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.
- 3.31 - Jeremy
- 3.33 - Ryan
- 3.34 - Lucy
- 3.35 - Ethan
- 4.1 -
- 4.2 - Tanner
12:45PM
Thanks for sharing things in Perusall. Here are the presenters for today.
- 3.31 - Alan
- 3.33 - Aaron
- 3.34 - Oscar
- 3.35 - Brad
- 4.1 -
- 4.2 -
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
- If we know $x=3u+2v$ and $y=-u+4v$, then areas in the $uv$-plane are multiplied by how much to obtain an area in the $xy$-plane.
Solution
Note that $dx = 3du+2dv$ and $dy=-1du+4dv$. We can write this as $$(dx,dy)=(3,-1)du+(2,4)dv.$$ The area of the parallelogram formed by the vectors $(3,-1)$ and $(2,4)$ is $$A=|12+2|=14.$$ The Jacobian of the transformation is 14, so areas in the $uv$-plane are multiplied by 14 to obtain areas in the $xy$-plane. In terms of integrals, we write $$A_{xy} = \iint_{R_{xy}}dxdy = \iint_{R_{uv}}14 dudv.$$ The 14 is our area stretch factor that appears when we change from one coordinate system to another.
- If we know $x=2u+3v$ and $y=4u+5v$, and a region in the $uv$-plane has area $A_{uv} = 5$ square units. What is the area of the transformed region in $xy$-plane?
Solution
Note that $dx = 2du+3dv$ and $dy=4du+5dv$. We can write this as $$(dx,dy)=(2,4)du+(3,5)dv.$$ The area of the parallelogram formed by the two vectors above is $A=|2\cdot 5-3\cdot 4|=2$. This means $$dA_{xy} = 2 dudv,$$ and so the area of the transformed region is twice the original area, hence $$A_{xy} = 5\cdot 2 = 10.$$
- Draw the two curves $r=2-2\cos\theta$ and $r=2\cos\theta$, and locate their point of of intersection.
Solution
We'll draw it together. To find the point of intersection, we need $2-2\cos\theta = 2\cos\theta$ which means $2 = 4\cos\theta$ or $\cos\theta = \frac{1}{2}$. This occurs at $\theta = \pi/3$, which gives $r=2\cos(\pi/3) =1$.
- Set up a double integral formula that gives the area of the region in the $xy$ plane that lies inside the curve $r=2-2\cos\theta$ and inside the curve $r=2\cos\theta$.
Solution
Notice that rays starting from the origin, and heading out the curves, hit the cardioid for $0\leq \theta\leq \pi/3$, and then swap to hitting the circle for $\pi/3\leq theta\leq \pi/2$. This means we'll need two different integrals. The solution is $$ \int_{0}^{\pi/3}\int_{0}^{2-2\cos\theta} r dr d\theta + \int_{\pi/3}^{\pi/2}\int_{0}^{2\cos\theta} r dr d\theta. $$
Group problems
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside the cardiod $r=2+2\cos\theta$.
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi/4$ and $0\leq r\leq 3\cos2\theta$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside one petal of the rose $r=3\cos2\theta$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside the curve $r=2-2\cos\theta$ and outside the curve $r=2\cos\theta$.
- Draw and shade the region in the $xy$-plane that lies inside the curve $r=3+2\cos\theta$ and outside the curve $r=1$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside the curve $r=3+2\cos\theta$ and outside the curve $r=1$.
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