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.
- 4.28.5 - No One
- 4.29 - Braydon, Marissa
- 4.30 - Tanner
- 4.31 - Santiago
- 4.32 - Zack
- 4.33 -
- 4.34 -
- 4.35 -
- 4.36 -
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.
- 4.28.5 -
- 4.29 - Trevor
- 4.30 - Alan
- 4.31 -
- 4.32 -
- 4.33 -
- 4.34 -
- 4.35 -
- 4.36 -
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 vector field $\vec F(x,y) = (5x+6y,xy^2)$, compute the derivative $D\vec F(x,y)$.
Solution
We compute the two partial derivatives, namely
- $\vec F_x = (5,y^2)$
- $\vec F_y = (6,2xy)$
These vectors are the columns of the derivative of $\vec F$, which means $$D\vec F(x,y) = \begin{bmatrix}5&6\\y^2&2xy\end{bmatrix}. $$
- For the function $f(x,y)=x^2+4xy+3y^2-10x-18y$, find the critical points by solving $Df(x,y)=\begin{bmatrix}0&0\end{bmatrix}$.
Solution
The only critical point is $(x,y)=(3,1)$.
The first derivative is $$Df(x,y) = \begin{bmatrix}2x+4y-10&4x+6y-18\end{bmatrix}.$$ To find the critical points, we need this derivative to be the zero matrix, which means we must solve the system of equations $$ 2x+4y-10 = 0,\quad 4x+6y-18=0.$$ This is equivalent to solving $2x+4y=10$ and $4x+6y=18$. Dividing both sides of the second equation by 2 gives the system $2x+4y=10$ and $2x+3y=9$. Subtracting the first equation from the second yields $y=1$, and substitution into either equation yields $x=3$.
- The function $f(x,y) = x^3+3xy+y^3$, the first and second derivatives are $$Df(x,y) = \begin{bmatrix}3x^2+3y&3x+3y^2\end{bmatrix}\quad\text{and}\quad D^2f(x,y) = \begin{bmatrix}\begin{matrix}6x\\3\end{matrix}&\begin{matrix}3\\6y\end{matrix}\end{bmatrix}. $$ A critical point of the function is $(x,y)=(-1,-1)$, which we can verify by computing $$Df(-1,-1) = \begin{bmatrix}3(-1)^2+3(-1)&3(-1)+3(-1)^2\end{bmatrix} = \begin{bmatrix}0&0\end{bmatrix}.$$ Does the function have a maximum or minimum at this critical point? Use the eigenvalues of $D^2f(-1,-1)$ to explain.
Solution
Note that $D^2f(-1,-1) = \begin{bmatrix}\begin{matrix}-6\\3\end{matrix}&\begin{matrix}3\\-6\end{matrix}\end{bmatrix}.$ We find the eigenvalues by solving $(-6-\lambda)(-6-\lambda)-(3)(3) = 0$. This is equivalent to solving $$0=\lambda^2+12\lambda+36-9 = \lambda^2+12\lambda+27 = (\lambda+9)(\lambda+3).$$ The eigenvalues of $D^2f(-1,-1)$ are $\lambda = -9$ and $\lambda = -3$, both of which are negative. We have already seen that negative eigenvalues mean the gradient field points inwards, meaning the critical point $(-1,-1)$ is the location of a local maximum. We can verify all these facts using the following Mathematica notebook.
- A rover travels along the path $2x-4y=12$. The elevation is given by $z=x+y^2-3$. Use Lagrange multipliers to find the $(x,y)$ location of the lowest elevation reached by the rover along this path.
Solution
The solution is $(x,y)=(4,-1)$.
To use Lagrange multipliers, remember you need to identify
- $f(x,y)$ - the thing you wish to optimize - $f(x,y)=x+y^2-3$
- $g(x,y)=c$ - the constraint - $g(x,y)=2x-4y=12$
- Then solve $\vec \nabla f = \lambda \vec \nabla g$ together with $g(x,y)=c$.
The equation $\vec \nabla f = \lambda \vec \nabla g$ gives $(1,2y) = \lambda(2,4)$, which means $1=2\lambda$ and $2y = 4\lambda$. The first equation means $\lambda = \frac{1}{2}$, which means $y=1$ from the second equation. Plugging $y=1$ into $2x-4y=12$ yields $x=4$.
We can check our work, as well as visualize what is happening, with the following Mathematica notebook.
Group problems
- Consider the function $f(x,y)= 2x^2+3xy+4y^2-5x+2y$.
- Find all critical points of $f$. [Check: $(x,y) = (2,-1)$.]
- Determine the eigenvalues of the second derivative at each critical point. [Check: The eigenvalues are $\lambda = 6\pm\sqrt{13}$, so $\lambda \approx 9.6$ or $\lambda \approx 2.4$.]
- At each critical point, do we have a local max, local min, or saddle?
- A rover travels along the curve $x^2y=16$ with $x>0$. The elevation near the rover is given by $z=-x^2-y^2$. Locate the $(x,y)$ coordinates where the rover reaches maximum height. [Check: You should obtain $y=2$ and two different $x$ values, giving two solutions.]
- Consider the function $f(x,y,z) = 3xy+z^2$. We'll be analyzing the surface at the point $P=(1,-3,2)$.
- If $dx=0.1$, $dy=0.2$ and $dz=0.3$, then what is $df$ at $P$.
- Find the directional derivative of $f$ at $P$ in the direction $(1,-2,2)$.
- Give an equation of the tangent plane to the level surface of $f$ that passes through $P$.
- Give an equation of the tangent plane to the level surface of $f$ that passes through $(a,b,c)$.
- Give an equation of the tangent plane to $xy+z^2=7$ at the point $P=(-3,-2,1)$. [Check: $(-2)(x-(-3))+(-3)(y-(-2))+2(1)(z-1)=0$. ]
- Give an equation of the tangent plane to $z=f(x,y)=xy^2$ at the point $P=(4,-1,f(4,-1))$. [Check: $z-4 = (-1)^2(x-4)+2(4)(-1)(y-(-1))$.]
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