Announcments
- Miller's Law and the Magical Number Seven, Plus or Minus 2 - Our working memory as humans is essentially 7 (plus or minus 2) things simultaneously. We can "chunk" ideas into groups to increase our RAM (this is why we have units). "Chunking" is a very personal thing and requires individual organization of content into meaningful "chunks". After you have completed 80% of the objectives on a quiz, one thing you might do for an SDL is to create a lesson plan that "chunks" together the ideas from the unit. Then share that lesson plan (with actual people, record yourself teaching it and start a YouTube channel, TikTok the unit in 15 second chunks, etc.).
- Zone of Proximal Development - Out of Class Group Study Suggestion - Your goal: Discover what your questions are. This requires we attempt something first so we know where our boundaries are. I highly recommend working with others, and as you do so, work on problems that no one yet has started. Each work on it by yourself, and then ask questions of each other as soon as you have them. If you meet during my office hour times, and you can't answer your questions together, swing by my office and invite to join you.
Brain Gains
- 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?).
- Draw $\ds \frac{u^2}{16}+\frac{v^2}{25}=1$ in the $uv$-plane.
Solution
This is an ellipse, centered at the origin, passing through $u=\pm4$ and $v=\pm5$.
- Draw $\ds \frac{(x-2)^2}{16}+\frac{(y-3)^2}{25}=1$.
Solution
This is the same ellipse as above, though this time in the $xy$-plane, translated so the center is at $(2,3)$.
- Give a change-of-coordinates (so equations for $x$ and $y$ in terms of $u$ and $v$) that would transform $\ds \frac{u^2}{16}+\frac{v^2}{25}=1$ into $\ds \frac{(x-2)^2}{16}+\frac{(y-3)^2}{25}=1$.
Solution
The two equations become the same under the substitution $u=x-2$ and $v=y-3$. Solving for $x$ and $y$ gives $$x=u+2, y=v+3.$$ This change-of-coordinates causes the graph in the $uv$-plane to move right 2 and up 3, which is precisely what we saw in our graphs.
Group problems
- Draw the vector field $\vec F(x,y) = \langle2x+y,x+2y\rangle$. (Based at $(x,y)$, draw the vector $\langle2x+y,x+2y\rangle$.)
- Draw the vector field $\vec F(x,y) = \langle y,-x\rangle$.
- Let $x=2u+3$ and $y=4v-5$. Complete the $u,v,x,y$ table below, and then construct a graph of both $u^2+v^2=1$ (in the $uv$-plane) and the corresponding equation in the $xy$-plane (called a Cartesian equation). $$ \begin{array}{c|c|c|c} u&v&x&y\\\hline 0&0&3&-5\\ 1&0&5&-5\\ 0&1&&\\ -1&0&&\\ 0&-1&&\\ \end{array} $$
- Draw the curve $u^2+v^2=1$ in the $xy$-plane using the change of coordinates $x=3u-1$ and $y=2v+4$ (make a table like above). Give a Cartesian equation of the curve (so the equation in the $xy$-plane.
- Draw $\ds \left(\frac{x}{3}\right)^2+\left(\frac{y}{5}\right)^2=1$.
- Draw $\ds \frac{(x-2)^2}{16}+\frac{(y-3)^2}{9}=1$.
- Draw $\ds \left(\frac{x}{16}\right)^2-\left(\frac{y}{9}\right)^2=1$ and $\left(\frac{y}{9}\right)^2-\left(\frac{x}{16}\right)^2=1$.
- 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.
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