- Miller's Law - Our working memory as humans is essentially 7 (plus or minus 2) things simultaneously. We can "chunk" ideas into groups to increase our RAM (that's what the lesson plans are for). "Chunking" is a very personal thing and requires individual organization of content into meaningful "chunks".
- Out of Class Group Study Suggestion - Make sure everyone works on the problem independently. Your goal: Discover what your questions are. You can't do this if you let someone else do the problem for you.
Rapid Recall
- If I know that $\vec F=\left<1,1,6\right>$ and $\text{proj}_{\vec d} \vec F = \left<-1,3,4\right>$, then state $\vec F_{\perp \vec d}$.
Solution
$\vec F_{\perp \vec d} = \vec F - \vec F_{\parallel \vec d} = \left<1,1,6\right>-\left<-1,3,4\right>=\left<2,-2,2\right>$. Note that $\vec F_{\parallel \vec d} \cdot \vec F_{\perp \vec d}=0$ which should be the case (why?).
- Give a nonzero vector that is orthogonal to $(3,4)$. Then give a nonzero vector that is orthogonal to $(1,2,3)$.
Solution
A vector orthogonal to $(3,4)$ is $(-4,3)$ or $(4,-3)$, or any multiple of these. A vector orthogonal to $(1,2,3)$ is $(0,-3,2)$, $(-3,0,1)$, $(-2,1,0)$, $(1,1,-1)$, any multiple of these, and tons more. All you need is a vector $(x,y,z)$ such that $$1x+2y+3z=0.$$ Does this look to anyone like a "sum of forces equals zero" equation, or maybe even a "sum of moments equals zero" equation?
- Draw $\ds \frac{x^2}{16}+\frac{y^2}{25}=1$.
Solution
This is an ellipse centered at the origin passing through $x=\pm4$ and $y=\pm5$.
- Draw $\ds \frac{(x-2)^2}{16}+\frac{(y-3)^2}{25}=1$.
Solution
This is the same ellipse as above, translated so the center is at $(2,3)$.
Group problems
- Let $\vec F = (0,-10)$ and $\vec d = (3,2)$. Compute both $\vec F_{\parallel \vec d}$ and $\vec N = \vec F_{\perp \vec d}$.
- Draw the curve $u^2+v^2=1$ in the $xy$ plane using the change of coordinates $x=3u-1$ and $y=2v+4$. Give a Cartesian equation of the curve.
- Draw $\left(\frac{x}{3}\right)^2+\left(\frac{y}{5}\right)^2=1$.
- Draw $\ds \frac{(x-2)^2}{16}+\frac{(y-3)^2}{9}=1$.
- Draw $\ds \frac{(x-2)^2}{16}-\frac{(y-3)^2}{9}=1$ and $\ds \frac{(y-3)^2}{9}-\frac{(x-2)^2}{16}=1$.
- Draw $\ds \frac{(x-1)^2}{16}+\frac{(y-5)^2}{9}=1$ and then draw $\ds \frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1$.
- Draw $\ds \frac{(x+2)^2}{9}+\frac{(y-4)^2}{25}=1$ and then draw $\ds -\frac{(x+2)^2}{9}+\frac{(y-4)^2}{25}=1$.
- 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.
|
Sun |
Mon |
Tue |
Wed |
Thu |
Fri |
Sat |