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anonymous

  • 5 years ago

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)

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  1. anonymous
    • 5 years ago
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    do you want a separate parametrization for each line segment?

  2. anonymous
    • 5 years ago
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    yes please

  3. anonymous
    • 5 years ago
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    \[x(t)=t, y(t)=0, \operatorname {for} 0\leq t\leq 2\]

  4. anonymous
    • 5 years ago
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    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

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

  6. anonymous
    • 5 years ago
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    parameterizations are not unique as long as the parametric representations trace out those line segments , you are good

  7. anonymous
    • 5 years ago
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    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?

  8. anonymous
    • 5 years ago
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    you will have two integrals, but yes, those should work

  9. anonymous
    • 5 years ago
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    ok, I'll try that. Can i verify my answer with you in a bit?

  10. anonymous
    • 5 years ago
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    sure

  11. anonymous
    • 5 years ago
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    great, thank you

  12. anonymous
    • 5 years ago
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    i get 17/3 as my solution

  13. anonymous
    • 5 years ago
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    i got 10 but let me check my work

  14. anonymous
    • 5 years ago
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    ok

  15. anonymous
    • 5 years ago
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    i reworked the problem and got 10 again

  16. anonymous
    • 5 years ago
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    hmm, for the c1 i get the integral to equal 0 so then the c2 second integral i got 17/3

  17. anonymous
    • 5 years ago
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    I got 2 for c1 and 8 for c2

  18. anonymous
    • 5 years ago
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    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

  19. anonymous
    • 5 years ago
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    yeah your right, my mistake,

  20. anonymous
    • 5 years ago
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    good catch

  21. anonymous
    • 5 years ago
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    what was your second?

  22. anonymous
    • 5 years ago
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    phew ok good, i thought i really didn't know what i was doing anymore.

  23. anonymous
    • 5 years ago
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    I still get 8

  24. anonymous
    • 5 years ago
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    for my second my parametric equations are x(t) = 2+t and y(t) = 2t

  25. anonymous
    • 5 years ago
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    so its integral from 0 to 1 again of (2+t)(2t)(1)dt + (2+t-2t)(2)dt

  26. anonymous
    • 5 years ago
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    you know what, you are right, I am trying to do too much in my head...

  27. anonymous
    • 5 years ago
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    Ah, yeah i gave up trying to do math in my head since I never get it right.

  28. anonymous
    • 5 years ago
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    said thing is I'll be teaching this stuff next week...ha! thanks for the practice

  29. anonymous
    • 5 years ago
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    oh no way? thats pretty neat. Thank you for helping me with my parameterization confusion.

  30. anonymous
    • 5 years ago
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    17/3 is correct

  31. anonymous
    • 5 years ago
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    awesome

  32. anonymous
    • 5 years ago
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    my students know I seldom add correctly....good luck

  33. anonymous
    • 5 years ago
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    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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