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.
- 3.24 - Spencer B
- 3.25 - Spencer H
- 3.28 (cd) - Jeremy (finish up)
- 3.29 - Kallan
- 3.30 - Kylar
- 3.31 - Jeremy
- 3.32 - Zack
- 3.33 - Next Time
12:45PM
Thanks for sharing things in Perusall. Here are the presenters for today.
- 3.25 - Brad and Chad
- 3.28 - Angel
- 3.29 - Forrest and Matty
- 3.30 - Alan and Ben
- 3.31 - Next Time
- 3.32 - Alan
- 3.33 -
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
- Draw the region described by $\pi/6\leq \theta\leq \pi/3$ and $4\leq r\leq 5$.
Solution
The problem didn't specify if you should draw in the $r\theta$ or $xy$ plane, which means (by convention) that we should draw in the $xy$-plane. We'll do this together. This region is the image of a typical polar rectangle in the $xy$-plane. Our goal is to understand the area stretch factor between polar and Cartesian coordinates.
- Find the area of a parallelogram whose edges are parallel to the vectors $(\cos\theta dr,\sin\theta dr)$ and $(-r\sin\theta d\theta,r\cos\theta d\theta)$.
Solution
$|r\cos^2\theta dr d\theta +r\sin^2\theta dr d\theta| = |r dr d\theta|$
- Compute the polar double integral $\ds \int_0^{2\pi}\int_{0}^{7} r\, dr d\theta$.
Solution
We compute $$\ds \int_0^{2\pi}\int_{0}^{7} r\, dr d\theta = \int_0^{2\pi}\frac{r^2}{2}\bigg|_0^{7} d\theta = \int_0^{2\pi}\frac{7^2}{2} d\theta = \frac{7^2}{2}\theta\bigg|_0^{2\pi} = \pi7^2. $$ This shows that that area inside a circle of radius $7$ is $\pi 7^2$. Changing the 7 to an an arbitrary radius $a$, the above work shows that the area inside a circle of radius $a$ is $\pi a^2$.
- For the curve $r=7$, compute $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$.
Solutions
Since $x=r\cos\theta = 7\cos\theta$ and $y=r\sin\theta=7\sin\theta$, we have $\frac{dx}{d\theta} = -7\sin\theta$ and $\frac{dy}{d\theta}=7\cos\theta$. The curve is a circle of radius $7$. The integral
Group problems
- Consider the polar curve $r=7$. We already showed that $\frac{dx}{d\theta} = -7\sin\theta$ and $\frac{dy}{d\theta} = 7\cos \theta$. Use the arc length formula $$\int_C\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta$$ to find the arc length for the portion of this curve with $0\leq \theta\leq \alpha$. Your result will show that the arc length on a circle of radius $7$ through an angle $\alpha$ is $s = 7\alpha$. This generalizes to a circle of radius $r$ through an angle $d\theta$ to give $s=rd\theta$.
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi$ and $2\leq r\leq 5$.
- Compute the double integral $\ds \int_{0}^{\pi}\left(\int_{2}^{5}rdr\right)d\theta$. [Check: $\frac{25\pi}{2} - \frac{4\pi}{2}$, the difference in areas of two semicircles.]
- 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 $r=2+2\cos\theta$.
- 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 $r=3\cos2\theta$.
- 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 $r=1$.
- 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 inside the curve $r=2\cos\theta$.
- 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 $r=2\cos\theta$.
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