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 9 - Prep

Task 9.1

We have shown that for a parametric curve given by $\vec r(t)$, for $a\leq t\leq b$, the length of the curve is given by $s = \int_a^b \left|\dfrac{d \vec r}{dt}\right|dt$. We can use a different variable for the parameter, say $\tau$, and then for the curve $\vec r(\tau )$, for $0\leq \tau \leq t$ we get the arc length parameter $$s(t) = \int_0^t \left|\dfrac{d \vec r}{d\tau}\right|d\tau.$$ This provides a function $s(t)$ that tells us the arc length traveled along a curve after $t$ units of time. Note that the derivative of $s(t)$ is (why?) $$\frac{ds}{dt} = \frac{d}{dt}\int_0^t \left|\dfrac{d \vec r}{d\tau}\right|d\tau = \left|\dfrac{d \vec r}{dt}\right|,$$ which is just the speed. By requiring that speed be positive on a curve, then the arc length parameter is an increasing function. This is the reason that the definition of a smooth curve requires that the a parametrization have a nonzero derivative.

For the helix $\vec r(t) = (3\cos t, 3\sin t, 4t)$, set up the intergral formula that gives the arc length parameter, and then simplify to show that the arc length parameter is $s(t) = 5t$.
If you've traveled 20 units along the helix $\vec r(t) = (3\cos t, 3\sin t, 4t)$, how much time has elapsed? In general, if you've traveled $s$ units, how much time has elapsed. You will have written $t$ as a function of the arc length $s$, obtaining the inverse function $t(s)$.
State the functions $\vec r(t)$ and $\vec r(t(s))$. Then compute both $\dfrac{d\vec r}{dt} = \dfrac{d\vec r(t)}{dt}$ and $\dfrac{d\vec r}{ds}=\dfrac{d\vec r(t(s))}{ds}$.
Rather than writing $\dfrac{d\vec r}{ds}$ in terms of $s$, it's common to replace $s$ with $s(t)$, giving $\dfrac{d\vec r}{ds}(t)$. Explain why, in general, we have $\dfrac{d\vec r}{ds}(t) = \dfrac{d\vec r/dt}{ds/dt}$.
For the curve $\vec r(t) = (a\cos t, a\sin t, bt)$, compute both $\dfrac{d\vec r}{dt}$ and $\dfrac{d\vec r}{ds}$.
Show that the magnitude of $\dfrac{d\vec r}{ds}$ is always 1. We call this the unit tangent vector and write $\vec T = \dfrac{d\vec r}{ds}.$

Task 9.2

In this task, we'll show that the product rule applies to the dot product, and then use that to prove that if a vector-valued function has constant length, then the derivative of the function is orthogonal to the original function.

Let $\vec r_1(t) = (f(t), g(t))$ and $\vec r_2(t) = (m(t), n(t))$. Prove that $$\frac{d}{dt}(\vec r_1\cdot \vec r_2) = \frac{d}{dt}(\vec r_1)\cdot \vec r_2+\vec r_1\cdot \frac{d}{dt}(\vec r_2).$$
Now suppose that $\vec r(t)$ has constant length (meaning $|\vec r(t)|=c$ for some constant $c$). Prove that $\vec r(t) \cdot \dfrac{d\vec r(t)}{dt} = 0$. [Hint, if we know $|\vec r(t)|=c$, what does this mean about $\vec r(t)\cdot \vec r(t)$? How are the dot product and length connected?]
Draw the curve $\vec r(t) = (4\cos t+2, 4\sin t-1)$. Show that the velocity has constant length and then verify that the derivative of the velocity (so acceleration) is orthogonal to the velocity.

The problem above gets used quite often in engineering applications. If a beam is attached to a system at a single point, then the beam can rotate about the point. Because the length of the beam is constant, then any rotational forces caused by the beam as it impacts other objects will always be normal to the beam.

Task 9.3

Let $\vec u = (a,b,c)$ and $\vec v = (d,e,f)$. Our goal is to find a single nonzero vector $(x,y,z)$ that is orthogonal to both $\vec u$ and $\vec v$, preferably with as few fractions as possible in the final answer.

