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.
- 4.3 (Matlab) - Braydon
- 4.7 - Jeremy, Gavin
- 4.8 - Karen
- 4.9 - Rachel
- 4.10 - Santiago
- 4.11 -
- 4.12 -
- 4.13 -
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.
- 4.6 - You can view Trevor's solutions in Perusall. No one else submitted anything for parts 2 and 3.
- 4.7 - Matty, Alan
- 4.8 - Trevor
- 4.9 - Cheyenne
- 4.10 - Oscar
- 4.11 -
- 4.12 -
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 function $f(x,y)=x^2y+5y$. Compute $\vec \nabla f(x,y)$.
Solution
- $\vec \nabla f(x,y)= (2xy, x^2+5)$.
You can do this several ways.
- One option is to first compute the differential $dz=2xydx+x^2dy+5dy$ and factor to get $dz=2xydx+ (x^2+5)dy$. Then extract the partials to get the gradient above. This option is sometimes much more time consuming that the next option.
- Another option is to compute the partials directly, without ever computing the differential.
- If $y$ is constant, then the derivative of $x^2y+5y$ with respect to $x$ is $f_x=2xy+0$. Note that $5y$ is a constant which is why the derivative was zero.
- If $x$ is constant, then the derivative of $x^2y+5y$ with respect to $y$ is $f_y=x^2(1)+5(1)$.
- For $f(x,y)=x^2y+5y$, compute $D_{ (-2,3) }f(1,1)$, the derivative of $f$ in the direction of $(-2,3)$. This is the same as the slope of the function at $(1,1)$ in the direction $(-2,3)$, which you can find using $\frac{dz}{\sqrt{dx^2+dy^2}}$.
Solution
Using directional derivative notation, we have
- $D_{ (-2,3) }f(1,1) = \vec \nabla f(1,1)\cdot \dfrac{ (-2,3) }{|(-2,3)|} = (2,6)\cdot \dfrac{ (-2,3) }{\sqrt{4+9}} = \dfrac{14}{\sqrt{13}}$.
Using differential notation, we have
- $\ds\frac{dz}{\sqrt{dx^2+dy^2}} = \frac{2xydx+ (x^2+5)dy}{\sqrt{dx^2+dy^2}}.$
Plugging in $(x,y)=(1,1)$ and $(dx,dy)=(-2,3)$ gives
- $\ds\frac{dz}{\sqrt{dx^2+dy^2}} = \frac{ (2)(-2)+ (6)(3) }{\sqrt{(-2)^2+(3)^2}} = \frac{14}{\sqrt{13}}.$
- For $z=f(x,y)=x^2y+5y$, give a Cartesian equation of the contour (level curve) that passes through the point $(1,1)$.
Solution
- We know $f(1,1) = 6$, so an equation is $6=x^2y+5y$. While not needed, you can solve for $y$ to obtain $y = \frac{6}{x^2+5}.$
- Draw the vertical cross section of the surface $z=x^2y+5y$ that occurs from letting $y=0$, then $y=1$, then $y=2$.
Solution
We just need to plot the three parabolas
- $z=0$,
- $z=x^2+5$, and
- $z=2x^2+10$.
The links below point to WolframAlpha.
- Draw the vertical cross section of the surface $z=x^2y+5y$ that occurs from letting $x=0$, then $x=1$, then $x=2$.
Solution
We draw the three lines
- $z=5y$,
- $z=6y$, and
- $z=9y$.
The links below point to WolframAlpha.
Group problems
- Let $g(x,y) =xy^3$.
- Compute $dg$.
- State $g_x$ and $\dfrac{\partial g}{\partial y}$. Then state $\vec \nabla g$.
- Find the directional derivative (slope) of $g$ at $P=(3,1)$ in the direction $(-3,2)$.
- Find the directional derivative of $g$ at $P=(3,1)$ in the direction $(2,-5)$.
- Consider the function $z=f(x,y)=x^2+y^2-4$.
- Construct a contour plot of $f$. So let $z=0$ and draw the resulting curve in the $xy$ plane. Then let $z=5$ and draw the resulting curve in the $xy$ plane. Then pick other values for $z$ and draw the resulting curve in the $xy$ plane. If you get a bunch of concentric circles, you're doing this right. On each circle you draw, write the height of that circle.
- Construct a 3D surface plot of the function.
- Consider the function $z=4-y^2$.
- Construct a 2D contour plot.
- Construct a 3D surface plot.
- Consider the function $f(x,y)=4-|x|$.
- Construct a 2D contour plot.
- Construct a 3D surface plot.
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