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anonymous
 5 years ago
how (stepbystep 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
 5 years ago
how (stepbystep 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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you want a separate parametrization for each line segment?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x(t)=t, y(t)=0, \operatorname {for} 0\leq t\leq 2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get that? I know you use the equation r(t) = (1t)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
 5 years ago
Best ResponseYou've already chosen the best response.0your first one is correct the second should be \[x(t)=2+t, y(t)=2t, \operatorname{for} 0\leq t\leq 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0parameterizations are not unique as long as the parametric representations trace out those line segments , you are good

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then when i evaluate the line integral of integral of (xydx + (xy)dy) where C consists of those line segments, i can use those parametric equations?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you will have two integrals, but yes, those should work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, I'll try that. Can i verify my answer with you in a bit?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i get 17/3 as my solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got 10 but let me check my work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i reworked the problem and got 10 again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm, for the c1 i get the integral to equal 0 so then the c2 second integral i got 17/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got 2 for c1 and 8 for c2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm ok so my c1 is integral from 0 to 1 (2t)(0)(2)dt + (2t0)(0)dt which is zero. since my parametric equations are x = 2t and y = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah your right, my mistake,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what was your second?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0phew ok good, i thought i really didn't know what i was doing anymore.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for my second my parametric equations are x(t) = 2+t and y(t) = 2t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so its integral from 0 to 1 again of (2+t)(2t)(1)dt + (2+t2t)(2)dt

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you know what, you are right, I am trying to do too much in my head...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, yeah i gave up trying to do math in my head since I never get it right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0said thing is I'll be teaching this stuff next week...ha! thanks for the practice

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh no way? thats pretty neat. Thank you for helping me with my parameterization confusion.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my students know I seldom add correctly....good luck

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah 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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