anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
do you want a separate parametrization for each line segment?
anonymous
  • anonymous
yes please
anonymous
  • anonymous
\[x(t)=t, y(t)=0, \operatorname {for} 0\leq t\leq 2\]

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anonymous
  • 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
  • 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
  • anonymous
parameterizations are not unique as long as the parametric representations trace out those line segments , you are good
anonymous
  • 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
  • anonymous
you will have two integrals, but yes, those should work
anonymous
  • anonymous
ok, I'll try that. Can i verify my answer with you in a bit?
anonymous
  • anonymous
sure
anonymous
  • anonymous
great, thank you
anonymous
  • anonymous
i get 17/3 as my solution
anonymous
  • anonymous
i got 10 but let me check my work
anonymous
  • anonymous
ok
anonymous
  • anonymous
i reworked the problem and got 10 again
anonymous
  • anonymous
hmm, for the c1 i get the integral to equal 0 so then the c2 second integral i got 17/3
anonymous
  • anonymous
I got 2 for c1 and 8 for c2
anonymous
  • 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
  • anonymous
yeah your right, my mistake,
anonymous
  • anonymous
good catch
anonymous
  • anonymous
what was your second?
anonymous
  • anonymous
phew ok good, i thought i really didn't know what i was doing anymore.
anonymous
  • anonymous
I still get 8
anonymous
  • anonymous
for my second my parametric equations are x(t) = 2+t and y(t) = 2t
anonymous
  • anonymous
so its integral from 0 to 1 again of (2+t)(2t)(1)dt + (2+t-2t)(2)dt
anonymous
  • anonymous
you know what, you are right, I am trying to do too much in my head...
anonymous
  • anonymous
Ah, yeah i gave up trying to do math in my head since I never get it right.
anonymous
  • anonymous
said thing is I'll be teaching this stuff next week...ha! thanks for the practice
anonymous
  • anonymous
oh no way? thats pretty neat. Thank you for helping me with my parameterization confusion.
anonymous
  • anonymous
17/3 is correct
anonymous
  • anonymous
awesome
anonymous
  • anonymous
my students know I seldom add correctly....good luck
anonymous
  • 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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