Parameterizing a curve (line integrals)

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SUMMARY

The discussion focuses on parameterizing a curve for line integrals using a vector-valued function r(t). The proposed piecewise function includes three segments defined for specific intervals of t: r(t) for 0 ≤ t < 1, 1 ≤ t < 2, and 2 ≤ t ≤ 3. The derivative r'(t) is noted to be undefined at t=2, raising questions about the appropriateness of this parameterization. The conversation also touches on path independence in vector fields, emphasizing the need to verify conditions for path independence to simplify the line integral calculation.

PREREQUISITES
  • Understanding of vector-valued functions
  • Knowledge of piecewise functions
  • Familiarity with line integrals
  • Concept of path independence in vector fields
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  • Study the properties of vector fields and conditions for path independence
  • Learn how to compute line integrals for piecewise functions
  • Explore the implications of undefined derivatives in vector calculus
  • Investigate alternative parameterizations for curves in the XY-plane
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Students and professionals in mathematics, particularly those studying calculus and vector analysis, as well as anyone working on problems involving line integrals and curve parameterization.

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Homework Statement


I have a vector valued function that I need to integrate over a curve C (which I know how to do). I need to create a vector valued function r(t) for any position on the curve C (see the picture). r(t) is a defined area in the XY-plane and I'm pretty sure it needs a piecewise function.


Homework Equations


See picture
http://img13.imageshack.us/img13/4368/graphgr.jpg

The Attempt at a Solution


This might work, but I'm afraid of the undefined derivative.

r(t)=
0[tex]\leq[/tex]t<1 9ti+0j
1[tex]\leq[/tex]t<2 9i+3(t-1)j
2[tex]\leq[/tex]t[tex]\leq[/tex]3 (9-9(t-2))i+(3-[tex]\sqrt{9(t-2)}[/tex]j

r'(t)=
0[tex]\leq[/tex]t<1 9i+0j
1[tex]\leq[/tex]t<2 0i+3j
2[tex]\leq[/tex]t[tex]\leq[/tex]3 -9i-3/(2[tex]\sqrt{t-2}[/tex])j



You'll notice that the last part of the r'(t) vector is undefined at t=2. Is this appropriate? How do you do it?
 
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Wait...is this a path independence thing? Can I get rid of the square root curve and use a straight line from (9,3) to (0,0)?
 
As to your parameterizations. You can work all three pieces separately as if they were separate problems so you don't need to make t itself increase through all three parts. And t doesn't have to go from 0 to 1. So think about these parameterizations:

C1 x = t, y = 0, t: 0 --> 9

C2 x = 9, y = t, t: 0 --> 3

On the square root you can think of it as x = y2 and do this:

C1 x = t2, y = t t: 3 --> 0 (note the direction)

As to your second question about independence of path. That isn't a question about the path, it is a question about the vector field. You didn't give us that but you can check it. If the conditions for independence of path hold, you don't have to do any work to solve the line integral. Do you see why?
 

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