Line Integrals

So on Tuesday, we continued the lesson on line intergrals. A few students (me being one of them) were caught cutting class on Monday. Luckily we weren't punished. At least not yet.
We reviewed the four step process for completing a line integral:
Start with the integral of f (x,y) ds
1) Find a parametric curve, including the limits of the curve
2) Re-parameterize f (x,y) as f (x(t),y(t))
3) Change ds using the arc length formula (which we learned previously)
4) Compute the integral

We also got into the fact that there's a physics aspect to the use of line integrals. (Sigh. Just when I thought I was done with physics)
Let's say you go a path on a 2-D xy coordinate plane with a vector field F (x,y). Think of the vector field as wind that you experience along your "path". We can use line integrals to figure out the net amount of force that is either helping or slowing you down along your "path". It's a cool beginning to the further expansion of line integrals.

     
Note:
The graph of the parametric curve where x (t) = sin (t)
                                                              y (t) = cos (t)
                                                              z (t) = t
 can be described in mutliple ways: a helix, a spring, a slinky, and the list continues on...

How do we compute line integrals?

So the three basic steps to evaluating line integrals is:
1. Figure out the curve C, including limits.
2. Reparameterize F in terms of t.
3. Evaluate.

Sometimes, you may get the same problems with switched x(t) and y(t) values and the answers may have opposite signs of each other. Why? Try drawing out the graph and think about orientation.

Not to say that Mr. Honner didn't teach us enough, but here's a video that might help you practice more with line integrals: http://www.youtube.com/watch?v=fjEvsinvtnw . Ready for that quiz on Friday?

How do we evaluate line/path integrals?

Today we began the lesson by reflecting on parameterization of curves. After such a parameterization, we found the arc length of the same curve. This leads into finding the mass of the curve, which takes both parameterization and arc length into consideration. Quite simply, mass is density times the change in arc length. Refer to the first few problems in the text book (p1069) to practice. More explanation can be found here: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/part-c-parametric-equations-for-curves/

If you are having trouble with some curves, check this out:

This is a new multivariable blogging site that we'll be updating very often about what was learned in class. If you're ever absent, feel free to visit this site to catch up on what we did. We hope this will help you.

On Friday, we learned about flow lines (stream lines). One of the questions in the Utexas homework assignment basically explains what these are: "The flow lines (or streamlines) of a vector field are the paths followed by a particle whose velocity field is the given vector field. Thus, the vectors in a vector field are tangent to the flow lines."

The path of a particle is given by the position function s(t) = < x(t), y(t) >, which are the parametric equations of the flow lines for a vector field F(x,y). You plug them into the separable differential equation dy/dt / dx/dt (or dy/dx) and find its antiderivative after separating all the "x'es" to one side and all the "y's" to another side to get the equation of the flow lines of the vector field. Don't forget to add the constant (C)!

For example, given F(x, y) = < y, x >, find the equation of its flow lines.
dy/dt = x; dx/dt = y --> dy/dx = x/y (separable differential equation)
After separating the "x'es" and "y's," you get ydy = xdx.
Antiderive each side to get ½ y² = ½ x² + C, which is equivalent to y²/2 - x²/2 = C. This is the equation of the given vector field's flow lines. Can you recognize what kind of graph that is?

Hope this was helpful.