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
- 4.8 - Karen
- 4.11 - Spencer B, Cecilia
- 4.12 - Ryan C
- 4.13 - Kallan
- 4.14 - Ethan
- 4.15 - Class
- 4.16 - Tanner
- 4.17 -
- 4.18 -
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.
- 4.8 - Trevor
- 4.11 -
- 4.12 -
- 4.13 -
- 4.14 -
- 4.15 -
- 4.16 -
- 4.17 -
- 4.18 -
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)
- Earlier in the semester we showed that $\vec u\cdot \vec v = |\vec u||\vec v|\cos\theta$, where $\theta$ is the angle between vectors $\vec u$ and $\vec v$. Find a unit vector $\hat w$ so that $(3,4)\cdot \hat w$ is as large as possible.
Solution
Using $\vec u\cdot \vec v = |\vec u||\vec v|\cos\theta$, we find that $(3,4)\cdot \hat w = (5)(1)\cos\theta$. This is largest when $\cos\theta = 1$, or $\theta = 0$. So the unit vector we need points in the same direction as $(3,4)$, which means $\hat w = \frac{1}{5}(3,4)$.
Let's look at 4.8 now. Recall that the slope of a function in the direction $\vec u$ is given by $$D_{\vec u}f(P) = \vec \nabla f(P)\cdot \hat u.$$ This is the dot product of the gradient with a unit vector. We have the tools needed to prove that the gradient points in the direction of greatest slope, one of our first optimization facts.
- Consider the function $z=\sin(x)+e^y$, where $x=3t$ and $y=t^2$. Compute $\frac{dz}{dt}$.
Solution
There are two ways to do this.
- Substitution gives $z=\sin(3t)+e^{t^2}$. Differentiation (using the chain rule) then gives $$\frac{dz}{dt}=\cos(3t)\frac{d}{dt}(3t)+e^{t^2}\,\frac{d}{dt}(t^2)=\cos(3t)3+e^{t^2}\,2t.$$
- Differentials give $dz = \cos(x)dx+e^ydy$, with $dx = 3dt$ and $dy=2tdt$. Substitution then gives $$dz = \cos(3t)3dt+e^{t^2}\,2tdt\quad\text{or}\quad\frac{dz}{dt}=\cos(3t)3+e^{t^2}\,2t.$$
In both cases, we obtained the same solution of $$\frac{dz}{dt}=\underbrace{\cos(3t)}_{f_x}\underbrace{3}_{\frac{dx}{dt}}+\underbrace{e^{t^2}}_{f_y}\underbrace{2t}_{\frac{dy}{dt}}.$$ Writing the solution above symbolically gives us the chain rule $$\frac{dz}{dt} = f_x\frac{dx}{dt}+f_y\frac{dy}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}.$$
- Suppose $dz = e^{x^2}dx+\cos(2y)dy$, $x=3t$, and $y=t^2$. Compute $\frac{dz}{dt}$.
Solution
This time we don't know what the function $z$ equals, so we cannot first substitute and then differentiate. We do know however that $f_x = e^{x^2}$ and $f_y = \cos(2y)$. We can compute differentials and then substitute.
- Note that $dx = 3dt$ and $dy = 2tdt$. Substitution then gives $$dz = e^{(3t)^2}3\,dt+\cos(2(t^2))2t\,dt \quad\text{and so}\quad \frac{dz}{dt} = e^{(3t)^2}3+\cos(2(t^2))2t.$$
Again, the the solution above symbolically gives us the same chain rule $$\frac{dz}{dt} = f_x\frac{dx}{dt}+f_y\frac{dy}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}.$$
- For $f(x,y)=x^2-y^2$, draw the level curve (contour) that passes through the point $(0,1)$.
Solution
- We compute $f(0,1) = -1$. We then need to draw the curve $-1=x^2-y^2$ or $1=y^2-x^2$. It's a hyperbola opening up and down along the $y$-axis.
Group problems
Please download the Mathematica Notebook ContourSurfaceGradient.nb. You do not need to become an expert at using Mathematica, but learning some basic commands can only help.
- Compute $f_x$ and $\frac{\partial f}{\partial y}$ for each of the following (try to do it without computing $df$ first).
- $f(x,y) = x^2y$
- $f(x,y) = 3xy+4y^2$
- $f(x,y) = \sin(xy^2)$
- Consider the elevation function $f(x,y)=e^x\sin y$ and the path $\vec r(t) = (t^2,t^3)$.
- Compute $f(\vec r(t))$ and then compute $\frac{df}{dt}$.
- Find $df$ in terms of $x$, $y$, $dx$, and $dy$. Then find $dx$ and $dy$ in terms of $t$ and $dt$.
- Use substitution to state $df$ in terms of $dt$ and divide by $dt$ to obtain $\frac{df}{dt}$. [You should have the same answer as the first part.]
- In your solution for $\frac{df}{dt}$, label each of $f_x$, $f_y$, $\frac{dx}{dt}$ and $\frac{dy}{dt}$ to verify that $\frac{df}{dt} = f_x\frac{dx}{dt}+f_y\frac{dy}{dt}$.
- Consider the function $z=4-y^2$.
- Construct a 2D contour plot by hand. So pick several values for $z$ and plot the resulting curves. If you end up with lots of horizontal lines in the $xy$-plane, you're doing this correctly. Write the height on each horizontal line you draw.
- Construct a 3D surface plot by hand.
- Construct both the above with software.
- Consider the function $f(x,y)=4-|x|$.
- 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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