Explain why we need to solve the system of equations $$ax+by+cz=0\quad\text{and}\quad dx+ey+fz=0.$$
To solve the system above, multiply the first equation by $d$ and the second equation by $-a$ (assume for a moment that both $a$ and $d$ are not zero). Then add the two equations together to eliminate $x$, which will allow you to solve for $y$ in terms of $z$. Finish by solving for $x$ in terms of $z$ (there are many ways to do this). You will have shown that every solution to this system can be written in the form $$(x,y,z) = \left(\left(\frac{bf-ce}{ae-bd}\right)z,\left(\frac{cd-af}{ae-bd}\right)z,z\right). $$
The above solution has some complicated fractions. Why is $(x,y,z) = (bf-ce, cd-af, ae-bd)$ a solution to the system?

From your work above you will have developed a formula for the cross product of two vectors. The cross product of the two vectors $\vec u = (u_1,u_2,u_3)$ and $\vec v = (v_1,v_2,v_3)$ is the vector $$\vec u\times \vec v = (u_2v_3-u_3v_2, u_3v_1-u_1v_3, u_1v_2-u_2v_1).$$

Let $\vec u=(1,-2,3)$ and $\vec v=(2,0,-1)$.
1. Compute $\vec u\times \vec v$ and $\vec v\times \vec u$. How are they related?
2. Compute and simplify both $\vec u \cdot (\vec u\times \vec v)$ and $\vec v \cdot (\vec u\times \vec v)$. Did you get zero for both? What fact about the cross product guarantees you get zero?

Task 9.4

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

Arc Length Parameter: section 3.3: checkpoint 3.10, exercises 127-128, 116, 119, 121, 126
Look at theorem 3.3 in section 3.2. Properties 4 and 7 were task 9.2. Prove any of the other properties that seem interesting to you.
Cross product: section 2.4: checkpoint 2.30, exercises 183-192

Day 9 - In class

Brain Gains (Rapid Recall, Jivin' Generation)

For the curve $r(t) = (t,\cos t, \sin t)$, compute $\ds \int_0^a \left|\frac{d\vec r}{dt}\right|dt$.
For the same curve $r(t) = (t,\cos t, \sin t)$, compute the arc length parameter $\ds s(t) = \int_0^t \left|\frac{d\vec r}{d\tau}\right|d\tau$.
Compute $\ds \frac{d}{dt}\left(\int_0^t x^3 dx\right)$.
Compute $\ds \frac{d}{dt}\left(\int_0^t \sqrt{1+\tau^2+\cos(\tau)} d\tau\right)$.
Compute $\ds \frac{ds}{dt}$, so compute $\ds \frac{d}{dt}\left(\int_0^t \left|\frac{d\vec r}{d\tau}\right|d\tau\right)$.
For the curve $\vec r(t) = (t,t^2)$, compute both $\dfrac{d\vec r}{dt}$ and $\dfrac{d\vec r}{ds}$, giving both in terms of $t$.
For the vectors $\vec u = (2,-1,5)$ and $\vec v = (-3, 4,6)$, compute the cross product $\vec u\times \vec v$.
Compute $\vec v\times \vec u$, and verify that $\vec v\times \vec u = - \vec u\times \vec v$.

Group Problems

For the curve $r(t) = (2\cos t, 3t, 2\sin t)$, compute the arc length parameter $s(t)$.
1. Compute $ds/dt$, $d\vec r/dt$, $|d\vec r/dt|$, and $d\vec r/ds$.
2. State the unit tangent vector $\vec T$ at time $t$, and identify how to obtain it from the parts above.
3. Compute $d\vec T/dt$, and show that this vector is orthogonal to $\vec T$.
For the curve $r(t) = (t^2, 3t)$, set up an integral that gives the arc length parameter $s(t)$.
1. Compute $ds/dt$, $d\vec r/dt$, $|d\vec r/dt|$, and $d\vec r/ds$.
2. State the unit tangent vector $\vec T$ at time $t$, and identify how to obtain it from the parts above.
3. Compute $d\vec T/dt$, and show that this vector is orthogonal to $\vec T$.
For the vectors $\vec u = (1,2,3)$ and $\vec v = (4, 5,6)$, compute the cross product $\vec u\times \vec v$ and $\vec v\times \vec u$. Then change the vectors to something else, and practice doing this multiple times.

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