# Parameterizing a curve (line integrals)

1. Dec 6, 2009

### filter54321

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
See picture
http://img13.imageshack.us/img13/4368/graphgr.jpg [Broken]

3. The attempt at a solution
This might work, but I'm afraid of the undefined derivative.

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

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

Last edited by a moderator: May 4, 2017
2. Dec 6, 2009

### filter54321

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. Dec 6, 2009

### LCKurtz

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?