trace out those line segments , you are good

- anonymous

so then when i evaluate the line integral of integral of (xydx + (x-y)dy) where C consists of those line segments, i can use those parametric equations?

- anonymous

you will have two integrals, but yes, those should work

- anonymous

ok, I'll try that. Can i verify my answer with you in a bit?

- anonymous

sure

- anonymous

great, thank you

- anonymous

i get 17/3 as my solution

- anonymous

i got 10 but let me check my work

- anonymous

ok

- anonymous

i reworked the problem and got 10 again

- anonymous

hmm, for the c1 i get the integral to equal 0 so then the c2 second integral i got 17/3

- anonymous

I got 2 for c1 and 8 for c2

- anonymous

hmm ok so my c1 is integral from 0 to 1 (2t)(0)(2)dt + (2t-0)(0)dt which is zero.
since my parametric equations are x = 2t and y = 0

- anonymous

yeah your right, my mistake,

- anonymous

good catch

- anonymous

what was your second?

- anonymous

phew ok good, i thought i really didn't know what i was doing anymore.

- anonymous

I still get 8

- anonymous

for my second my parametric equations are x(t) = 2+t and y(t) = 2t

- anonymous

so its integral from 0 to 1 again of (2+t)(2t)(1)dt + (2+t-2t)(2)dt

- anonymous

you know what, you are right, I am trying to do too much in my head...

- anonymous

Ah, yeah i gave up trying to do math in my head since I never get it right.

- anonymous

said thing is I'll be teaching this stuff next week...ha! thanks for the practice

- anonymous

oh no way? thats pretty neat. Thank you for helping me with my parameterization confusion.

- anonymous

17/3 is correct

- anonymous

awesome

- anonymous

my students know I seldom add correctly....good luck

- anonymous

Ah yes, it is frustrating when one can do the entire problem and in the end mess it up because of addition.
Thank you,
good luck to you when teaching it next week!

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