How do we parameterize surfaces?

Today was like a review of the lesson we learned back in September/October. Parameterizations are given and you have to find how the x, y, and z vector-valued functions, which are in terms of u and v, relate in order to draw the surface. Doesn't this sound like finding traces and level curves?

The concept with gridlines was also introduced. First, the region that the surface covers is drawn. Then horizontal (y= __) and vertical lines (x= __) are drawn within the region (like a grid) and the equation of the level curve is found after plugging in the horizontal and vertical line values. After putting together several level curves, the curve can be seen and sketched.

Here's a cool site I found that kind of explains how artists apply surface parameterization to create their art: http://www.thegnomonworkshop.com/tutorials/paramaterization/parameterization.html.

Review: Green's Theorem

Today, we formed tripods to work on a worksheet on Green's Theorem. The problems on it were straightforward with basic applications of the theorem, but for question 2a, I got a negative answer. Does anyone know why? Or got a different answer? Just a reminder, Green's Theorem is:

Let R be a simply-connected region (no holes; doesn't cross itself) with a piecewise smooth (differentiable everywhere) boundary C (continuous, with a finite number of cusps) with positive orientation (counter clockwise orientation). If P and Q have continuous partial derivatives, then

How do we apply Green's Theorem?

Today we started the lesson by evaluating a line integral. As Mr. Honner reminded us, this is a crucial skill that we should remember, even though Green's theorem makes it unattractive in solving. We then brought back the idea of conservative vector fields. What happens when F is conservative? Quite simply, the integral equals zero. We can actually use dN/dx - dM/dy (written as dM/dx - dL/dy in yesterday's blog), to measure how "un-conservative" a vector field is. What happens if dN/dx - dM/dy is equal to 1? This actually equals the area of that region! Typical F(x,y) you may want to use (for simplicity) include <0,x> and <-y,0>.

If you are lost, visit IIT's courseware at http://www.youtube.com/watch?v=1aS7nTIYMx0 or MIT's courseware at http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/part-c-greens-theorem/

Both are great at explaining not only Green's theorem, but also typical (and not so typical) applications of it outside of the classroom.

Also from now on, a little homework section will be included, in order to let you know what the night's homework was. If you were absent, this would be a great way to briefly catch up with what was going on in class and get the homework.

Homework: p1099 # 7, 14, 20, 26, 32, 42

How do we turn line integrals into region integrals?

Today we went over line integrals and a new method of evaluating them. Usually, we find the integral and break it down into a piecewise curve. However, if we think of this curve as a region, we can then use a double integral to find it out. Green's theorem can be defined as follows:

Let R be a simply-connected region with a piecewise smooth boundary C, with positive orientation. If L and M have continuous partial derivatives, then 

• simply-connected meaning region has no holes in it and doesn't cross itself
• piecewise smooth meaning the region is continuous with a finite number of cusps
• positive orientation meaning counter clockwise

This provides us with a much easier method to solve some problems. Go try it out on some old homework! An efficient method of turning 40 minutes of long winded problems into 5 minutes of integration!

Review Day: Line Integrals

Today we went over several problems involving Line Integrals. We re-emphasized the need to remember how to solve line integrals through other methods that don't involve the fundamental theorem of line integrals. Tomorrow will be the test so there won't be a blog.

What is path independence?



Today in class we went over the implications set forth by the Fundamental Theorem of Line Integrals. For conservative functions, this theorem actual makes otherwise very long problems very easy by simply finding the potential function and then subtracting the function of the beginning point from the function of the end point. We observed that the following are equivalent (refered as T.F.A.E.)

F is conservative <=> F dot dr is path independent <=> Integral over c of F dot dr = 0 for every closed curve

This gives rise to a new integral: a closed-curve (or closed-loop) integral, which can be seen to the left.

We also tested path independence, which is proved by the Fundamental Theorem of Line Integrals: no matter what path you take, the amount of work done is soley based on the beginning and end points (as long as F is conservative)

What is the fundamental Theorem of Line Integrals?

In today's class we explored the implications set forth by the Theorem of Line Integrals, which states that if F(x,y) is conservative and M, N are continuous, then the Integral over C of F dot dr is f(r(b))-f(r(a)) where a<t<b is the bound. What this means is that if the function given is conservative, you can simply find the potential function, then plug in the extreme ends of the bounds in order to evaluate or find work.

One interesting comment that was brought up by Wilson was, can we reverse the direction of piecewise parametrization? Such as going from (1,0) to (0,0) instead of (0,0) to (1,0) in a triangle that goes (0,0) to (1,0), (1,0) to (1,1) and (1, 1) to (0,0)? We will explore this in the next few days, but this led to a larger observance: the direction doesn't matter if the function is conservative and ends up in the same point, work will always be zero.

Here is the inversion seen in the triangle discussed before:

If the function is conservative, the work will be the same for both paths, as long as they both end up at (1,1)