I-Learn, Class Pictures, Learning Targets, Text Book Practice
Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus

We didn't quite finish up 9.1 last time, so feel free to work on 9.1 as well as the problems below.

Day 10 - Prep

Task 10.1

Given a curve with parametrization $\vec r(t)$, we have already seen that a unit tangent vector is given by $\ds \vec T = \frac{d\vec r}{ds} = \frac{d\vec r/dt}{|d\vec r/dt|}$. Note that this vector has constant length of 1, which means that it's derivative, so $\frac{d\vec T}{dt}$, is orthogonal to $\vec T$. This vector describes how the direction of motion changes. The vector $\ds\vec N = \frac{d\vec T/dt}{|d\vec T/dt|}$ provides a unit vector, we call the principle unit normal vector, that describes the direction in which an object is turning. The cross product $\vec B = \vec T\times \vec N$ we call the binormal vector. These three vectors, namely $\vec T$, $\vec N$, and $\vec B$, provide what we call the Frenet or TNB frame, and are commonly used when describing motion.

For the curve $\vec r(t) = (3\cos t, 3\sin t, 4t)$, compute $\vec T$, $\vec N$, and $\vec B$. Show how you obtained each step in your computations.
- The definitions of $\vec T$ and $\vec N$ both made them one unit long. How long is $\vec B$?
For the curve $\vec r(t) = (t,0,t^2)$, compute $\vec T$, $\vec N$, and $\vec B$. Show how you obtained each step in your computations. If things get ugly quite quickly, because of a quotient rule, then you're on the right path.

For a visual representation of the Frenet Frame, please visit this Geogebra site. It's possible to create a very similar visual in Mathematica or Python (something you could aim for with a self-directed learning project).

Task 10.2

Given a parametric curve with parametrization $\vec r(t)$, the curvature vector is the rate of change of the direction of motion with respect to arc length, so $\ds \vec \kappa = \frac{d\vec T}{ds}$. We compute the derivative with respect to arc length so that we obtain a physical property of the curve, rather than a property that relates to how quickly we traverse the curve. The curvature is the magnitude of the curvature vectors, so $$\kappa = |\vec \kappa| = \left| \frac{d\vec T}{ds} \right|.$$ The radius of curvature is the quantity $1/\kappa$.

Explain why $\ds \kappa = \frac{|d\vec T/dt|}{|d\vec r/dt|}$.
For the circle $\vec r(t) = (5\cos(2t), 5\sin(2t))$, compute the curvature $\kappa$ and radius of curvature $1/\kappa$.
For the helix $\vec r(t) = (3\cos(t), 3\sin(t),4t)$, compute the curvature and radius of curvature.
For the parabola $\vec r(t) = (t,t^2)$, at $t=0$ compute the curvature $\kappa(0)$ and radius of curvature.
Draw the parabola from the previous part. How would you interpret the radius of curvature at $t=0$ in this context?

Task 10.3

Let $P=(a,b,c)$ be a point on a plane in 3D. Let $\vec n=(A,B,C)$ be a normal vector to the plane (so the angle between the plane and $\vec n$ is 90$^\circ$). Let $Q=(x,y,z)$ be another point on the plane.

What is the angle between $\vec {PQ} = (x-a,y-b,z-c)$ and $\vec n=(A,B,C)$?
Explain why an equation of the plane through $P$ with normal vector $\vec n$ is $$A(x-a)+B(y-b)+C(z-c)=0.$$
Consider the three points $R=(1,0,0)$, $S=(2,0,-1)$, and $T=(0,1,3)$. Give an equation of the plane which passes through these three points. [You already have a point on the plane. With three points, you can get two vectors that are in the plane. How can you get a vector that is normal to the plane?]

Task 10.4

The last problem for prep each day will point to relevant problems from OpenStax. Spend 30 minutes working on problems from the sections below.

Return to any of the previous day's OpenStax problems to locate extra practice.

