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\noindent This unit covers the following ideas. In
preparation for the quiz and exam, make sure you have a lesson
plan containing examples that explain and illustrate the
following concepts.
- Develop formulas for the velocity and position of a projectile, if we neglect air resistance and consider only acceleration due to gravity. Show how to find the range, maximum height, and flight time of the projectile.
- Develop the $TNB$ frame for describing motion. Make sure you can explain why $\vec T$, $\vec N$, and $\vec B$ are all orthogonal unit vectors, and be able to perform the computations to find these three vectors.
- Explain the concepts of curvature $\kappa$, radius of curvature $\rho$, center of curvature, and torsion $\tau$. Make sure you can describe geometrically what theses quantities mean.
- Find the tangential and normal components of acceleration. Show how to obtain the formulas $a_T=\frac{d}{dt}|\vec v|$ and $a_N=\kappa |\vec v|^2=\frac{|\vec v|^2}{\rho}$, and explain what these equations physically imply.

- YouTube playlist for 07 - Motion and The TNB Frame.
- A PDF copy of the finished product (so you can follow along on paper).

{|c|c|c|} \hline Quantity & Symbol &
Formula\\\hline\hline Position (“r”adial
vector) & $\vec r$ & $\vec r(t) =
(x(t),y(t),z(t))$\\\hline Velocity & $\vec v$ &
$\ds \vec v(t) = \frac{d\vec r}{dt}$\\\hline Speed & $v
= \ds\frac{ds}{dt}$ & $\ds v(t) = |\vec v(t)|$\\\hline
Acceleration & $\ds \vec a$ & $\ds \vec a(t) =
\frac{d \vec v}{dt}= \frac{d^2\vec r}{dt^2}=
\frac{d}{dt}\frac{d\vec r}{dt}$\\\hline Unit Tangent Vector
& $\vec T$ & $\ds\frac{d\vec r}{ds} = \frac{d\vec
r/dt}{ds/dt} = \frac{\vec r^\prime(t)}{|\vec
r^\prime(t)|}$\\\hline Curvature Vector & $\vec \kappa
$& $\ds\frac{d\vec T}{ds} =\frac{d\vec T/dt}{ds/dt} =
\frac{d\vec T/dt}{|\vec v|} = \frac{\vec T^\prime(t)}{|\vec
r^\prime(t)|} $\\\hline Curvature (a scalar)& $ \kappa
$&$\ds \left|\frac{d\vec T}{ds}\right|
=\left|\frac{d\vec T/dt}{ds/dt}\right| = \frac{\left|d\vec
T/dt\right|}{|\vec v|}= \frac{|\vec T^\prime(t)|}{|\vec
r^\prime(t)|} $ \\\hline Curvature of $y=f(x)$& $
\kappa(x) $&$\ds \kappa(x) =
\frac{|f”(x)|}{(1+(f')^2)^{3/2}}. $ \\\hline
Principal unit normal vector & $ \vec N$& $\ds
\frac{d\vec T/dt}{|d\vec T/dt|} = \frac{\vec
T^\prime(t)}{|\vec
T^\prime(t)|}=\frac{1}{\kappa}\frac{d\vec T}{ds} =
\frac{1}{\kappa |\vec v|}\frac{d\vec T}{dt}$\\\hline
Binormal vector & $ \vec B$& $ \vec T\times\vec
N$\\\hline Radius of curvature & $ \rho$ &
$1/\kappa$\\\hline Center of curvature & & $\vec
r(t)+\rho(t)\vec N(t)$ \\\hline Torsion & $ \tau $
& $\ds \pm\left|\frac{d\vec B}{ds}\right|$ (pick the
sign) or $\ds-\frac{d\vec B}{ds}\cdot \vec N $\\\hline
Tangential Component of acceleration & $ a_T$ &
$\ds \vec a \cdot \vec T = \frac{d}{dt}|\vec v|$\\\hline
Normal Component of acceleration & $ a_N$ & $\ds
\vec a \cdot \vec N = \kappa \left(\frac{ds}{dt}\right)^2 =
\kappa |\vec v|^2$\\\hline Acceleration (sum the
components)& $ \vec a$ & $\vec a = a_T\vec
T+a_N\vec N = \left(\frac{d}{dt}|\vec v|\right) \vec T
+\left(\kappa |\vec v|^2\right) \vec N $\\\hline

\caption{This table summarizes the key ideas in this
unit. Most of our work in this chapter will be to explain the
connections between these variables.\label{motion table}}
If $y'(t) = 3t^2+12e^{2t}$ (the velocity) and $y(0)=2$ (initial
height), then what is $y(t)$? See footnote \footnote{Integrate
to get $y(t) = t^3+6e^{2t}+C$. Since $y(0)=2$, we know
$2=0+6(1)+C$, which gives $C=-4$. So the height is $y(t) =
t^3+6e^{2t}-4$. } for an answer.

To solve the next problem, we need to remember that
acceleration is the derivative of velocity, and that velocity is
the derivative of position. These facts hold true for
vector-valued functions as well.
Consider a rocket in space (so we can neglect air resistance
and gravity). The rocket's boosters apply an acceleration $\vec
a(t) = (2t,-8)$ m/s$^2$. The rocket's initial velocity is $\vec
v(0) = (4,5)$ m/s. The initial position is $\vec r(0) = (1,16)$
m. Use this information to determine the position of the object
after 2 seconds, and after 3 seconds. [Hint: Integrate each
component to get velocity, then repeat to get position. Don't
forget the 4 arbitrary constants you get from integration. Use
the initial velocity and initial position to determine these
constants.]

