Day 54 in MAT229

Last time: Section 13.1/13.2

Next time: Section 13.3/13.4

Today:

Announcements:
- Reminder: Homework problems:
  - pp. 665-, #22, 30, 38, 59, 60 (due Tuesday, 12/9).
  - pp. 673-674, #13, 18, 22, 26, 45, 50, 51 (due Thursday, 12/11).
  - pp. 680-683, #2, 7, 8, 14, 16, 41, 42, 60, 70, 71 (due Friday, 12/12).
- A problem or two from 13.4/13.5 will appear on the final.
Vector Geometry
- Last time we defined a product of two vectors (resulting in a scalar, or number) as follows:
  
  It turns out that
  
  where
  
  Otherwise, the angle is acute.
  
  We put it all together to get this result, which shows us how to use the dot product to define components:
  
  Examples:
  - #69, p. 682
  - #79, p. 683
- We then went on to consider parametric equations of lines, starting from uniform circular motion. As we saw in that case, ${\underline{\bf{v}}=\frac{d\underline{\bf{r}}}{dt}}$
  and
  
  ${\underline{\bf{a}}=\frac{d\underline{\bf{v}}}{dt}}$
  But that ${\underline{\bf{v}}\cdot\underline{\bf{r}}=0}$ and that ${||\underline{\bf{v}}||=||\underline{\bf{r}}||}$ .
  We "cut the string" holding the particle (or planet!) in uniform circular motion, and found the equation of the line along which it would move to be the following:
  
  Examples:
  - #46, p. 674
  - #52, p. 674
- Section 13.4: The Cross Product
  - The next product we'll find convenient to define is the cross product, which is a product of two vectors resulting in a third vector, which juts into the space perpendicular to which the two vectors live. It is only useful in three-space, which makes it somewhat unusual (the dot product exists and is useful in any dimensional space).
    Because the cross product is linear in the components, we can define it on the unit vectors in three space, and then deduce it using the component-wise definition of a three-vector.
    
    ${\hat{i}\times\hat{j}=\hat{k}}$ ${\hat{j}\times\hat{k}=\hat{i}}$ ${\hat{k}\times\hat{i}=\hat{j}}$
    
    When you reverse direction, however, you get ${\hat{i}\times\hat{k}=-\hat{j}}$ ${\hat{j}\times\hat{i}=-\hat{k}}$ ${\hat{k}\times\hat{j}=-\hat{i}}$
    
    Examples:
    - #20, p. 691
    - #28, p. 691
    - The force ${\bf{F}}$ on a proton moving at velocity ${\bf{v}}$ (in m/s) in a uniform magnetic field ${\bf{B}}$ (in teslas) is ${\bf{F}=q(\bf{v}\times\bf{B})}$ (in Newtons), where ${q=1.6\times{10^{-19}}}$ Coulombs:
      
      #36, p. 692
    - Torque: #64, p. 693

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