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. Remember, I start picking presenters at 8am. If you upload something after that time, I may not see it.
- 6.17 -
12:45PM
Thanks for sharing things in Perusall. Here are the presenters for today. Remember, I start picking presenters at 12 noon. If you upload something after that time, I may not see it.
- 6.17 -
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.
Brain Gains (Recall/Generation)
- For the surface $\vec r(u,v) = (u\cos v,u\sin v,v)$ for $0\leq v\leq 6\pi$ and $2\leq u\leq 4$, set up an integral formula to compute the surface area (recall that $d\sigma = \left|\frac{\partial \vec r}{\partial u} \times \frac{\partial \vec r}{\partial v}\right|dudv$).
Solution
The partial derivatives are $$\frac{\partial \vec r}{\partial u} = (\cos v,\sin v,0),\quad \frac{\partial \vec r}{\partial v} = (-u\sin v,u\cos v,1), $$ A normal vector $\vec n$ is the cross product of these two, so $$\vec n = (\sin v, -\cos v,u\cos^2v+u\sin^2v) = (\sin v, -\cos v,u).$$ The surface area is hence $$\sigma =\iint_S d\sigma = \int_{2}^{4}\int_{0}^{6\pi}\sqrt{\sin^2v+\cos^2v+u^2}dvdu = \int_{2}^{4}\int_{0}^{6\pi}\sqrt{1+u^2}dvdu .$$
- Give an equation of the tangent plane to the surface above at $(u,v) = (3,\pi)$. The cross product of the partial derivatives of will get you a normal vector to the surface.
Solution
The partial derivatives $$\frac{\partial \vec r}{\partial u} = (\cos v,\sin v,0),\quad \frac{\partial \vec r}{\partial v} = (-u\sin v,u\cos v,1) $$ are tangent to the surface. Their cross product will be a normal vector to the plane. The cross product of these two is $$\vec n(u,v) = (\sin v, \cos v,u\cos^2v+u\sin^2v) = (\sin v, -\cos v,u).$$ We now use $(3,\pi)$ to find $(x,y,z) = \vec r(3,\pi) = (-3,0,\pi)$ and $\vec n(3,\pi) = (0,1,3)$. A plane with normal vector $(0,1,3)$ that passes through $(-3,0,\pi)$ has an equation $$0(x+3)+(1)(y-0)+3(z-\pi)=0.$$
- Recall for a two dimensional region that $\bar x = \dfrac{\iint_R xdA}{\iint_R dA}$. If you know that the area of a region is $A = 3$ and the centroid is $(5,7)$, compute $\iint_R 2xdA$.
Solution
Since $\bar x = \dfrac{\iint_R xdA}{\iint_R dA}$, we know that $\iint_R xdA = \bar x \iint_R dA = \bar x A$. In other words, we can replace $\iint_R xdA$ with $\bar x A$. We then compute $$\iint_R 2xdA = 2\iint_R xdA = 2\bar x A = 2(5)(3) = 30.$$
- A curve $C$ traverses around a region $R$ with $A = 3$ and the centroid is $(5,7)$. Compute the work done by $\vec F = (2x+3y,x^2+y^2)$ along $C$ (Recall Green's theorem: $\oint Mdx+Ndy = \iint_R N_x-M_y dA$).
Solution
Green's theorem will apply here, because we have a simple closed curve. This gives $$\iint_R N_x-M_y dA =\iint_R 2x-3 dA =2\iint_R xdA - \iint_R 3 dA = 2\bar xA - 3A = 2(5)(3)-3(3) = 21.$$
Group problems
- Compute the work done by $\vec F = (-3y,3x)$ to move an object counterclockwise once along the circle $\vec r(t) = (5\cos t, 5\sin t),$ using Green's theorem $\iint_R N_x-M_y dA$.
- Compute the work done by $\vec F = (y^2,3x)$ to move an object counterclockwise once along the circle $\vec r(t) = (5\cos t, 5\sin t),$ using Green's theorem $\iint_R N_x-M_y dA$.
- Compute the work done by each vector field below to move an object counterclockwise once along the triangle with corners $(0,0)$, $(2,0)$, and $(0,3)$.
- $\vec F = (2x-y,2x+4y)$
- $\vec F = (x^2+y^2,x+y)$
- Consider the surface parametrized by $\vec r(u,v) = (u\cos v, u\sin v, u^2)$ for $0\leq u\leq 2$ and $0\leq v\leq 2\pi$.
- Draw the surface.
- Compute $d\sigma = \left |\dfrac{\partial \vec r}{\partial u}\times\dfrac{\partial \vec r}{\partial u}\right|dudv$.
- Set up an integral formula to compute the surface area $\sigma$.
- Set up an integral formula to compute $\bar z$ for this surface.
- Consider the parametric surface $\vec r(u,v) = (u, u\cos v,u\sin v)$ for $0\leq v\leq \pi$ and $0\leq u\leq 4$.
- Draw the surface.
- Compute $d\sigma = \left |\dfrac{\partial \vec r}{\partial u}\times\dfrac{\partial \vec r}{\partial u}\right|dudv$.
- Set up an integral formula to compute $\bar x$ for this surface.
- Give an equation of the tangent plane to the surface at $(u,v) = (1,\pi/2)$.
- Let $\vec F = (5y,5x)$.
- Find a potential for $\vec F$, or explain why none exists.
- Compute the work done by $\vec F$ to go once counter-clockwise along the circle $\vec r(t) = (3\cos t, 3\sin t)$.
- Let $\vec F = (-5y,5x)$.
- Find a potential for $\vec F$, or explain why none exists.
- Compute the work done by $\vec F$ to go once counter-clockwise along the circle $\vec r(t) = (3\cos t, 3\sin t)$.
- Compute the derivative of each vector field $\vec F$ below (obtaining a square matrix). Then find a potential for $\vec F$ or explain why the vector field has no potential.
- $\vec F = (2x-y,-x+4y)$
- $\vec F = (y^2+2x,2xy)$
- $\vec F = (x+yz,xz+4yz,xy+2y^2)$
- $\vec F = (x+yz,4yz,xy+2y^2)$
- $\vec F = (x+yz,xz+4yz,xy)$
- $\vec F = (yz,xz+4yz,xy+2y^2)$
- Compute the divergence and curl of each vector field.
- $\vec F = (x+yz,xz+4yz,xy+2y^2)$
- $\vec F = (x+yz,4yz,xy+2y^2)$
- $\vec F = (x+yz,xz+4yz,xy)$
- $\vec F = (yz,xz+4yz,xy+2y^2)$
- If a vector field has a potential, then what is the curl of that vector field?
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