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.
- 4.2 - Nathan B
- 4.3 - Parker, Braydon
- 4.4 - Jordan
- 4.5 - Olivia
- 4.6 - Makenzy
- 4.7 - Jeremy, Gavin
- 4.8 -
12:45PM
Thanks for sharing things in Perusall. Here are the presenters for today.
- 4.2 - Carter
- 4.3 - Chad, Michael
- 4.4 - Forrest
- 4.5 - Brian
- 4.6 - Trevor
- 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.
Brain Gains (Rapid Recall)
- A rover is located on a hill whose elevation is given by $z=f(x,y) = xe^{3xy}$. Compute the differential $dz$ and write it in the form $dz=(?)dx+(?)dy.$
Solution
The differential is $$dz = dxe^{3xy}+xe^{3xy}(3ydx+3xdy) = (e^{3xy}+3xye^{3xy})dx+(3x^2e^{3xy})dy.$$
- In your solution above, circle $f_y$, put a box around $\ds\frac{\partial f}{\partial x}$, and state the gradient $\vec \nabla f(x,y)$.
Solution
The requested quantities are given below.
- $\frac{\partial f}{\partial x} = e^{3xy}+3xye^{3xy}$
- $f_y = 3x^2e^{3xy}$
- $\vec \nabla f(x,y) = (e^{3xy}+3xye^{3xy},3x^2e^{3xy})$
- Suppose at a given point $(a,b)$, the differential $dz$ of a function (representing elevation on a hill) is $dz = 3 \, dx - 4 \, dy$. Find a nonzero vector $(dx,dy)$ such that $dz=0$.
Solution
Two possible answers are $(dx,dy) = (4,3)$ or $(dx,dy) = (-4,-3)$. All possible answers are multiples of these. The gradient of the function at $ (a,b) $ is $\vec \nabla z = (3,-4)$. Any vector $(dx,dy)$ orthogonal to this gradient will result in the differential being zero.
- In the example above, what is the slope of hill in the direction chosen above to make $dz=0$?
Solution
Since the differential is zero, then the slope would be zero as well. The slope is the rise $dz$ over the run $\sqrt{(dx)^2+(dy)^2}$. Walking orthogonal to the gradient always results in a slope of zero.
Conclusion: If you're on a hill and always walk orthogonal to the gradient, then your height won't change. You'll be on a level curve.
- A right circular cylindrical rod with radius $r$ and height $h$ has volume given by $V=\pi r^2 h$. Compute the differential $dV$ in terms of $r$, $h$, $dr$, and $dh$.
Solution
The differential is $$dV = 2\pi rh dr+\pi r^2dh.$$ If we haven't had someone present #4.4 yet, let's do that now.
Group problems
- A rover is located on a hill whose elevation is given by $z=f(x,y) = 3xy+y^2$.
- Compute the differential $dz$, and write it in the form $dz=(?)dx+(?)dy.$
- State $\dfrac{\partial f}{\partial x}$ and $f_y$. Then give $\vec \nabla f$.
- State the differential at the point $P=(1,1)$ (the spot where the rover currently resides).
- 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 $(3,4)$?
- 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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