- I-Learn, Class Pictures, Learning Targets, Text Book Practice
- Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus
Brain Gains (Rapid Recall, Jivin' Generation)
We start each class with a few brain pushups. Both recall and generation serve the same purpose, namely helping your brain form and grow pathways to connect information.
- True/False: It is OK if I don't get everything right the first time I attempt it.
Answer:
- The solution is a possible class theme song: Try Everything
- Give the component form of a vector that points from $(0,3)$ to $(4,0)$.
Answer:
There are lots of ways to write the answer. Note that the vector points 4 right, and 3 down (obtained visually, or subtracting). Two answers are $(4,-3)$ and $\langle 4, -3\rangle$.
- Find the distance between $(1,2,0)$ and $(0,0,2)$.
Answer:
A vector between the two is $\vec v = (-1,-2,2)$. The length, magnitude, or norm, of this vector is $|\vec v| = ||\vec v|| = \sqrt{(-1)^2+(-2)^2+2^2} = 3$. Note that either single or double bars may be used to signify the magnitude of a vector.
- Construct a rough sketch of the points $(1,0,3)$ and $(0,2,3)$.
Answer:
We'll discuss this one.
- Using the vectors on the screen, draw $\vec u+\vec v$ and $\vec u-\vec v$.
Answer:
We'll discuss this one.
- Give the component form of a vector of length 2 that is parallel to $\left<3,4\right>$.
Answer:
A unit vector is $\frac{1}{5}(3,4)$, so the desired vector is twice this, namely $$\frac{2}{5}(3,4) = \left(\frac{6}{5},\frac{8}{5}\right).$$
Group Problems
Remember to pass the chalk after each problem.
- In 3D, plot the points $(1,2,3)$ and $(-2,4,-3)$.
- Find the distance between the two points $(3,5,-2)$ and $(-1,6,4)$. Then find the distance between the point $(3,5,-2)$ and an arbitrary point $(x,y,z)$.
- Find the component form of the vector that starts at $(4,-3)$ and ends at $(2,-2)$.
- Give a unit vector that points in the same direction as the previous. Then give a vector of length 3 that points in the same direction.
- Plot the vector valued function $\vec r(t) = (-2,1)t+(4,-3)$ for $0\leq t\leq 3$.
- Give the component form of a vector that points from $(1,2,3)$ to $(-2,4,9)$.
- Give a vector equation of the line that passes through the point $(1,2,3)$ and $(-2,4,9)$.
- Use the law of cosines ($c^2=a^2+b^2-2ab\cos\theta$ or $\vec u\cdot \vec v = |\vec u||\vec v|\cos\theta$) to find the angle between each pair of vectors below.
- $(-2,1)$ and $(1,3)$.
- $(2,3)$ and $(-1,4)$
- $(\pi,e)$ and $(\sqrt{17},c)$
- $(1,2,3)$ and $(-7,2,1)$
- $(1,2,3)$ and $(x,y,z)$.
- Draw the parametric curve $x=2+3\cos t$, $y=5+2\sin t$. Make a $t,x,y$ table of points, and then graph the $(x,y)$ coordinates.
- Draw $x=3-2\cos t$, $y=4+5\sin t$ in the $xy$-plane.
- Draw $x=2t^2-5$, $y=3t-4$ in the $xy$-plane.
Presentations
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