9:00 AM Jamboard Links
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12:45 PM Jamboard Links
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Presenters
9AM
Thanks for sharing things in Perusall. Here are the presenters for today.
- 3.20 - Ethan B
- 3.24 - Next Time (View Ethan's work if you need help.)
- 3.25 - Next Time (View Ethan's work if you need help.)
- 3.26 - Kallan P
- 3.27 - Tanner H
- 3.28 - Jeremy B
- 3.29 - Next Time (View Kallan's solution if you need help.)
12:45PM
Thanks for sharing things in Perusall. Here are the presenters for today.
- 3.20 - Matty D
- 3.21 - Cheyenne
- 3.22 - Hamilton
- 3.23 - Forrest
- 3.24 - Trevor
- 3.25 - Skip
- 3.26 - Alan
- 3.27 - Joshua
Learning Reminders
- We are in the 13th week of the semester. If you are on track for an A, then ideally you're finishing your SDL project for the 5th unit, and proposed something for the 6th unit.
- There are only 2 weeks left, which means you can submit at most 2 more SDL projects.
- The final SDL project (6th) can be over any topic from the entire semester. You can use it to expand what we do in the 6th unit, or you may choose to revisit something from a prior unit that you would like to spend more time with.
Rapid Recall
- When we add up lots of little areas, so $\int_R dA$, what do we get?
Solution
Total area.
- Finish the following statement: "Adding up lots of little changes in $x$ along a curve $C$, so $\int_C dx$, gives __________."
Solution
The solution is, "Adding up lots of little "changes in $x$" along a curve $C$, so $\int_C dx$, gives "the total change in $x$", or $x_{final}-x_{initial}$."
- Adding up lots of little masses gives the total mass. $m=\int_C dm$
- Adding up lots of little areas gives the total area. $A=\int_C dA$
- Adding up lots of little length gives the total length. $s=\int_C ds$
- Adding up lots of little charges gives the total charge. $Q=\int_C dQ$
- Adding up lots of little forces gives the total force. $F=\int_C dF$
- Adding up lots of little work gives the total work. $W=\int_C dW$
- Adding up lots of little widths gives the total width. $width=\int_C dx$
- Adding up lots of little heights gives the total height. $height=\int_C dy$
- Adding up lots of little "changes in $x$" gives the total "change in $x$." The words and the concepts generalize perfectly. Unfortunately, the notation does not generalize perfectly in this instance (as we think of $x$ as both a number and a vector in the same phrase).
- Adding up lots of little "changes in $y$" gives the total "change in $y$." $\text{total change in y}=\int_C dy$
- Adding up lots of little "changes in time" gives the total "change in time." $\text{total change in time}=\int_C dt$
- Find the area of a parallelogram whose edges are given by the vectors $(x,y)$ and $(p,q)$.
Solution
$|xq-yp|$
- Shade the region whose area is given by the double integral $\ds \int_{-2}^{1}\left(\int_{x}^{2-x^2}dy\right)dx$.
- Shade the region (in the $xy$-plane) described by $0\leq \theta\leq \pi/2$ and $2\leq r\leq 4$.
Group problems
- Draw the region in the $xy$ plane described by $\pi/2\leq \theta \leq \pi$ and $0\leq r\leq 5$.
- Compute the integral $\ds\int_{0}^{5}rdr$.
- Compute the double integral $\ds \int_{\pi/2}^{\pi}\left(\int_{0}^{5}rdr\right)d\theta$. Verify you get $\frac{25\pi}{4}$, the area of a quarter circle of radius 5.
- Draw the region in the plane described by $-3\leq y\leq 2$ and $y\leq x\leq 6-y^2$.
- Compute the integral $\ds\int_{y}^{6-y^2}dx$ (assume $y$ is a constant).
- Compute the double integral $\ds \int_{-3}^{2}\left(\int_{y}^{6-y^2}dx\right)dy$.
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi$ and $2\leq r\leq 5$.
- Compute the double integral $\ds \int_{0}^{\pi}\left(\int_{2}^{5}rdr\right)d\theta$.
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi/3$ and $0\leq r\leq 2\sin3\theta$.
- Compute the double integral $\ds \int_{0}^{\pi/3}\left(\int_{0}^{2\sin 3\theta}rdr\right)d\theta$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside the cardiod $r=2+2\cos\theta$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside one petal of the rose $r=3\cos2\theta$.
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