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.
- 3.14 - Nathan T
- 3.15 - Jae
- 3.16 - Lucy
- 3.17 - Jeremy
- 3.18 - Tanner
- 3.19 - Denali
- 3.20 -
- 3.21 -
12:45PM
Thanks for sharing things in Perusall. Here are the presenters for today.
- 3.14 - Trevor
- 3.15 - Carter
- 3.16 - Michael
- 3.17 - Next Time
- 3.18 - Joshua
- 3.19 - Joshua
- 3.20 -
- 3.21 -
- 3.22 -
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
- Graph the curve $r=3-2\cos\theta$ in the $xy$-plane (in the $r\theta$-plane can be useful).
Solution
We'll do this one together.
- Given that $\ds \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta-r\sin\theta$ and $\ds \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta+r\cos\theta$, write a vector equation, in the $xy$-plane, for the line tangent to the curve $r=3-2\cos\theta$ at $\theta=\pi/2$ .
Solution
Note that when $\theta=\pi/2$, we have $r=3$. The $(x,y)$ coordinate is $(0,3)$. Since $\frac{dr}{d\theta}=-2\sin\theta$, we get $$\ds \frac{dx}{d\theta} = (2\sin\theta)\cos\theta-(3-2\cos\theta)\sin\theta\text{ and }\ds \frac{dy}{d\theta} = (2\sin\theta)\sin\theta+(3-2\cos\theta)\cos\theta.$$ At $\theta=\pi/2$, the above gives $$\frac{dx}{d\theta}\bigg|_{\theta = \pi/2}=(2)(0)-(3)(1) = -3\text{ and }\frac{dy}{d\theta}=(2)(1)+(3)(0) = 2.$$ Putting this all together, we have $$ (x,y) = (-3,2)t+(0,3)\text{ or }\vec r(t) = (-3,2)t+(0,3).$$
- Write an integral formula for the arc length of the curve $r=3-2\cos\theta$.
Solution
Arc length in general is $\int_\alpha^{\beta}\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta$. From our graph, we discover that we need $0\leq\theta\leq \beta$ for our bounds to trace over the curve once. The values were computed above for $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$. This gives the arc length as $$s=\int_0^{2\pi}\sqrt{\left((2\sin\theta)\cos\theta-(3-2\cos\theta)\sin\theta\right)^2+\left((2\sin\theta)\sin\theta+(3-2\cos\theta)\cos\theta\right)^2}d\theta.$$
- Find the area of a parallelogram whose edges are the vectors $(a,b)$ and $(c,d)$.
Solution
From problem 20, we learn the answer is $A=|ad-bc|$.
- Find the area of a parallelogram whose corners are the points $(1,0)$, $(0,3)$, $(5,4)$, and $(6,1)$.
Solution
We need two vectors $(a,b)$ and $(c,d)$ for the edges to use $A=|ad-bc|$. From $(1,0)$ to $(0,3)$ we get $(a,b) = (-1,3)$. From $(0,3)$ to $(5,4)$ we get $(c,d) = (5,1)$. We then have $$A = |ad-bc| = |(-1)(1) - (3)(5)| = |-16| = 16.$$
Group problems
- Consider the polar curve $r=\frac{2\theta}{\pi}$. Graph is a spiral See Desmos.
- Compute $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$. (Hint: They appear in the integral in the next part.)
- What quantity does the following integral calculate? $$ \int_0^{4\pi}\sqrt{\left(\frac{2}{\pi}\cos\theta-\frac{2\theta}{\pi}\sin\theta\right)^2+\left(\frac{2}{\pi}\sin\theta+\frac{2\theta}{\pi}\cos\theta\right)^2}d\theta$$
- Find the slope $dy/dx$ at $\theta=\pi$. [Check: $\frac{dy}{dx} = \frac{-2}{-2/\pi} = \pi$.]
- Give an equation of the tangent line at $\theta = \pi$.
- Consider the curve $v=u^2$ and use the change-of-coordinates $x=2u+v$, $y=u-2v$.
- Draw the curve in both the $uv$-plane and the $xy$-plane.
- Compute $dx$ and $dy$ in terms of $u$ and $du$ (we know $dv = 2udu$ since $v=u^2$).
- Find the slopes $\frac{dv}{du}$ and $\frac{dy}{dx}$ at $u=-2$.
- Give a vector equation of the tangent line to the curve in the $uv$-plane at $u=-2$. [Check: $(u,v) = (1,-4)t+(-2,4)$.]
- Give a vector equation of the tangent line to the curve in the $xy$-plane at $u=-2$. [Check: $(x,y) = (-2,9)t+(0,-10)$.]
- Let $v=u^3$ and use the coordinates $x=2u+v$, $y=u-2v$.
- Draw the curve in both the $uv$-plane, and the $xy$-plane (make a $(u,v)$ and $(x,y)$ table).
- Find $dx$ and $dy$ in terms of $u$ and $du$.
- Find the slope $dy/dx$ at $u=1$.
- Give a vector equation of the tangent line to the curve in the $xy$ plane at $u=1$.
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