Question

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)

Answer 1

do you want a separate parametrization for each line segment?

Answer 2

yes please

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

parameterizations are not unique

as long as the parametric representations trace out those line segments , you are good

Answer 7

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?

Answer 8

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

Answer 9

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

Answer 10

sure

Answer 11

great, thank you

Answer 12

i get 17/3 as my solution

Answer 13

i got 10 but let me check my work

Answer 14

ok

Answer 15

i reworked the problem and got 10 again

Answer 16

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

Answer 17

I got 2 for c1 and 8 for c2

Answer 18

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

Answer 19

yeah your right, my mistake,

Answer 20

good catch

Answer 21

what was your second?

Answer 22

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

Answer 23

I still get 8

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

17/3 is correct

Answer 31

awesome

Answer 32

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

Answer 33

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!