Suppose we fire a projectile (like a pumpkin) from a
cannon. The projectile leaves the cannon with an initial speed
$v_0$, at an angle of $\alpha$ above the $x$-axis. All the motion
in this problem occurs with a plane, and we'll use $x$ and $y$ to
represent motion in that plane. Our goal is to find the velocity
$\vec v(t)$ and position $\vec r(t)$ of the projectile at any
time $t$. We need some assumptions prior to solving.
- Assume the only force acting on the object is the force due to gravity. We will neglect air resistance.
- The force due to gravity is the mass of the projectile multiplied by the acceleration of gravity. The mass of the object will not be important in our work here, though in future classes you may study how mass affects energy computations.
- The projectile is shot over a small enough range that we can assume gravity only pulls the object straight down.
- Most branches of science use the letter $g$ to represent the magnitude of the vertical component of acceleration, so we can write the acceleration of the projectile as $$\vec a(t) = (0,-g) \quad \quad \text{or}\quad \quad \vec a(t)= 0{\bf i}-g{\bf j}.$$
- Our text uses the approximations $g\approx 9.8$ m/s$^2$ or $g\approx32$ ft/s$^2$.

The function $\vec r(t) = (2t+3, 4t^2+7t+5)$ is a
parametrization of a plane curve. Give a Cartesian equation of
the curve. See \footnote{Since $t=\dfrac{x-3}{2}$, a Cartesian
equation is $y =
4\left(\dfrac{x-3}{2}\right)^2+7\left(\dfrac{x-3}{2}\right)+5$.
} for an answer.

Watch a
YouTube video.

% \marginparbmw{You can practice finding position from
velocity and acceleration with problems 13.2: 11-18, and
especially 13.2: 29.} Suppose a projectile is fired from the
point $(x_0,y_0)$ with an initial velocity $\vec
v(0)=(v_{x_0},v_{y_0})$, and that gravity is the only force
acting on the object. This means the acceleration on the object
is $\vec a(t) = (0,-g)$.
- Show that the velocity at any time $t$ is $\vec v(t) = (c_1,-gt+c_2)$. What are $c_1$ and $c_2$? Explain
- Show that the position at any time $t$ is $\vec r(t) =
(v_{x_0}t+c_3, -\frac{1}{2}gt^2+v_{y_0}t+c_4)$. What are
$c_3$ and $c_4$?
%
- Give parametric equations $x=x(t)$ and $y=y(t)$ that give the horizontal and vertical position of the projectile at time $t$.

- Eliminate the parameter $t$ to give a Cartesian equation of the projectile's path. This will prove that the path of the particle is parabolic.

- The range is the horizontal distance $R$ traveled by the projectile.
- The flight time is how long the projectile is in the air. It is the time $t$ at which $\vec r(t)=(R,0)$.
- The maximum height is the largest $y$ value obtained by the projectile.

Watch a
YouTube video.

% Answer the following questions. Assume we fire a
projectile from the origin, which means the acceleration,
velocity, and position are $$ \vec a(t) = (0,-g),\quad \vec
v(t) = (v_{x_0}, -gt+ v_{y_0}),\quad \vec r(t) = (v_{x_0}t,
-\frac12 gt^2+ v_{y_0}t) .$$
- %
- What should the velocity vector equal when the object has reached the maximum height?
- What's the time to max height? What's the flight time?
- Show why the maximum height is $\ds y_{\max}=\frac{v_{y_0}^2}{2g}$ and the range is $\ds R=\frac{2v_{x_0}v_{y_0}}{g}$.
- If the initial speed is $v_0$, with a firing angle of $\alpha$ above the horizontal, rewrite $v_{x_0}$ and $v_{y_0}$ in terms of $v_0$ and $\alpha$, and then state the range in terms of $v_0$ and $\alpha$.

This problem was created around the opening ceremony of the
Barcelona Spain Olympics. Antonio Rebollo was the archer, but
he didn't try to hit the flame at the peak of the flight. You
can watch a YouTube
video of the opening ceremony by following the link.

% \marginparbmw{See 13.2: 19-28 for more practice.} An
archer stands at ground level and shoots an arrow at an object
which is 90 feet away in the horizontal direction and 74 ft
above ground. The arrow leaves the bow at about 6 ft above
ground level (not the origin). The archer wants the arrow to
hit the target at the peak of its parabolic path. For the
purposes of this problem, Let $g = 32 \text{ft}/\text{s}^2$.
What initial speed $v_0$ and firing angle $\alpha$ are needed
to achieve this result? [Hint: This is much easier to solve if
you first find $v_{x_0}$ and $v_{y_0}$, the horizontal and
vertical components of the velocity. You may want to move the
origin as well, so that you can use the formulas from above.]
A horse runs once around an elliptical track, which is
parametrized by $\vec r(t) = (3\cos t,4\sin t)$. Set up, do not
solve, an integral formula that tells us the distance the horse
traveled. What's the displacement? See \footnote{The velocity
is $\vec v(t) = (3\sin t, -4\cos t)$. The speed is $v(t) =
\sqrt{9\sin^2t+16\cos^2t}$. The distance traveled is the arc
length $\ds s=\int_0^{2\pi}
\left(\sqrt{9\sin^2t+16\cos^2t}\right)dt$. Since the horse's
initial and final position are equal, the displacement is zero.
Arc length does not equal displacement. } for an answer.

Let's now develop a formula for the arc length of a space
curve, a curve in 3D. We can always parameterize a space curve
with $\vec r(t) = (x,y,z)$ (one input, 3 outputs).
Watch a
YouTube video.

