9:00 AM Jamboard Links
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12:45 PM Jamboard Links
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Presenters
- 24 - Trevor
- 25 - Michael
- 26 - Alan
- 27 - Joshua
- 1 - Rick, Chad
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 force given by $\vec F = (y,-x+y)$ acts on an object as it moves along the curve $\vec r(t) =(t^2,t)$ for $-1\leq t\leq 2$. In the integral $\int_C Mdx+Ndy$, state $M$, $N$, $dx$, and $dy$ all in terms of $t$.
Solution
We have $$\int_{-1}^{2} \underbrace{[t]}_{M}\underbrace{[2t]}_{dx/dt}+\underbrace{[-(t^2)+(t)]}_{N}\underbrace{[1]}_{dy/dt}dt.$$
- Set up an integral to find the length of the curve with parametrization $\vec r(t) =(t^2,t)$ for $-1\leq t\leq 2$.
Solution
The velocity is $\vec v(t) = (2t,1)$. The speed is $v = \sqrt{(2t)^2+1^2}$. This means the arc length differential is $$\underbrace{ds}_{\text{little distance}} = \underbrace{\sqrt{(2t)^2+1^2}}_{\text{speed}}\underbrace{dt}_{\text{little time}} = \sqrt{4t^2+1}dt.$$ The arc length is hence $$s=\int_C ds = \int_{-1}^2\sqrt{4t^2+1}dt.$$
- A wire lies along the curve $\vec r(t) =(t^2,t)$ for $-1\leq t\leq 2$. The density (mass per length) of the wire at a point $(x,y)$ on the curve is given by $\delta(x,y) = y+2$, which means little bits of mass are equal to $dm = \delta(x,y) ds$. Set up an integral formula that gives the total mass of the wire.
Solution
The mass is $$s=\int_C(y+2) ds = \int_{-1}^2(\underbrace{t}_{y}+2)\underbrace{\sqrt{4t^2+1}dt}_{ds}.$$ We used the fact that $y=t$ from $\vec r(t) =(t^2,t)$ to replace $y+2$ with $(t+2)$.
Group problems
After each problem, or each part, remember to take turns letting people write.
- Consider the curve $C$ parametrized by $\vec r(t) = (t^2, t^3)$ for $0\leq t\leq 2$.
- Give a vector equation of the tangent line to the curve at $t=1$.
- Set up an integral that gives the length of this curve. Just set it up.
- A wire lies along the curve $C$. The density (mass per length) of the wire at a point $(x,y)$ on the curve is given by $\delta(x,y) = y+2$, which means little bits of mass along little length $ds$ are given by $dm = \delta ds$. Set up an integral formula that gives the total mass of the wire.
- The wire contains charged particles. The charge density (charge per length) at a point $(x,y)$ on the curve is given by the product $q(x,y)=xy$, which means little bits of charge along little length $ds$ are given by $dQ = q ds$. Set up an integral formula that gives the total charge on the wire.
- Consider the ray from the origin through the point $P=(-2,2)$. What's the angle between this ray and the positive $x$ axis? What the distance from the origin to $P$?
- Plot the polar points with $(r,\theta)$ given by $(2,0)$, $(2,\pi/6)$, $(-2,\pi/6)$, $(4,\pi/2)$, $(-4,\pi/2)$.
- Give a polar equation of the curve $2x+3y=4$. (So substitute $x=r\cos\theta$ and $y=r\sin\theta$, and then solve for $r$.)
- Give a Cartesian equation of the polar curve $r=\tan\theta\sec\theta$. (Use $x=r\cos\theta$ and $y=r\sin\theta$.)
- We know $x=r\cos\theta$ and $y=r\sin\theta$. Compute $dx$ in terms of $r, \theta,dr, d\theta$. (If you need to, assume that everything depends on $t$, compute derivatives, then multiply by $dt$.)
- Plot the curve $r=3-2\sin\theta$.
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