Parameterizing a curve (line integrals)

In summary, the problem involves integrating a vector valued function over a curve C and creating a vector valued function r(t) for any position on the curve. This may require a piecewise function and consideration of path independence. The attempt at a solution includes a parameterization for each piece of the curve and a question about path independence.
  • #1
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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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  • #2
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)?
 
  • #3
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?
 

Related to Parameterizing a curve (line integrals)

1. What is parameterization of a curve?

Parameterization of a curve is the process of representing a curve in terms of one or more parameters. This allows for a more precise and efficient way to describe and analyze the curve.

2. Why is parameterization important in line integrals?

Parameterization is important in line integrals because it helps to define the path of integration along a curve. It also allows for easier calculation of the line integral by breaking it down into smaller, more manageable segments.

3. How do you parameterize a curve?

To parameterize a curve, you need to determine the independent variable(s) that best describe the curve. These variables can be time, distance, or any other measurable quantity. Then, you need to express the curve's coordinates in terms of the chosen variable(s).

4. What are the benefits of parameterization?

Parameterization offers several benefits in line integrals, such as simplifying calculations, allowing for more accurate representations of curves, and providing a way to analyze complex curves. It also helps in finding solutions to various real-world problems in fields such as physics, engineering, and mathematics.

5. Can any curve be parameterized?

Yes, any curve can be parameterized as long as it can be expressed in terms of one or more variables. However, some curves may have multiple possible parameterizations, and the choice of parameterization can affect the ease of calculation and accuracy of the analysis.

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