A space ship travels through the galaxy. Let $\vec r(t) =
(x,y,z)$
Technically, we should write $\vec r(t) = (x(t),y(t),z(t))$.
However, we already know that $x$, $y$, and $z$ depend on
$t$, hence we'll just leave the dependence on $t$ off.

% be the position of the space ship at time $t$, with the
earth at the origin $(0,0,0)$.
- What are the velocity and speed of the space ship at time $t$? You answers should involve some derivatives (such as $\frac{dx}{dt}$).
- If the space ship travels for a really small time $dt$, then the speed is about constant. Since distance is speed times time, about how much distance (we'll call it $ds$) will the space ship travel in this short amount of time?
- As the ship travels from time $t=a$ to time $t=b$, explain why the distance traveled (the arc length of the path followed) is $$s=\int_a^b |\vec r '(t)|\ dt = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}\ dt .$$

We've encountered the polar curve $r = 1-\sin\theta$ before (we
called it a cardioid, and it looked like heart). Recall that we
can switch from polar to Cartesian using the coordinate
transformation $x=r\cos\theta$ and $y=r\sin\theta$.

We'd like to avoid paths that contain a cusp, because at a
cusp the direction of motion changes rather abruptly. This can
happen physically, but it requires the speed of an object to
reach zero, the object stops moving, and then the path changes
direction. The fact that the speed reaches zero will mean we
can't divide by it in our work that follows. To avoid this, we
make a definition that requires the path is differentiable, and
the velocity is never zero.
- Draw the curve.
- A parametrization of this curve is $\vec r(\theta) = ((1-\sin\theta)\cos\theta, ?)$. This parametrization is completely differentiable. Find $\dfrac{d\vec r}{d\theta}$.
- You should notice a sharp cusp in the graph. At what $\theta$ does this cusp occur? What is the value of the derivative $\dfrac{d\vec r}{d\theta}$ at this value of $\theta$.

Let $\vec r(t)=(x,y,z)$ be a parametrization of a space curve
$C$. We say that $\vec r$ is smooth if $\vec r$ is
differentiable, and the derivative is never the zero vector. If
$\vec r$ is a smooth parameterization, then we call $C$ a
smooth curve.

Watch a
YouTube Video.

% \marginparbmw{See 13.3: 1-10 for more practice.}%
Consider the helical space curve $C$ with parameterization
$\vec r(t)=(\cos t, \sin t, t)$.
- Is C a smooth curve?
- Find the length of this space curve for $t\in[0,2\pi]$. Compute any integrals.
- Now find the length of the space curve from $t=0$ to time $t=t$.
- Give a vector of length 1 that is tangent to the curve at $t=2\pi$.

Compute $\ds \int_0^t 3x^2 dx$ and $\ds \int_0^t 3p^2 dp$ and
$\ds \int_0^t 3\tau^2 d\tau$. Does it matter what you call the
variable inside the integral? Then compute $\ds\frac{d}{dt}
\int_0^t 3\tau^2 d\tau$. See \footnote{ The first integral is
$x^3|_0^t = t^3$. The other two are the same. You can change
the variable inside the integral whenever you want. For this
reason, some people call it a dummy variable. The last part is
$\ds\frac{d}{dt} \int_0^t 3\tau^2 d\tau = \frac{d}{dt} t^3 =
3t^2$, we just replaced $\tau$ with $t$ in $3\tau^2$.} for an
answer.

\label{fundamental theorem of calculus as it applies to arc
length parameter}

We'll call $\ds s(t)=\int_0^t \left|\frac {d\vec
r}{d\tau}\right|\ d\tau$ the arc length parameter. It tells us
how far we've have traveled after $t$ seconds. We can now predict
distance traveled from time elapsed. Because $s(t)$ is an
increasing function, we can also invert this process and give
time elapsed from distance traveled. This means we could compute
derivatives with respect to $s$ instead of $t$. When we take a
derivative with respect to $s$, we ask how much a curve changes
if we increase length by 1 unit, instead of increasing time by 1
unit. We'll write $$\ds\frac{d\vec r}{ds} =\ds\frac{d\vec
r/dt}{ds/dt} = \frac{d\vec r/dt}{|d\vec r/dt|} = \frac{\vec
v}{v}.$$
You can remember $\ds\frac{ds}{dt} = \left|\frac {d\vec
r}{dt} \right|$ as follows. We use the differential $ds$ to
represents a change in distance, and $dt$ represents a change
in time. So the speed of an object is the change in distance
$ds$ over the change in time $dt$.

% Let $\vec r(t)=(x,y,z)$ be a parametrization of a
smooth space curve. Let $\ds s(t)=\int_0^t \left|\frac {d\vec
r}{d\tau} \right|\ d\tau$. Explain why $\ds\frac{ds}{dt} =
\left|\frac {d\vec r}{dt} \right|$, the speed. Then explain why
$s(t)$ is an increasing function. [Hint: Look up the
fundamental theorem of calculus. To answer why is $s$
increasing, what does “smooth” mean?]
\marginparbmw{See 13.3: 11-14 for more practice.} Consider
again the helical space curve $\vec r(t)=(\cos t, \sin t, t)$.
We've shown that $s(t) = t\sqrt{2}$.

The previous problem motivates the following definition.
- Solve for $t$ in terms of $s$ (so find the inverse of $s(t)$). If you've traveled 4 units of distance, how much time has elapsed.
- Compute $D\vec r(t)$ and $Dt(s)$. You should have a 2 by 1 matrix, and a 1 by 1 matrix.
- Use the chain rule to compute the derivative of $\vec r(t(s))$.
- Compute the length $\left|\ds\frac{d\vec r}{ds}\right|$.