Day 10 - In class

Brain Gains (Rapid Recall, Jivin' Generation)

1. For $\vec r(t) = (\cos t, \sin t, t)$, compute $\vec T = \dfrac{d\vec r}{ds}$, $\vec N = \dfrac{d\vec T/dt}{|d\vec T/dt|}$, and $\vec B = \vec T\times \vec N$.

Let's view an animation.

https://www.geogebra.org/m/kaGKS9BB

2. For the circle $\vec r(t) = (3\cos t, 3\sin t)$, compute the curvature $\kappa = \left|\dfrac{d\vec T}{ds}\right|$ and radius of curvature $\dfrac{1}{\kappa}$.

3. Find a vector that orthogonal to the two vectors $(-1,2,5)$ and $(3,0,4)$.

4. Give an equation of the plane that passes through the three points $(2,0,0)$, $(0,3,0)$, $(0,0,5)$.

Group Problems

1. For $\vec r(t) = (4\cos t, 4\sin t, 3t)$, compute $\vec T = \dfrac{d\vec r}{ds}$, $\vec N = \dfrac{d\vec T/dt}{|d\vec T/dt|}$, $\vec B = \vec T\times \vec N$, and $\kappa$.

2. Give an equation of a plane that passes through the points $(2,0,1)$, $(1,-2,0)$, and $(3,3,2)$.

3. For $\vec r(t) = (t^2, 0,t)$, at $t=1$ please compute $\vec T = \dfrac{d\vec r}{ds}$, $\vec N = \dfrac{d\vec T/dt}{|d\vec T/dt|}$, $\vec B = \vec T\times \vec N$, and $\kappa$.

Day 11 - Prep

Task 11.1

We'll use Mathematica to create a program to compute all the quantities related to the TNB frame and curvature.

Download the Mathematica file TNB-intro.nb.
Read through the introductory examples, evaluating each block of code.
Adapt the code to compute the TNB frame and curvature for a given curve.

Task 11.2

Recall that the curvature vector is $\kappa = \dfrac{d\vec T}{ds}$, with curvature being a the length of this vector. This vector tells us how much the direction of motion ($\vec T$) changes, as we increase the distance moved along the curve. As such, a tight corner will result in a large change of direction and hence a large curvature. Large corners will result in a small curvature.

The radius of curvature at a point, namely $1/\kappa$, provides the radius of a circle that approximates the shape of curve at that point. Large turns results in a large radius of curvature, while tight turns results in a small radius of curvature. This circle lies in the plane formed by $\vec T$ and $\vec N$ (so a normal vector to this plane is $\vec B$). We call this plane the osculating plane. The center of the circle can be found by following $\vec N$ from the point on the curve.

For the curve $\vec r(t) = (3\cos t, 3\sin t, 4t)$, we have already computed $\vec T$, $\vec N$, $\vec B$, and $\kappa$. At $t=\pi/2$, evaluate these quantities.
Give an equation of the the osculating plane at $t=\pi/2$. You'll need to identify a normal vector to the plane, and a point on the plane.
Explain why the center of curvature is given by $\vec r + \frac{1}{\kappa}\vec N$.
Give the location of the center of curvature for $\vec r(t) = (3\cos t, 3\sin t, 4t)$ at $t=\pi/2$.

Task 11.3

There are many ways to compute the TNB frame and curvature. In this problem, we'll develop a few others.

Explain why $ \vec N = \dfrac{\vec r^{\prime\prime}_{\perp \vec r^{\prime}}}{ |\vec r^{\prime\prime}_{\perp \vec r^{\prime} }| } $.
Explain why $ \vec B = \dfrac{\vec r^{\prime}\times \vec r^{\prime\prime}}{|\vec r^{\prime}\times \vec r^{\prime\prime}|}$.
For a function of the form $\vec r(x) = (x, f(x))$, show that $\kappa = \dfrac{|f''(x)|}{(1+(f'(x))^2)^{3/2}}$.

Task 11.4