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.
- 5.01 - Kai
- 5.03 - Logan, Spencer H
- 5.04 - Rachel, Lucy
- 5.05 - Gavin
- 5.06 -
- 5.07 -
- 5.08 -
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.
- 5.01 - Cheyenne, Jeremy, Forrest
- 5.02 - Jacob
- 5.03 - Michael R
- 5.04 - Christian
- 5.05 -
- 5.06 -
- 5.07 -
- 5.08 -
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 (Rapid Recall)
- The elevation is $z=y+x^2$, along a path $y-2x=5$. Find the $(x,y)$ location of any maxes or mins.
Solution
With $f(x,y) = y+x^2$ and $g(x,y)=y-2x=5$, we have $\vec \nabla f = (2x,1)$ and $\vec \nabla g = (-2,1)$. The system we must solve is $$2x=\lambda(-2), 1=\lambda 1, y-2x=5.$$ The middle equation gives $\lambda =1$, which when plugged into the first equation gives $x=-1$. In the last equation, this give $y=3$.
- The elevation is $z=y+x^2$, along the path $\vec r(t) =(t,2t+5)$. Find the $(x,y)$ location of any maxes or mins.
Solution
Since $\vec r(t) =(t,2t+5)$, we know $x=t$ and $y=2t+5$. Substitution gives $z=(2t+5)+(t)^2$. We compute $\frac{dz}{dt} = 2+2t$. This derivative equals zero when $t=-1$, so we know $(x,y) = \vec r(-1) = ((-1),2(-1)+5) = (-1,3)$. Note that this is almost identical to the last problem.
- For the function $f(x,y)=x^2+xy+y^2-2y$, determine the location of any maxes, mins, or saddles, and classify each location appropriately using eigenvalues.
Solution
The gradient is $\vec \nabla f(x,y) = (2x+y,x+2y-2)$. The gradient equals zero when $2x+y=0$ and $x+2y-2=0$. The first equation tells us $y=-2x$. Plugging this into the second equation gives $x+2(-2x)=2$, or $x=-2/3$. Back substitution tells us $y=4/3$. The only critical point is hence $(x,y) = (-2/3,4/3)$.
The second derivative is $D^2f(x,y) = \begin{bmatrix}2&1\\1&2\end{bmatrix}$. The matrix does not change at the critical point, so this is also $D^2f(-2/3,4/3)$. The eigenvalues are the solution to the equation $(2-\lambda)(2-\lambda)-1=0$. Note that $$(2-\lambda)(2-\lambda)-1 = \lambda^2-4\lambda+3 = (\lambda-3)(\lambda-1).$$ The eigenvalues are $\lambda = 3$ or $\lambda = 1$. Since both are positive, the gradient points away from the critical point. This means the critical point corresponds to a minimum.
Group problems
- Two objects lie on the $z$-axis. The first object has a mass of 2 kg and is located at the point $z=3$ (or rather its center of mass is at that point). The second object has a mass of 4 kg, and after being placed on top of the first object, its center-of-mass is located at the point $z=6$. Find the center-of-mass of the combined system.
- A box lies inside the rectangle $ [-2,6]\times [1,5] $ (so $-2\leq x\leq 6$ and $1\leq y \leq 5$ ).
- What is the center-of-mass $(\bar x,\bar y)$ of the rectangle? (Where is the geometric center?)
- Compute the integral formula $\ds\frac{\int_{-2}^{6}\int_1^5 x dydx}{\int_{-2}^{6}\int_1^5 1 dydx}.$ [Check: 64/32=2. This gives $\bar x$. ]
- Compute the integral formula $\ds\frac{\int_{-2}^{6}\int_1^5 y dydx}{\int_{-2}^{6}\int_1^5 1 dydx}.$ [Check: 96/32=3. This gives $\bar y$. ]
- Compute the integral formula $\ds\frac{\int_1^5 \int_{-2}^{6}x dxdy}{\int_1^5 \int_{-2}^{6}1 dxdy}$, to verify that swapping the order of integration still yields $\bar x = 2$.
- Draw the region described by the bounds of each integral.
- $\ds\int_{0}^{2}\int_{2x}^{4}dydx$
- $\ds\int_{0}^{4}\int_{0}^{y/2}dxdy$
- $\ds\int_{0}^{3\pi/2}\int_{0}^{2+2\cos\theta}rdrd\theta$
- $\ds\int_{-3}^{3}\int_{0}^{9-x^2}\int_{0}^{5}dzdydx$
- $\ds\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{\sqrt{1-x^2}}dydxdz$
- Set up an integral formula to compute each of the following:
- The mass of a disc that lies inside the circle $x^2+y^2=9$ and has density function given by $\delta = x+10$
- The $x$-coordinate of the center of mass (so $\bar x$) of the disc above.
- The $z$-coordinate of the center-of-mass (so $\bar z$) of the solid object in the first octant (all variables positive) that lies under the plane $2x+3y+6z=6$.
- The $y$-coordinate of the center-of-mass (so $\bar y$) of the same object.
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