\label{def unit tangent vector} Let $\vec r(t)$ be a
parametrization of a smooth space curve. We define the unit
tangent vector $\vec T(t)$ to be the derivative of $\vec r$
with respect to arc length, which means $$\vec T =
\ds\frac{d\vec r}{ds}=\ds\frac{d\vec r/dt}{ds/dt} = \frac{d\vec
r/dt}{|d \vec r/dt|} = \frac{\vec v}{|\vec v|}.$$ This is
exactly the same as unit vector in the same direction as the
velocity.

As we progress through this unit, one of our key goals is
to learn the new notation. We've got position $\vec r$, velocity
$\vec v$, speed $v$ or $ds/dt$, acceleration $\vec a$, the unit
tangent vector $\vec T$, the derivative of position with respect
to arc length $d\vec r/ds$. The last two are the exact same since
$\vec T = d\vec r/ds$. Did you also notice that $ds/dt$ and $v$
are both the speed? We'll need to start realizing that the same
quantity can be developed in many ways.
\marginparbmw{See 13.3: 1-10 for more practice.}% Suppose an
object moves along the space curve given by $\vec r(t)=(a\cos
t,a\sin t,b t)$.

As we progress in this chapter, we'll be computing more
derivatives with respect to $s$, instead of $t$. Did you notice
in the previous problem that to compute a derivative with respect
to $s$, you just compute the regular derivative with respect to
$t$, and then divide by the speed. Please do the following review
problem to make sure you've got down what $\frac{d}{ds}$ means.
- Find the object's velocity and speed. What is $ds/dt$?
- Compute $\frac{d\vec r}{ds}$, the derivative of $\vec r$ with respect to arc length. Leave your answer in terms of $t$.[Hint: Divide the top and bottom by $dt$ and then compute $d\vec r/dt$ and $ds/dt$.]
- State the unit tangent vector $\vec T(t)$.

Suppose $\vec r(t)=(3\cos t,3\sin t,4t)$. Compute $v$, $ds/dt$,
$d\vec r/ds$, $\vec T$, and $d\vec T/ds$. See \footnote{ We
have $\vec v = \dfrac{d\vec r}{dt} = (-3\sin t, 3\cos t, 4)$.
The speed is $\dfrac{ds}{dt}=|\vec v| =
\sqrt{9\sin^2t+9\cos^2t+16}=5$. We then compute $\dfrac{d\vec
r}{ds}=\dfrac{d\vec r/dt}{ds/dt} = \dfrac{1}{5}(-3\sin t, 3\cos
t, 4)$, which equals $\vec T$. We finally compute $$
\dfrac{d\vec T}{ds}=\dfrac{d\vec
T/dt}{ds/dt}=\left(\dfrac{1}{ds/dt}\right)\dfrac{d\vec T}{dt} =
\left(\dfrac{1}{5}\right)\dfrac{1}{5}(-3\cos t, -3\sin t, 0) =
\dfrac{1}{25}(-3\cos t, -3\sin t, 0). $$ } for an answer.

Sammy sits on a merry go round. He sits 3 feet from the center
of the merry ground, and lets his big sister spin him around.
We can parameterize Sammy's path with the vector equation $\vec
r(t) = (3\cos t, 3\sin t)$, where $t$ is in seconds. The point
$(0,3)$ corresponds to the time $t=\pi/2$.

- Draw the curve for $0\leq t\leq 2\pi$. Compute $\vec v$ and $\vec a$. At $(0,3)$ (so $t=\pi/2$), draw these two vectors. Are these two vectors orthogonal?
- Compute $\vec T$ and $\dfrac{d\vec T}{dt}$. At $(0,3)$, draw these two vectors. Are these two vectors orthogonal?
- Compute $\dfrac{d\vec T}{ds}$ and $\left|\dfrac{d\vec T}{ds}\right|$. How is the length of $\dfrac{d\vec T}{ds}$ related to the circle?

Sammy's older sister grabs another friend to help push the
merry-go-round. The spinning speed doubles. If we replace $t$
with $2t$ in our parameterization, we get the exact same path,
but traverse it twice as fast. Sammy's path is now parametrized
by $\vec r(t) = (3\cos 2t, 3\sin 2t)$.

Look back at the previous two problems. We computed $\vec
v$, $\vec a$, $\vec T$, $\dfrac{d\vec T}{dt}$, $\dfrac{d\vec
T}{ds}$, and $\left|\dfrac{d\vec T}{ds}\right|$ in each. Which of
these changed when we changed the speed? Which stayed the same?
The next problem has you develop this.
- Draw the curve. How long does it take to go around once? Note that $t=\pi/4$ now corresponds to $(0,3)$.
- Compute $\vec v$, $\vec a$, $\vec T$, $\dfrac{d\vec T}{dt}$, $\dfrac{d\vec T}{ds}$, and $\left|\dfrac{d\vec T}{ds}\right|$. At $(0,3)$, add the vectors to your picture.
- Compute $\dfrac{d\vec T}{ds}$ and $\left|\dfrac{d\vec T}{ds}\right|$. How is the length of $\dfrac{d\vec T}{ds}$ related to the circle?

We started with the parameterization $\vec r(t) = (3\cos t,
3\sin t)$, and replaced $t$ with $2t$ to double the speed.

