You can access this page from the homepage in I-Learn.
Pre-Class Chatter
Feel free to turn on your mic and chat with people or use the zoom chat. Or chat with your group in your Jamboard.
9:00 AM Jamboard Links
| G1 | G2 | G3 | G4 | G5 | G6 | G7 |
|
|
|
|
|
|
|
12:45 PM Jamboard Links
| G1 | G2 | G3 | G4 | G5 | G6 |
|
|
|
|
|
|
- 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".
- 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, invite me to join you via Zoom.
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?).
- A rover is travelling on the line $\vec r(t) = (a,b)t+(c,d)$. Let $\vec v =(a,b)$ and let $P=(c,d)$. An anomaly is located at $Q=(m,n)$. Which of the following gives the distance between the rover's location and the anomaly as a function of time?
- $|\vec Q-\vec v|$
- $|\vec Q-\vec r|$
- $|\vec Q-\vec P|$
Solution
The rover's location is $\vec r$, and the anomaly is at $Q$, so a vector from the position to the anomaly is $\vec s = \vec Q-\vec r$. The length of this vector is $|\vec Q-\vec r|$.
- Continuing from before, let $\vec s$ be the vector from the rover's current position to the anomaly, so $\vec s = \vec Q-\vec r$. Which of the following gives the angle between the direction the rover is traveling and the anomaly.
- $\ds\cos^{-1}\left(\frac{\vec v\cdot \vec s}{|\vec v||\vec s|}\right)$
- $\ds\cos^{-1}\left(\frac{\vec r\cdot \vec s}{|\vec r||\vec s|}\right)$
- $\ds\cos^{-1}\left(\frac{\vec P\cdot \vec s}{|\vec P||\vec s|}\right)$
Solution
The rover's direction of motion is $\vec v$, and a vector from the position to the anomaly is $\vec s = \vec Q-\vec r$. The angle between these is $$\ds\cos^{-1}\left(\frac{\vec v\cdot \vec s}{|\vec v||\vec s|}\right).$$
- 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
- Ben - Remember to send everyone a link to this page in chat.
- In BreakOut rooms - Take turns writing on your Jamboard. Give everyone a chance to be the scribe. The point is to discuss the problems below, and help one another.
- 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.
|
Sun |
Mon |
Tue |
Wed |
Thu |
Fri |
Sat |