New groups today
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.
- 4.1 - Rachel
- 4.2 - Nathan B, Jared
- 4.3 - Parker
- 4.4 - Gavin
- 4.5 - Ethan
- 4.6 - Kallan
- 4.7 - Jeremy
- 4.8 -
12:45PM
Thanks for sharing things in Perusall. Here are the presenters for today.
- 4.1 - Jacob
- 4.2 - Carter
- 4.3 - Reed
- 4.4 -
- 4.5 -
- 4.6 -
- 4.7 -
- 4.8 -
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.
Class Discussion
- We'll start with a class discussion related to the following Rover Activity (9am - 12:45pm).
- Open your chat, and be prepared to give answers, when I say, at the same time.
Brain Gains (Rapid Recall)
- A rover moves in a straight line to a new spot that is 3 m north and 4 m west of its current location. The rover's height drops 2 m along the way. What is the average slope of the hill in the direction the rover moved?
Solution
The displacement of the rover has a magnitude of $\sqrt{\Delta x^2+\Delta y^2} = \sqrt{3^2+(-4)^2} = 5$ meters in the $xy$-plane. This means the average slope is $$m = \frac{\text{rise}}{\text{run}} = \frac{-2}{5}.$$ Note that the slope is unitless, as the meters cancel.
- Let $z= x^3+4xy$. We can think of this as an elevation function that tells us the height $z$ of a hill at a point $(x,y)$. Compute $dz$ in terms of $x$, $y$, $dx$, and $dy$.
Solution
We have $$dz = 3x^2 dx + (4x)(dy)+(4dx)(y).$$ If needed, start with implicit differentiation and then strip off the $dt$'s.
- Write your answer above in the form $dz = (?_1) dx+(?_2) dy$.
Solution
By collecting the terms with a $dx$ in them, and then the terms with a $dy$ in them, we obtain $$dz = (3x^2+4y)dx+(4x)dy.$$
- Write your answer above as the dot product of two vectors, namely $dz= (?????)\cdot (dx,dy)$.
Solution
From inspection, we obtain $$dz = (3x^2+4y,4x)\cdot (dx,dy).$$ You can check this is correct by performing the dot product above and comparing it to the previous problem. This problem is essentially pattern recognition, and trying to find two vectors that give the desired dot product.
- What is the gradient of $z$, written $\vec \nabla z(x,y)$, or just $\vec\nabla z$ when the input variables are clear from context?
Solution
The gradient of $z$ is the vector $$\vec \nabla z = (3x^2+4y,4x).$$ Just strip the vector of differentials $(dx,dy)$ off of $dz$. This means we always have $$dz = \vec \nabla z\cdot (dx,dy).$$
In first semester calculus, we had $dy = f' dx$. The gradient $\vec \nabla z$ is the generalization of the derivative to higher dimensions. Note that the gradient here has to parts, which we call partial derivatives.
- What is the partial derivative of $z$ with respect to $x$, written $\ds\frac{\partial z}{\partial x}$ or $z_x$?
Solution
The partial derivative of $z$ with respect to $x$ is the quantity $$\frac{\partial z}{\partial x} = 3x^2+4y.$$ This is the slope $\frac{dz}{dx}$ if you let $dy=0$ in $dz = (3x^2+4y)dx+(4x)dy.$ It represents the slope of the function if you change $x$ but hold $y$ constant (hence $dy=0$). We use a stylized "d" (written $\partial$ in $\frac{\partial z}{\partial x}$) to help us remember that just $x$ is changing. The subscript notation can be used (so $z_x$) when no ambiguity arises from other subscripts.
- What is the partial derivative of $z$ with respect to $y$, written $\ds\frac{\partial z}{\partial y}$ or $z_y$?
Solution
The partial derivative of $z$ with respect to $y$ is the quantity $$\frac{\partial z}{\partial x} = 4x.$$ This is the slope $\frac{dz}{dx}$ if you let $dy=0$ in $dz = (3x^2+4y)dx+(4x)dy.$ It represents the slope of the function if you change $y$ but hold $x$ constant (hence $dx=0$). We use a stylized "d" (written $\partial$ in $\frac{\partial z}{\partial y}$) to help us remember that just $y$ is changing. The subscript notation can be used (so $z_y$) when no ambiguity arises from other subscripts.
Group problems
- A rover is located on a hill whose elevation is given by $z=f(x,y) = 3x^2+2xy+4y^2$.
- Compute the differential $dz$ in terms of $x$, $y$, $dx$, and $dy$, and write it in the form $dz=(?_1)dx+(?_2)dy.$
- Write your answer above as the dot product of two vectors, i.e., $dz = (??, ??)\cdot(dx, dy).$
- State the gradient of $f$, so $\vec \nabla f$.
- State $\dfrac{\partial f}{\partial x}$ and $f_y$.
- State the differential at the point $P=(1,1)$ (the spot where the rover currently resides) [Check: $dz = 8dx+10dy$].
- What is the slope of the hill at $P=(x,y)=(1,1)$ in the direction $(dx,dy)=(1,0)$?
- What is the slope of the hill at $P=(1,1)$ in the direction $(0,1)$?
- What is the slope of the hill at $P=(1,1)$ in the direction $(3,4)$? [Check: $\frac{64}{5}$. The rise is $dz = 64$, with a run of $5$.]
- The sides of a rectangle $x=3$ ft and $y=2$ ft, with tolerances $dx = .1$ ft and $dy = 0.05$ ft. Use differentials to estimate the tolerance on the area $A=xy$ that results from the given tolerances on $x$ and $y$. So compute $dA$ and then plug in $x$, $y$, $dx$, $dy$, as given. [Check: $dA = 0.35.$]
- Let $g(x,y) =x^2y$.
- Give $g_x$ 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 $(2,-5)$.
- The sides of a box are supposed to be $x=3$ ft by $y=2$ ft by $z=1$ ft, with tolerances $dx = .1$ ft by $dy = 0.05$ ft by $dz=0.02$ ft. Use differentials to estimate the tolerance on surface area $A=2xy+2yz+2xz$ that results from the given tolerances.
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