- Now replace $t$ with $5t$ and compute $\vec v$, $\vec a$, $\vec T$, $\dfrac{d\vec T}{dt}$, $\dfrac{d\vec T}{ds}$, and $\left|\dfrac{d\vec T}{ds}\right|$. Evaluate each at $t=\frac{\pi}{10}$, the time corresponding to point $(0,3)$.
- Replace $t$ with $\omega t$ and find $\vec v$, $\vec a$, $\vec T$, $\dfrac{d\vec T}{dt}$, $\dfrac{d\vec T}{ds}$, and $\left|\dfrac{d\vec T}{ds}\right|$. Evaluate each at $t=\frac{\pi}{2\omega}$, the time corresponding to point $(0,3)$.
- At $(0,3)$, which of these 6 quantities remain the same?
%
- In part 2, show that $|\vec a| = |\vec v|^2\left|\dfrac{d\vec T}{ds}\right|.$ So tripling the speed causes the acceleration to increase by a factor of \rule{1in}{.5pt}.

Gauss developed and expanded many of the ideas in this section
while on a map making mission for a king. He wanted to create
extremely high quality 2D images of the 3D world we are in. I
will be seeking for a good reference to put here for further
study.

We started with motion on a circle with a relatively simple
speed. We then doubled the speed, multiplied the speed by 5, and
then chose a variable speed. You should have noticed that at the
same point on the curve, regardless of speed, the quantities
$\vec T$, $\dfrac{d\vec T}{ds}$, and $\left|\dfrac{d\vec
T}{ds}\right|$ remained the same. Will this pattern continue if
we were to change from constant speeds to variable speeds? What
if we left the path of a circle, and moved to any smooth curve?
How far can we take this pattern? Is it true always? If so, then
the quantities $\vec T$, $\dfrac{d\vec T}{ds}$, and
$\left|\dfrac{d\vec T}{ds}\right|$ are pretty important
quantities that describe the curve we're on. Because
$\dfrac{d\vec T}{ds}$ and $\left|\dfrac{d\vec T}{ds}\right|$ show
up a lot, let's give them a definition.
Let $\vec r(t)$ be a smooth curve (so that $\vec v$ is never
zero).

Let's apply this new definition to a circle of any radius.
We'll quickly see that the curvature and radius of a circle are
related. The curvature vector $d\vec T/ds$ tells us how quickly
$\vec T$ changes as we increase in length, which is a measure of
how sharply we turn a corner. If our curve is a circle of radius
$3$, then the curvature is $1/3$. A larger circle should result
in smaller curvature.
- The vector $\vec \kappa = \dfrac{d\vec T}{ds}$ we'll call the curvature vector. It measures how quickly the unit tangent vectors changes as we increase in length (not time).
- The number $\kappa = \left|\dfrac{d\vec T}{ds}\right|$ we'll call the curvature.

Consider the curve $\vec r(t)=(a\cos t, a\sin t)$.

For any curve, we could approximate how rapidly the curve
turns at a point by drawing a circle that best approximates the
curve (kind of like a Taylor polynomial, only now we'll use a
circle.) We want the circle to meet the curve $\vec r$
tangentially, and we want the curvature of the circle to match
the curvature of the curve. Since the curvature of the circle
must match the curvature of the curve, we know they are inverse
related. This gives the following definition.
- Draw the curve, and state the radius $\rho$ of the best approximating circle.
- Find the curvature $\vec \kappa$ by performing a computation.
- What relationship exists between $\rho$ and $\kappa$? If the radius $\rho$ were to increase, what would happen to $\kappa$?

Watch a
YouTube Video.

% When the curvature $\kappa$ of a smooth curve is
nonzero, we'll define the radius of curvature, written $\rho$,
to be the reciprocal $\rho = \dfrac{1}{\kappa}$. The curvature
and radius of curvature are inversely related.
Suppose $f(x) = \sin(x^2)$. Compute $f'(x)$ and $f”(x)$.
See \footnote{ We have $f'(x) = 2x\cos(x^2)$ and $f”(x) =
-4x^2\sin(x^2)+2\cos(x^2)$. }.

Show that $(-2,3,1)$ and $(4,2,2)$ are orthogonal. See
\footnote{The dot product of these two vectors is
$(-2,3,1)\cdot(4,2,2) = -8+6+2=0$. Because the dot product is
zero, the vectors are orthogonal. }.

The big kids pushing Sammy on the merry-go-round decide to
constantly increase the spinning rate. To parametrize Sammy's
path, let's replace $t$ with $t^2$ in the original
parametrization to obtain $\vec r(t) = (3\cos t^2, 3\sin t^2)$.
Fill in the missing values below, and then answer the questions
that follow. $$

{|c|c|}\hline \vec v & ( -6 t \sin \left(t^2\right),6 t
\cos \left(t^2\right) ) \\\hline ds/dt & 6 t \\\hline
\vec a & %(-6 \sin \left(t^2\right)-12 t^2 \cos
\left(t^2\right),6 \cos \left(t^2\right)-12 t^2 \sin
\left(t^2\right)) \\\hline \vec T & \left(-\sin
\left(t^2\right),\cos \left(t^2\right)\right) \\\hline d\vec
T/dt & \left(-2 t \cos \left(t^2\right),-2 t \sin
\left(t^2\right)\right)\\\hline \vec \kappa = d\vec T/ds
& %\left\{-\frac{1}{3} \cos \left(t^2\right),-\frac{1}{3}
\sin \left(t^2\right)\right\} \\\hline \kappa =|d\vec T/ds|
& \frac{1}{3} \\\hline

