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.
- 4.22 - Kai
- 4.23 - Makenzy
- 4.24 - Karen
- 4.25 - Olivia, Braydon
- 4.26 -
- 4.27- Lucy
- 4.28 -
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.
- 4.21 - Matty, Alan
- 4.22 - Reed, Forrest
- 4.23 -
- 4.24 - Aaron
- 4.25 -
- 4.26 -
- 4.27 - Oscar
- 4.28 -
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)
- For the function $f(x,y) = 3x^2+5xy+y^2$, compute $f_y$ and then both $f_{yx}$ and $\frac{\partial^2f}{\partial y^2}$.
Solution
We have
- $f_y = 0+5x+2y$,
- $f_{yx} = 0+5+0$, and
- $\frac{\partial^2f}{\partial y^2} = 0+0+2$.
- Let $f(x,y)=2x^2+4y$, and $g(x,y)=2x+y$. Solve the system $\vec \nabla f = \lambda \vec \nabla g$ together with $g(x,y)=3$.
Solution
We have $\vec \nabla f = (4x,4)$ and $\vec \nabla g = (2,1)$. The equation $\vec \nabla f = \lambda \vec \nabla g$ means $$(4x,4) = \lambda(2,1) = (2\lambda,1\lambda).$$ This gives us the two equations $4x=2\lambda$ and $4 = \lambda$. The second equation tells us $\lambda=4$. The first equation tells us $x=2\lambda/4 = 2$. Substitution of $x=2$ into $2x+y=3$ tells us $y=-1$.
- The surface $x^2+3y^2-4z=-5$ passes through the point $P=(2,1,3)$. Give an equation of the tangent plane to this surface at $P$. Hint: Use differentials.
Solution
Differentials tell us $$2xdx+6ydy-4dz=0.$$ We know $x=2$, $y=1$, and $z=3$. We also know that if $Q=(x,y,z)$ is another point on the plane, then the change from $P$ to $Q$ is $dx = x-2$, $dy=y-1$, and $dz=z-3$. Substitution (plug it in, plug it in) gives the equation of the tangent plane as $$2(2)(x-2)+6(1)(y-1)-4(z-3)=0.$$
- Consider the function $f(x,y,z) = 4x^2+4y^2+z^2$. The level surface that passes through $(1/2,0,\sqrt{3})$ is the ellipsoid $4=4x^2+4y^2+z^2$ (because $f(1/2,0,\sqrt{3})=4$). Draw this ellipsoid, which we can rewrite at $1=x^2+y^2+\frac{z^2}{4}$.
Solution
We start by drawing several cross sections, found by holding a variable constant.
- Let $x=0$ to get an ellipse in the $yz$-plane that opens up and down 2, and left and right 1.
- Let $y=0$ to get an ellipse in the $xz$-plane that opens up and down 2, and left and right 1.
- Let $z=0$ to get a circle of radius 1 in the $xy$-plane.
Putting those three curves together gives an ellipsoid that is 2 units tall in the $z$ direction, with a circle of radius 1 intersecting the ellipsoid in the $xy$-plane.
- A parallelogram has edges $(5-\lambda, 2)$ and $(3, 4-\lambda)$. Find $\lambda$ so that the area of the parallelogram is zero.
Solution
The area is $$A=|(5-\lambda)(4-\lambda)-(2)(3)| = |\lambda^2-9\lambda+20-6| = |\lambda^2-9\lambda-14| = |(\lambda - 7)(\lambda - 2)|.$$ This equals zero when $\lambda = 7 $ or $\lambda =2$.
We call these the eigenvalues of the matrix $\begin{bmatrix}5&3\\2&4\end{bmatrix}$. To find the eigenvalues of a 2 by 2 matrix, we
- subtract $\lambda$ from the upper left and lower right entries in the matrix, and
- then we find the values of $\lambda$ that give zero area for a parallelogram formed by the columns of this matrix.
This will always result in solving a quadratic equation.
Group problems
- Find the eigenvalues of the following matrices (take turns).
- $\begin{bmatrix}2&4\\4&2\end{bmatrix}$, $\begin{bmatrix}2&3\\1&4\end{bmatrix}$, $\begin{bmatrix}1&6\\4&3\end{bmatrix}$, $\begin{bmatrix}3&2\\1&2\end{bmatrix}$.
- Check: 6,-2; 5,1; 7,-3; 4,1
- Consider the function $f(x,y,z) = 4x^2+4y^2+z^2$. We'll be analyzing the surface at the point $P=(1/2,0,\sqrt{3})$.
- Compute the gradient $\vec\nabla f(x,y,z)$, and then give $\vec\nabla f(P)$.
- Compute the differential $df$, and then the differential at $P$. [Check: For the latter, $df = 4dx+0dy+2\sqrt{3}dz$]
- For a level surface, the output remains constant (so $df=0$). If we let $(x,y,z)$ be a point on the surface really close to $P$, then we have $dx=x-1/2$, $dy=y-0$ and $dz = z-?$. Plug this information into the differential at $P$ to obtain an equation of the tangent plane to the surface.
- Give an equation of the tangent plane to the level surface of $f$ that passes through $(1,2,-3)$. [Check: $0=8(x-1)+16(y-2)-6(z+3)$.]
- Give an equation of the tangent plane to the level surface of $f$ that passes through $(a,b,c)$.
- Consider the function $f(x,y)=2-|x|$.
- Construct a 2D contour plot. Label your contours with their corresponding height.
- Construct a 3D surface plot.
- Construct both the above with software.
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