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.
- 6.8 -
- 6.10 -
- 6.11 -
- 6.12 -
- 6.13 -
- 6.14 -
- 6.15 -
- 6.16 -
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.
- 6.8 -
- 6.10 -
- 6.11 -
- 6.12 -
- 6.13 -
- 6.14 -
- 6.15 -
- 6.16 -
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)
- Draw the curve $\vec r(t) = (3\cos t,3\sin t,t)$ for $0\leq t\leq 6\pi$.
Solution
It's a helix, radius is 3, spiraling counterclockwise (when viewed from above) as it wraps around the $z$-axis for $0\leq t\leq 6\pi$.
- Draw the surface $\vec r(t,v) = (3\cos t, 3\sin t, v)$ for $0\leq t\leq 2\pi$ and $1\leq v\leq 4$.
Solution
It's a right circular cylinder of radius 3, whose center lies along the $z$-axis for $1\leq z\leq 4$.
- Draw the surface $\vec r(t,v) = (v\cos t, v\sin t, 3)$ for $0\leq t\leq \pi$ and $1\leq v\leq 4$.
Solution
It's half a washer from radius 1 to 4, whose center lies at $(0,0,3)$.
- For the vector field $\vec F = (x+2y+3z, 4x+5y+6z, 7x+8y+9z)$, compute
- $D\vec F$,
- $\vec \nabla \cdot \vec F$, and
- $\vec \nabla \times \vec F$.
Solution
We'll do this together in class.
- Find a potential for the vector field $\vec G = \frac{(-x,-y,-z)}{(x^2+y^2+z^2)^{3/2}}$. Hint, start by computing $\int \frac{-x}{(x^2+y^2+z^2)^{3/2}}dx$.
Solution
Using the substitution $u=x^2+y^2+z^2$, we have $$\int \frac{-x}{(x^2+y^2+z^2)^{3/2}}dx = \frac{-1}{2}\int (u)^{-3/2}du =u^{-1/2} = (x^2+y^2+z^2)^{-1/2}.$$ A similar computation yields $$\int \frac{-y}{(x^2+y^2+z^2)^{3/2}}dy = (x^2+y^2+z^2)^{-1/2}\quad\text{and}\quad \int \frac{-z}{(x^2+y^2+z^2)^{3/2}}dz = (x^2+y^2+z^2)^{-1/2}.$$ A potential for $\vec G$ is $$g(x,y,z) = (x^2+y^2+z^2)^{-1/2} = \frac{1}{\sqrt{x^2+y^2+z^2}}.$$
- Find the work done by $\vec G = \frac{(-x,-y,-z)}{(x^2+y^2+z^2)^{3/2}}$ on an object as it moves from $(1,2,2)$ to $(0,3,4)$.
Solution
A potential for $\vec G$ is $$g(x,y,z) = (x^2+y^2+z^2)^{-1/2} = \frac{1}{\sqrt{x^2+y^2+z^2}}.$$ Work done is the difference in potential, which means $$W = g(0,3,4)-g(1,2,2) = \frac{1}{5}-\frac{1}{3} = -\frac{2}{15}.$$ The object moved from 3 units away from the origin to 5 units away from the origin, and negative work was done.
As a side note, the gravitational vector field is $\vec F = \frac{Gm_1m_2(-x,-y,-z)}{(x^2+y^2+z^2)^{3/2}}$, just a constant multiple of the one we worked with above. Electrostatics also uses a very similar vector field.
- Compute the work done by the vector field $\vec F = (4x+2xy,x^2+2y)$ along the curve $C$ parametrized by $\vec r(t) = (3t-1,-5t+2)$ for $0\leq t\leq 1$. [Hint: First find a potential.]
Solution
The vector field has a potential as the derivative $D\vec F =\begin{bmatrix}- &2x \\2x &-\end{bmatrix}$ is symmetric.
- A potential for the vector field is $f(x,y) = 2x^2+x^2y+y^2$ (note $\int 4x+2xy dx = 2x^2+x^2y +C(y)$ and $\int x^2+2y dy = x^2y+y^2+D(x)$).
- The start point is $\vec r(0) = (-1,2)$ and the end point is $\vec r(1) = (2,-3)$.
The work done by $\vec F$ is the difference in potential, which gives $$\int_C\vec F\cdot d\vec r = f(2,-3) - f(-1,2)=(8-12+9)-(2+2+4) = 5-8 = 3.$$
Group problems
- Recall $\vec \nabla = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$. For the vector field $\vec F(x,y,z) = (3xy,2x+4y+5z,y+xz^2)$ do the following.
- Compute the derivative $D \vec F$ (you'll get a 3 by 3 matrix).
- Compute the divergence $\vec \nabla \cdot \vec F$.
- Compute the curl $\vec \nabla \times \vec F$. [Check: $\left(-4, -z^2, 2 - 3x\right)$.]
- Draw each curve or surface given below.
- $\vec r(t) = (3\cos t,3\sin t)$ for $0\leq t\leq 2\pi$.
- $\vec r(u,v) = (3\cos u,3\sin u,v)$ for $0\leq u\leq 2\pi$ and $0\leq v\leq 5$.
- $\vec r(u,v) = (4\cos u,v, 3\sin u)$ for $0\leq u\leq \pi$ and $0\leq v\leq 7$.
- $\vec r(t) = (3\cos t,3\sin t,4t)$ for $0\leq t\leq 6\pi$. (Check: Helix)
- $\vec r(u,v) = (u\cos v,u\sin v,u)$ for $0\leq v\leq 2\pi$ and $0\leq u\leq 4$. (Check: Cone)
- $\vec r(u,v) = (u\cos v,u\sin v,v)$ for $0\leq v\leq 6\pi$ and $2\leq u\leq 4$. (Check: Spiral stair case)
- $\vec r(t) = (0,t,9-t^2)$ for $0\leq t\leq 3$.
- $\vec r(u,v) = (u\cos v,u\sin v,9-u^2)$ for $0\leq v\leq 2\pi$ and $0\leq u\leq 3$.
- 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.
- $\vec F = (2x-y,-x+4y)$
- $\vec F = (y^2+2x,2xy)$
- $\vec F = (x+yz,xz+4yz,xy+2y^2)$
- $\vec F = (x+yz,4yz,xy+2y^2)$
- $\vec F = (x+yz,xz+4yz,xy)$
- $\vec F = (yz,xz+4yz,xy+2y^2)$
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