$$
- Show that $\vec v$ and $\vec a$ are not orthogonal, but that $\vec T$ and $d\vec T/dt$ are.
- To get to $(0,3)$, we could have $t^2=\pi/2$ or $t^2=2\pi+\pi/2$ or many other values. Show that $\vec T$ is the same at $(0,3)$, regardless of what time we pass through $(0,3)$. Show the same is true for $d\vec T/ds$.
- When the speed is $\frac{ds}{dt}=6$ (so $t=1$), how long are $d\vec T/dt$ and $d\vec T/ds$? When the speed is $\frac{ds}{dt}=12$ (so $t=2$), how long are $d\vec T/dt$ and $d\vec T/ds$? When the speed is $\frac{ds}{dt}=30$ (so $t=5$), how long are $d\vec T/dt$ and $d\vec T/ds$? If the speed were some random number $v$, how long are $d\vec T/dt$ and $d\vec T/ds$?

\label{curvature observations} The previous problem showed the
following patterns.

Let's see if this pattern continues when we swap to a
different curve. Rather than try to connect the curve to some
physical real world example, let's look at a curve we are
familiar with, a parabola. If the pattern holds there as well,
then we may have found a key pattern that works with all curves.
Then we can take our new knowledge and apply it to ANYTHING (like
space flight, roller coasters, missiles, and anything that
moves). If you want to perform the computations below by hand,
you must master the product, quotient, and chain rule, as well as
working with rational exponents in algebra. Feel free to try the
computations by hand (if you can get them, then well done).
Please use Sage to do the computations below.
- The vectors $\vec v$ and $\vec a$ are not always orthogonal.
- The vectors $\vec T$ and $\dfrac{d\vec T}{dt}$ are always orthogonal.
- The vectors $\vec T$ and $\dfrac{d\vec T}{ds}$ are independent of speed. They depend only on the shape of the curve, not the speed at which you traverse the curve.
- If we multiply $\dfrac{d\vec T}{ds}$ by $\dfrac{ds}{dt}$, we'll get $\dfrac{d\vec T}{dt}$. The two vectors point in the same direction, which from here on out we'll call $\vec N$, the direction normal to the motion.

Use the Sage link.

Consider the curve $\vec r(t) = (t, t^2)$. The
computations get intense, so let's use a computer algebra
system, such as Sage (follow
this link) to help us. If we don't use a computer, we could
spend hours on this problem. The computer will do all the
computations, and give graphs, in seconds.
- Use this Sage link to compute $\vec v$, $\vec a$, $\dfrac{ds}{dt}$, $\vec T$, $\dfrac{d\vec T}{dt}$, $\dfrac{d\vec T}{ds}$, $\left|\dfrac{d\vec T}{ds}\right|$, and the unit vector $\vec N$ that is orthogonal to $\vec T$. Evaluate each at $t=1$ and $t=\sqrt{3}/2$ (use “sqrt(3)/2”). Record your answers in the provided table.
- Let's now double the speed at which we traverse along the curve. Replace $t$ with $2t$, and then repeat the problem above. However, instead of putting in $t=1$ and $t=\sqrt{3}/2$, we now need to use $t=1/2$ and $t=?$. Use Sage to record your answers in the provided table.
- After completing the table, does Observation TODO still hold?
- How are $\vec T$ and $\vec N$ related? Conjecture a
pattern.
{|c|c|c|c|c|}\hline \multirow{2}{*}{Value}&\multicolumn{2}{|c|}{$\vec r(t)=(t,t^2)$}&\multicolumn{2}{|c|}{$\vec r(t)=(2t,(2t)^2)$}\\ &\quad \quad at $t=1$ \quad \quad & \quad at $t=\sqrt3/2$\quad \quad &\quad \quad at $t=1/2$\quad \quad & \quad at $t=?$\quad\quad\quad \quad \\\hline $\vec r$ & $(1,1)$ & $(3/4,3/4)$ & $(1,1)$ & $(3/4,3/4)$ \\\hline $\vec v$ & $(1,2)$& & & \\\hline $\vec a$ & $(0,2)$& & & \\\hline $\frac{ds}{dt} $ &$\sqrt{5}$ & & & \\\hline $\vec T$ & $(\frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5})$ & & & \\\hline $\frac{d\vec T}{dt} $ &$(-\frac{4\sqrt{5}}{25},\frac{2\sqrt{5}}{25})$ & & & \\\hline $\vec\kappa=\frac{d\vec T}{ds} $ & $(-\frac{4}{25},\frac{2}{25})$& & & \\\hline $\kappa=\left|\frac{d\vec T}{ds}\right| $ &$\frac{2\sqrt{5}}{25}$ & & & \\\hline $\vec N$ &$(-\frac{2\sqrt{5}}{5},\frac{1\sqrt{5}}{5})$ & & & \\\hline

Consider the helix $\vec r(t)=(4\cos t,4\sin t, 3t)$.

The observation still holds. It looks like we've found some
key facts that pertain to any curve. We only have 3 examples to
justify our conclusion, but it's enough to make some definitions
and then seek for a concrete proof that holds for every smooth
curve. Let's prove one of the facts above, namely that $\vec T$
and its derivative will always be orthogonal. The key here is
that $\vec T$ is a vector of constant length.
Because the speed is constant, many of the computations are
simplified. You can use the Sage link to check your work.

