Math 214 - Multivariable Calculus - 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?
See http://www.inquirybasedlearning.org/.

Mastery Learning

What is mastery learning?

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HomePage Schedule	Class Day 1 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? See http://www.inquirybasedlearning.org/. Mastery Learning What is mastery learning? Edit Page History Source Attach File Backlinks List Group Page last modified on September 12, 2022, at 01:53 PM
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Day 1

Brain Gains

Let's Work

How Have I Learned Math?

Inquiry Based Learning

Mastery Learning