Rapid Recall
- Given $f(x) = x e^{-x}$, find the slope of the curve $y=f(x)$ at $x=2$.
Solution
The slope is given as a function of $x$ by the derivative $f'(x) = e^{-x}-x e^{-x}$ or $f'(x) = (1-x)e^{-x}$. Thus, at $x=2$, the slope is $\boxed{f'(2)=-e^{-2}}$.
- Write the equation of a line through $(3,1)$ with slope $-4/3$.
Solution
Slope is defined as change in $y$ divided by change in $x$. Let $(x,y)$ represent any point on the line. Then we can write the slope as $\displaystyle m =\frac{y-1}{x-3}=\frac{-4}{3}$. This is easily rearranged into so-called point-slope form: $\boxed{\displaystyle (y-1)=\frac{-4}{3}(x-3)}$.
- If $v(t)=r'(t) = -32 t + 100$ describes the speed (change in position over time) of a particle, find the particle's displacement (total change in position) between $t=1$ and $t=4$.
Solution
Applying the fundamental theorem of calculus gives $\displaystyle d = \int_1^4 r'(t)\,dt = r(4)-r(1)$. The antiderivative of $r'(t)$ is $r(t) = -16 t^2 +100 t + C$, so the displacement is $\boxed{d = -16(16-1)+100(4-1)=300-240=60}$
Let's Work
Grab a partner. Then as a group of two, join with another group of 2. As a group of 4, claim some board space. Write your names on the board. Then alternately take turns acting as scribe for the group. Each time you finish a problem, pass the chalk. If you get stuck on a problem, remember that you are the scribe for your group and they can help you.
- In the small town of Coriander, the library can be found by starting at the center of the town square, walking 25 meters north ($\vec a$), turning 90 degrees to the right, and walking a further 60 meters ($\vec b$).
- Draw a figure showing the displacement vectors $\vec a$ and $\vec b$, as well as their sum $\vec v = \vec a+\vec b$.
- How far is the library from the center of the town square.
- Let $\vec i$ represent walking 1 unit east and $\vec j$ represent walking 1 unit north. We call these unit vectors because their length is 1 unit. Express $\vec a$, $\vec b$, and $\vec v$ in terms of $\vec i$ and $\vec j$.
- It turns out that magnetic north in Coriander is approximately 14 degrees east of true north. The directions above won't actually get you to library if you use a compass. Instead, you must walk 39 meters in the direction of magnetic north ($\vec A$), and then turn 90 degrees to the right and walk another 52 meters ($\vec B$).
- Draw a figure showing the displacement vectors $\vec A$ and $\vec B$, as well as their sum $\vec v = \vec A+\vec B$.
- How far is the library from the center of the town square.
- Let $\vec I$ represent a unit vector pointing towards magnetic east, and let $\vec J$ represent a unit vector representing magnetic north. Express $\vec A$, $\vec B$, and $\vec v$ in terms of $\vec I$ and $\vec J$.
- The above two computations are partly both incorrect, as they both forgot to take into account the fact that the library is actually 5 meters higher ($\vec c$) in elevation than the center of the town square.
- Construct a new drawing that relates $\vec a$, $\vec b$, and $\vec c$ to the context of this problem.
- How far is the center of the town square from the library.
- Let $\vec k$ represent a unit vector that points upwards toward the sky. How can you use this to revisit the problems above?
- Find the distance between $(2,1,0)$ and $(0,-1,1)$.
The problems above are an adaptation of the work from the physics department at Oregon State University. See http://physics.oregonstate.edu/portfolioswiki/doku.php?id=activities:main&file=vcnorth
How Have I Learned Math?
- Think back over your math career. Could you describe a typical math class?
- What were your responsibilities?
- What was the teacher's role?
- Where/When did most of your learning take place?
Inquiry Based Learning
- When you see the title above, what do you think it has to do with this class?
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