Compute $\vec v$, $\vec T$, $\dfrac{d\vec T}{dt}$, and
$\dfrac{d\vec T}{ds}$. Then do the following:
- Show that $\vec T$ and $\dfrac{d\vec T}{dt}$ are orthogonal. (Do this for any time $t$.)
- At $t=2\pi$, compute $ds/dt$, $\vec T$, $\dfrac{d\vec T}{dt}$, and $\dfrac{d\vec T}{ds}$.
- Let's slow down how quickly we travel along the curve. Replace $t$ with $t/5$. Then at $t=10\pi$, give $ds/dt$, $\vec T$, $\dfrac{d\vec T}{dt}$, and $\dfrac{d\vec T}{ds}$ for the new curve $\vec r_2(t) =(4\cos(\frac{t}{5}),4\sin(\frac{t}{5}), 3(\frac{t}{5}))$.
- Give a unit vector that points in the same direction as $\dfrac{d\vec T}{dt}$ or $\dfrac{d\vec T}{ds}$.

\label{vector valued functions of constant length} If a vector
valued function $\vec r(t)$ has constant length, then the
vector $\vec r$ and its derivative $\ds\frac{d\vec r}{dt}$ are
always orthogonal. Vector valued functions of constant length
are orthogonal to their derivative.

TODO">

The above fact is so crucial, that we'll repeat what it
says.
Watch a
YouTube Video.

% Prove the theorem above. Here are some hints [as an
alternative to watching the YouTube video].
- We know that $\vec r(t)$ has constant length, so we write $|\vec r|=c$ for some constant $c$.
- We need to get from a magnitude to a dot product. Look in your text for a way to relate magnitude to the dot product. See problem TODO.
- After writing $|\vec r(t)|=c$ in terms of a dot product (squaring both sides may help), take the derivative of both sides. Apply the product rule to the dot product.

If the vector $\vec v(t)$ has constant length, then the vector
and its derivative $\frac{d\vec v}{dt}$ are orthogonal.

\label{T and N are orthogonal}

Based on our answer above, let's make the following
definition of the principle unit normal vector. The key idea is
that this vector points in the direction of normal acceleration.
Watch a
YouTube Video.

% Let $\vec r$ be a smooth parametrization of a curve.
How long is the {\it unit} tangent vector $\vec T(t)$? Explain
why $\vec T$ is orthogonal to $\dfrac{d\vec T}{dt}$. Then give
a formula for computing a unit vector that is orthogonal to
$\vec T(t)$.
Suppose $\vec r(t)$ is a smooth parametrization of a space
curve, where the unit tangent vector is $\vec T(t)$. We define
the principle unit normal vector $\vec N(t)$ to be the vector
$$\vec N(t) = \ds\frac{d\vec T/dt}{|d\vec T/dt|},$$ provided of
course that $|d\vec T/dt|\neq 0$. From problem TODO we know that $\vec
T$ and $\vec N$ are orthogonal.

The vector $\vec N$ is a unit vector in the same direction
as the curvature vector. Since $\dfrac{d\vec T}{dt}$ and
$\dfrac{d\vec T}{ds}$ point in the same direction, we could have
written $\vec N = \dfrac{d\vec T/ds}{|d\vec T/ds|} = \dfrac{\vec
\kappa}{\kappa}$. The vectors $\vec T$ and $\vec N$ give us
exactly the pieces needed to describe velocity and acceleration.
We'll visit this more in a bit. When we are in 3D, these two
vectors describe a plane. The binormal vector of the TNB frame is
the normal vector to this plane.
If $\vec r$ is a parametrization of a smooth space curve with
unit tangent vector $\vec T$ and principle unit normal vector
$\vec N$, then we define the binormal vector $\vec B$ to be the
cross product $$\vec B = \vec T\times \vec N.$$ It's normal to
both $\vec T$ and $\vec N$, hence called the \textit{bi}normal
vector.

We now have the entire $TNB$ frame. This gives us a moving
collection of unit vectors that act like an $xyz$ coordinate
system. Many of you will use this frame a ton in your dynamics
course. The TNB frame shows up in physical chemistry as well. A
key fact to remember is that all three vectors are unit vectors,
and they are each orthogonal to the other.
Find the area of a parallelogram with corners $(0,0,0)$,
$(1,0,2)$, and $(1,1,3)$, and $(2,1,5)$. See \footnote{ We find
area of a parallelogram by the vectors that form the edges.
Since one point is the origin (subtracting it from the 2nd and
3rd points won't change anything), we compute the cross product
$(1,0,2)\times (1,1,3) = (-2,-1,1) $. The magnitude of the
cross product is the area $A = \sqrt{4+1+1}=\sqrt{6}$. }.

Answer the following questions (this will review your knowledge
of the dot and cross products).

Let's carry through a single problem where we compute $\vec
T$, $\vec N$, and $\vec B$.
- What is $\vec T\cdot \vec N$? Explain. Then explain why $\vec T\cdot \vec B=0$ and $\vec N\cdot \vec B=0$.
- Both $\vec T$ and $\vec N$ are unit vectors. They are also orthogonal. Why is $\vec B$ is a unit vector? [What's the connection between the cross product and area.]
- We defined $\vec B=\vec T\times \vec N$. This means that $\vec N\times \vec T=-\vec B$. Do we have $\vec T\times \vec B$ equal to $\vec N$ or $-\vec N$? Explain. [Hint: What does the right hand rule say? Compare to $\hat i\times \hat k$. ]

\label{helix example of T N and B} \marginparbmw{See 13.4: 9-16
and 13.5: 9-16 (the relevant parts) for more practice.}%
Consider the helix $\vec r(t) = (3\cos t,3\sin t, 4t)$. Find
the unit tangent vector $\vec T(t)$, principle unit normal
vector $\vec N(t)$, and the binormal vector $\vec B(t)$.

