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.
- 5.05 - Lucy
- 5.06 - Nathan B
- 5.07 - Gavin
- 5.08 - Cecillia
- 5.09 -
- 5.10 - Ethan
- 5.11 -
- 5.12 -
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.
- 5.05 - Jacob, Michael R
- 5.06 - Jeremy
- 5.07 - Cheyenne
- 5.08 -
- 5.09 -
- 5.10 - Alan
- 5.11 -
- 5.12 -
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)
- Consider the two vectors $\vec u = (a,b,c)$ and $\vec v = (d,e,f)$. Compute $\vec u\cdot \vec v$. What must this equal if the two vectors are orthogonal?
Solution
We have $\vec u\cdot \vec v = ad+be+cf$. If the vectors are orthogonal, then this dot product equals zero.
- A wire lies along the curve $\vec r(t) =(t^3, 3t^2)$ for $0\leq t\leq 7$ and has a uniform density of $\delta = 5$, set up an integral formula that gives the mass of this curve.
Solution
Remember that mass is found by adding up little masses, so $m = \int_C dm$. A little mass $dm$ is equal to density $\delta$ multiplied by little length $ds$, which gives $dm = \delta ds$. We found $ds$ earlier in the semester by multiplying speed by a little time. The velocity is $\frac{d\vec r}{dt} = (3t^2,6t)$ which means the speed is $v(t) = \sqrt{(3t^2)^2+(6t)^2}$. The gives $$m = \int_C dm= \int_C \delta ds = \int_C 5 v(t) dt = \int_0^7 5 \sqrt{(3t^2)^2+(6t)^2}dt.$$
- Consider the region $R$ that is bounded by the lines $y=0$, $x=4$, and $y=x/2$. The density (mass per area) is given by $\delta(x,y)$.
- Draw the region.
- Set up a double integral to compute the mass using $\ds\int_{?}^{?}\int_{?}^{?}\delta(x,y) dydx$.
- Set up a double integral to compute the mass using $\ds\int_{?}^{?}\int_{?}^{?}\delta(x,y) dxdy$.
Solutions
The graph is a triangle underneath the line $y=x/2$ for $0\leq x\leq 4$. The requested integrals are
- $\ds\int_{0}^{4}\int_{0}^{x/2}\delta(x,y) dydx$.
- $\ds\int_{0}^{2}\int_{2y}^{4}\delta(x,y) dxdy$.
Group problems
- Consider the integral $\ds\int_{0}^{3}\int_{0}^{x}dydx$.
- Shade the region whose area is given by this integral.
- Compute the integral.
- Now compute $\ds\int_{0}^{x}\int_{0}^{3}dxdy$. Why do you not get a number?
- Adjust the bounds on the integral above (keeping the order $dxdy$) so that they describe the same region as the first part. In other words, fill in the question marks below so that the integral's bounds describe the same region as the first part of this problem. $$\ds\int_{?}^{?}\int_{?}^{?}dxdy$$
- A parallelogram has edge lengths of a and b. The acute angle in the parallelogram is $\theta$. Explain why the area of the parallelogram is $ab\sin\theta$. [Hint: Draw a picture, label the edges, add an extra line to form a right triangle with $\theta$ as one of the angles.]
- Draw the region described by the bounds of each integral. (Use the Mathematica notebook Integration.nb to check your work.)
- $\ds\int_{0}^{3\pi/2}\int_{0}^{2+2\cos\theta}rdrd\theta$
- $\ds\int_{-3}^{3}\int_{0}^{9-x^2}\int_{0}^{5}dzdydx$
- $\ds\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{\sqrt{1-x^2}}dydxdz$
- 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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