Upcoming 314 Review for Dr. Villamizar's class

Hi all,

The math lab will be hosting a review for Dr. Villamizar's upcoming 314 exam. Here is the information for the review,

Date: Thursday, Oct. 7, 2010
Time: 7pm - 9pm
Place: TMCB 104

The TA teaching this review will be Jared McBride. He's super-smart and knows his stuff. So get a break after Dr. Villamizar's review and then get ready to have a lot of fun with Jared. He will cover material from the book as well as anything Dr. Villamizar provides.

-Math Lab Blogger

110 Test 2 Review Info

Dear Math 110ers-

There are two reviews coming up for Test 2, tonight and tomorrow night. Here are the details:

Date: Wed Sept 29th
Time: 7-9 PM
Location: JKB 1104
TA: To Be Determined
Tests Covered: Test 2 from W ‘10

-AND-

Date: Thu Sept 30th
Time: 7-9 PM
Location: JKB 1104
TA: To Be Determined
Tests Covered: Test 2 from F ‘09

These tests can be found online at:

http://math.byu.edu/~wright/Math%20110/Math110.html

Good luck with Midterms!

-Math Lab Blogger

Here's a sample problem of the partial derivative chain rule using a table. Just click to enlarge.

Math 313 - Understanding a Basis

Dear Math 313 and 302 students,


I know how difficult it can be to understand the concept of a basis. Hopefully this will help.

A basis is simply the set containing the fewest necessary vectors possible to represent a space.

If that didn't make any sense at all, then picture a blank x-y axis. Let the two lines (the x-axis and the y-axis) be vectors. You can notice two things:

1. These two vectors are all you ever need to represent all of 2-dimensional space
   (Think about it-- all you need an x-value and a y-value and you can represent any point)
   
2. The two vectors are linearly independent.
   (Try and represent the point (0,5) with only the x-axis - kinda hard to do, right?)

Therefore these two vectors represent a basis for 2-dimensional space.


Now, let's generalize this to 3-dimensional space. Imagine the axes for 3-d space. Now let those axes be vectors. Again:

1. The three vectors are all you ever need to represent all of 3-d space (x,y,z)
2. The vectors are linearly independent.

Therefore these three vectors represent a basis for 3-dimensional space.


Are you seeing the pattern? Now let's pretend that in 3-d space, you didn't have the z-axis but had a vector from (0,0,0) to (1,1,0). Do the x-axis, y-axis, and the new vector form a basis for 3-dimensional space?

If you answered no, then you'd be correct. Why?

Can you represent the point (0,0,1)? Nope-- the vectors are linearly dependent, so you can't represent all of 3-dimensional space. Therefore the vectors do not form a basis for 3-d space.

Got it yet? I hope so. As a final reminder, the requirements to be a basis for, say, n-dimensional space are:

1. You must have n n-dimensional vectors (n vectors with n entries)
2. Those vectors must be linearly independent

If your vectors fulfill that requirement, you've got yourself a basis!


-Math Lab Blogger

Welcome!

Dear 119 Students:

Welcome to the Math Lab Blog. Feel free to leave us a comment or ask us a question about HW, reviews, or tests. We love you and we wish you the best of luck in your classes!

-Math Lab Blogger

Welcome!

Dear 113 Students-

Welcome to the Math Lab's Blog!! Please feel free to leave us comments and ask for help on problems. We'll do our best to answer your questions promptly. We love you and we wish you the best of luck in your classes!

-Math Lab Blogger

Welcome!

Dear 112 Students-

Welcome to the Math Lab's Blog!! Please feel free to leave us comments and ask for help on problems. We'll do our best to answer your questions promptly. We love you and we wish you the best of luck in your classes!

-Math Lab Blogger