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.
- 5.19 -
- 5.21 -
- 5.22 -
- 5.23 -
- 5.24 -
- 5.25 -
- 5.26 -
- 5.27 -
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.
- 5.19 -
- 5.21 -
- 5.22 -
- 5.23 -
- 5.24 -
- 5.25 -
- 5.26 -
- 5.27 -
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)
- is orthogonal to both $\vec u$ and $\vec v$,
- has a magnitude $|\vec u\times\vec v|$ equal to the area of the parallelogram formed by $\vec u$ and $\vec v$, and
- can be computed by the formula $$\vec u\times\vec v=(u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1).$$
Let $P=(3,0,0)$, $Q=(0,2,0)$, and $R=(0,0,1)$.
- Find a vector orthogonal to both $\vec {PQ}=\left<-3,2,0\right>$ and $\vec {PR}=\left<-3,0,1\right>$.
Solution
$$\left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\ -3 & 2& 0 \\ -3& 0 & 1\end{array} \right|=\left<2,3, 6\right>.$$
- Find the area of the triangle $\Delta PQR$.
Solution
We need half the area of the parallelogram formed from the vector above. We compute $$|\left<2,3, 6\right>| = \sqrt{4+9+36} = 7,$$ which means the area is 7/2.
- Suppose that $S=(x,y,z)$ is a point in the plane that passes through $P$, $Q$, and $R$, and note that $\vec {PS} = \left<x-3,y-0,z-0\right>.$ First compute $(\vec {PQ}\times \vec {PR})\cdot \vec {PS}$. Then explain why this dot product equals zero.
Solution
We have $$(\vec {PQ}\times \vec {PR})\cdot \vec {PS}=\left<2,3, 6\right>\cdot \left<x-3,y-0,z-0\right> = 2(x-3)+3(y-0)+6(z-0).$$ The dot product equals zero because $(\vec {PQ}\times \vec {PR})$ is orthogonal to any vector in the plane formed by $\vec {PQ}$ and $\vec {PR}$. This means an equation of the plane through $P$, $Q$, and $R$ is $$2(x-3)+3(y-0)+6(z-0)=0.$$
- Convert the integral $\int_0^{6\sqrt{2}/2} \int_{y}^{\sqrt{36-y^2}} x \, dx \, dy$ into polar coordinates.
Solution
A solution is $\int_{0}^{\pi/4} \int_0^{6} (r\cos\theta) \, r \, dr\,d\theta$.
- Draw the region in the first octant (all variables positive) that is bounded by the plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{5}=1$. Then set up an integral to compute the volume of this region.
Solution
The region is a triangular pyramid. The plane passes through $(2,0,0)$, $(0,3,0)$, and $(0,0,5)$, so it is the region in space underneath this triangle. One integral (of 6 possible) that gives the volume is $$\int_{0}^{2}\int_{0}^{3(1-\frac{x}{2})}\int_{0}^{5(1-\frac{x}{2}-\frac{y}{3})}dzdydx.$$
Group problems
- Set up an integral formula to compute each of the following: (Use the Mathematica notebook Integration.nb to check your bounds.)
- 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 volume of the solid object in the first octant (all variables positive) that lies under the plane $2x+3y+6z=6$ (so $\frac{x}{3}+\frac{y}{2}+\frac{z}{1}=1$).
- The $y$-coordinate of the center-of-mass (so $\bar y$) of previous object.
- A wire lies along the curve $C$ parametrized by $\vec r(t) = (t^2+1, 3t, t^3)$ for $-1\leq t\leq 2$.
- Compute $ds$. (Remember - a little distance equals the product of the speed and a little time.)
- Set up integral formulas to find $\bar x$, then $\bar y$, then $\bar z$, for the centroid of $C$.
- Consider $\int_{0}^{4}\int_{x}^{4}\cos(y^2)dydx$.
- Draw the region described by the bounds.
- Swap the order of the bounds on the integral (use $dxdy$ instead of $dydx$).
- Actually compute the integral.
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