- I-Learn, Class Pictures, Learning Targets, Text Book Practice
- Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus
Day 1 - Prep
There was no prep for the first day of class. Welcome to Math 214.
Day 1 - In class
Brain Gains

- 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 $$f'(2)=-e^{-2}.$$
- Write an 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 $\displaystyle m =\frac{y-1}{x-3}=\frac{-4}{3}$. An equation of the line in point-slope form is $$\displaystyle (y-1)=\frac{-4}{3}(x-3).$$
- Find a nonzero vector that is orthogonal to $(3,-4)$.
Solution
One option is $(4,3)$.
- Write an equation of a line through $(3,1)$ with normal vector $(4,3)$.
Solution
For any point $(x,y)$ on the line, we know $(x-3,y-1)$ is a vector in the line, and hence orthogonal to $(4,3)$. The dot product of these two vectors must be zero, which means $4(x-3)+3(y-1)=0$ is an equation of the line. Note that this is the same line $\displaystyle (y-1)=\frac{-4}{3}(x-3)$ as before.
- Write a vector equation of a line through $(1,2,3)$ and $(-2,4,0)$.
Solution
A vector from $(1,2,3)$ to $(-2,4,0)$ is $\vec v = (-3,2,3)$. Using point $(1,2,3)$ as a starting point, we can use $$(x,y,z) = (1,2,3)+(-3,2,3)t$$ as a vector equation for the line through these points. There are many more correct answers.
- 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 $\displaystyle d = \int_1^4 r'(t)\,dt = r(4)-r(1)$. All antiderivatives of $r'(t)$ are of the form $r(t) = -16 t^2 +100 t + C$ for some constant $C$, so the displacement is $$d = \left(-16(4)^2 + 100 (4)\right)- \left(-16(1)^2 + 100 (1)\right)=144 - 84 = 60.$$ You don't have to simplify answer in this class (we're focusing on calculus, not arithmetic), so $$d = \left(-16(4)^2 + 100 (4)\right)- \left(-16(1)^2 + 100 (1)\right)$$ is sufficient.
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 $\vec j$.
- A car travels along the path parametrized by $\vec r(t) = 2\vec a+ t\vec b$ for $-1\leq t\leq 2$. Construct a plot that contains the city center, the library, and the path of the car for $-1\leq t\leq 2$.
- 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 $(0,-1,1)$.
- Find a vector that is orthogonal to $(-2,-2,1)$. How do you know it is orthogonal?
- Find a vector that is orthogonal to both $(2,1,0)$ 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?
Mastery Learning
- What is mastery learning?
Day 2 - Prep
Task 2.1
Start by looking up the terms vector-valued function and vector parameterization of a curve.
- Write definitions in your own words for the terms above.
- For each vector parameterization below, construct a graph of the curve. [Hint: make a table of points if needed, including $t$, $x$, and $y$, and then just plot the $(x,y)$ coordinates).
- $\vec r(t) = \left< 2t+1, 4-3t\right>$ for $0\leq t\leq 2$.
- $(x,y) = (\cos t, \sin t)$ for $0\leq t\leq 3\pi/2$.
- $(x,y)(t) = (\sin t, \cos t)$ for $0\leq t\leq \pi$.
- $\langle x,y,z\rangle = (2\cos t, 2\sin t, t)$ for $0\leq t\leq 4\pi$.
Task 2.2
Start by locating a definition of the dot product of two vectors and what it means for two vectors to be orthogonal, as well as the dot product form of the law of cosines.
- Compute the dot product of the two vectors $\vec a = 3\vec i-4\vec j+2\vec k$ and $\vec b = (-1,3,6)$.
- Find the angle between $\vec a$ and $\vec b$.
- Give a value $k$ so that the vectors $\vec a = 3\vec i-4\vec j+2\vec k$ and $\vec c = \langle 2, -1, k\rangle$ are orthogonal.
- A car is moving in the direction $\vec v = (-5,7)$. The car makes a 90 degree turn to the left. Give a vector that is parallel to this new direction of motion.
Task 2.3
Start by looking up the definition of a unit vector. Consider the two points $P = (1, 2, 3)$ and $Q = (2, −1, 0)$.
- Write the vector $\vec {PQ} $ in component form $(a, b, c)$.
- Find the length of vector $\vec {PQ} $.
- Find a unit vector in the same direction as $\vec {PQ} $.
- Find a vector of length 7 units that points in the same direction as $\vec {PQ} $.
Task 2.4
The last problem for prep each day will point to relevant problems from OpenStax. Spend 30 minutes working on problems from the sections below.
- In section 1.1, complete checkpoints 1.1, 1.2, and 1.3. Use the corresponding examples, if needed, to help you.
- In section 2.3, complete an exercise for each group in 123-144, and then try a few problems in 149-154.
|
Sun |
Mon |
Tue |
Wed |
Thu |
Fri |
Sat |
