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.
- 3.9 - Jared, Cecilia
- 3.10 - Jae
- 3.11 - Spencer B
- 3.12 - Kai, Tanner
- 3.13 - Gavin
- 3.14 -
- 3.15 -
- 3.16 -
12:45PM
Thanks for sharing things in Perusall. Here are the presenters for today.
- 3.10 - Rick M
- 3.11 - Nicholas B
- 3.12 - Jeremy J, Reed H
- 3.13 - Alan L
- 3.14 - Joshua S
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
- A curve passes through the point $(3,5)$. Differentials at this point are $dx = 7dt$ and $dy=11dt$. Give an equation of the tangent line to the curve at this point.
Solution
One option is to give a vector equation such as $$ \begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}7\\11\end{pmatrix}t+\begin{pmatrix}3\\5\end{pmatrix} \quad\text{or}\quad \begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}7t+3\\11t+5\end{pmatrix} .$$ Another option is to use point-slope form and give $$y-5 = \frac{11}{7}(x-3).$$ Another option is to give parametric equations $$x=7t+3, y=11t+5.$$ All these answers are acceptable. All will generalize to higher dimensions in different ways.
- For the curve $r=5\cos 2\theta$, graph the curve in the $r\theta$ plane.
Solution
We'll do this together.
- For the curve $r=5\cos 2\theta$, graph the curve in the $xy$ plane.
Solution
We'll do this together.
- For the curve $r=5\cos 2\theta$, compute $\frac{dx}{d\theta}$.
Solution
We know that $x=r\cos\theta$, and since $r=5\cos 2\theta$, we have $x = 5\cos 2\theta \cos\theta$. From here we just use the product rule to obtain $$\begin{align} \frac{dx}{d\theta} &= \frac{d}{d\theta}(5\cos 2\theta) \cos\theta+5\cos 2\theta \frac{d}{d\theta}(\cos\theta)\\ &=(-10\sin 2\theta) \cos\theta+5\cos 2\theta (-\sin\theta). \end{align}$$
Group problems
- Plot the curve $r=4-4\cos\theta$ in both the $r\theta$-plane, and the $xy$-plane. [Hint: Make an $(r,\theta)$ table, but pick values for $\theta$ that make $\cos\theta$ easy to compute. Did you get a heart shaped object?]
- Plot the curve $r=3\sin2\theta$ in both the $r\theta$-plane, and the $xy$-plane. [Hint: Make an $(r,\theta)$ table, but pick values for $\theta$ that make $\sin2\theta$ easy to compute. Did you get a clover?]
- Consider the polar curve $r=\frac{2\theta}{\pi}$.
- Draw the curve in the $xy$-plane for $0\leq \theta\leq 4\pi$. Check your work with 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 $\theta=\pi$.
- 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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