Once the speed is no longer constant, things get a lot
messier than the previous problem. Ask me in class to show you
what happens with the computations when you consider something
like $r(t)=(t,t^2,t^3)$. Things get ugly really fast.
Fortunately, when you're working with a curve that lies in a
plane, there are some simplifications that occur.
\marginparbmw{See 13.4: 7-8 for more practice, and perhaps a
hint.}% Suppose you have already computed the unit tangent
vector for a curve in the plane and found at a specific time it
equals $\vec T=(a,b)$.

- State a nonzero vector that is orthogonal to $(a,b)$. (Guess one, and use the dot product to check. If you're struggling because of the variables, find a vector orthogonal to $(2,3)$.)
- Let $\vec r(t) = (t,t^2)$. We then have $\frac{d\vec r}{dt} = (1,2t)$ and $\vec T(t) = \frac{(1,2t)}{\sqrt{1+4t^2}}$. Without computing any more derivatives, guess the principle unit normal vector $\vec N(t)$?
- Draw a picture of the curve. At $t=1$ add to your picture the tangent vector $(1,2)/\sqrt{5}$ and your guessed normal vector. (If your guess was off by a sign, tell us how to modify your guess.)
- Why does $\vec B=(0,0,1)$?

From the problem above, we learn the following fact. If the
tangent vector to a planar curve is $\vec T(t) = (a(t),b(t))$,
then the principle unit normal vector is either $\vec
N(t)=(-b(t),a(t))$ or $\vec N(t)=(b(t),-a(t))$. You just
reverse the components, and then negate one of them. To
determine which one to negate, draw a picture.

Consider the curve $\vec r(t)=(t^2,t)$. Compute $\vec T(t)$.
Reverse the order and negate one of the component to find $\vec
N(t)$. To know if you guessed the write component to negate,
draw the curve and on your graph include these vectors at
$t=1$. Finally, state $\vec B(t)$.

\marginparbmw{See 13.4: 1-4 for more practice. Use the previous
problems.}% Consider the curve $y=\sin x$, parametrized by
$r(t)=(t,\sin t)$. We know that $\vec T(t) = \dfrac{(1,\cos
t)}{\sqrt{1+\cos^2t}}$. Draw the curve from $-\pi$ to $\pi$.
Then on your graph draw $\vec T$ and $\vec N$ at $t=\pi/2$,
$\pi/4$, $-\pi/4$.

You've now developed the TNB frame for describing motion.
Engineers will see this again when they study dynamics.
Mathematicians who study differential geometry will use these
ideas as well. Any time you want to analyze the forces acting on
a moving object, the TNB frame may save the day. Chemists will
encounter the TNB frame briefly when they study P-chem and the
motion of subatomic particles.
- What is $\vec T(t)$ at each of $t=\pi/2$, $\pi/4$, $-\pi/4$?
- Show how to get $\vec N(t)$ at each of $t=\pi/2$, $\pi/4$, $-\pi/4$?
- What is $\vec B(t)$ at each of $t=\pi/2$, $\pi/4$,
$-\pi/4$?
What happens if $t=0$?

Consider the curve $\vec r(t)=(t,\sin 3 t)$. We can compute
$\vec v(t) = (1,3\cos t)$ and $\vec T(t) = \frac{(1,3\cos
3t)}{\sqrt{1+9\cos^2(3t)}}$. Using the quotient rule, we find
that $$\ds\frac{dT}{dt} = \frac{\sqrt{1+9\cos^2(3t)}(?,?) -
(1,3\cos3t)(?)}{1+9\cos^2(3t)}.$$

- Fill in the blanks in the quotient rule above.
- % Show at $t=\pi/6$ that $d\vec T/dt = (0,-9)$. Then state $\vec \kappa$, $\kappa$, and $\rho$ at $t=\pi/6$.
- Draw the curve $\vec r(t)$ and on your curve at $t=\pi/6$ add the circle of curvature.
- Where is the center of curvature (the center of the best approximating circle) at $t=\pi/6$.

If you are standing at $(2,1,-3)$ and you wish to move 6 units
in the direction of the unit vector $(1/3, 2/3, -2/3)$, where
are you? See \footnote{We want to start at $(2,1,-3)$ and move
$6(1/3, 2/3, -2/3) = (2,4,-4)$. We just add the vectors, giving
$(4,5,-7)$.} for an answer.

% Consider the helix $\vec r(t)=(t,\sin t,\cos t)$. Find
the curvature and radius of curvature $t=\pi/2$. Then draw the
curve, and draw the circle of curvature at $t=\pi/2$. Finally,
find the center of curvature at $t=\pi/2$. Guess the center of
curvature at $t=\pi$? [Hint: If you're struggling with how to
get from the curve to the center of curvature, please do the
review problem above.]

When a civil engineering team builds a road, they have to
pay attention to the curvature of the road. If the curvature of
the road is too large, accidents will happen and the civil
engineering team will be liable. How do they make sure the
curvature never gets to large? They use the circle of curvature.
When they want to cause a road to turn, they'll find the center
of curvature, send a surveyor out to the center, and then have
the surveyor make sure that the road follows the circle of
curvature for a short distance. They actually pace out the circle
of curvature and then build the road along this circle for a
hundred feet or so. Then, they recompute the radius of curvature
(if they need the direction to change again), and pace out
another circle. In this way, they can guarantee that the
curvature never gets large. In the next section we'll see how
curvature is directly related to normal acceleration (which is
what causes semis to tip, and vehicles to slide off icy roads.)
Consider the force vector $\vec F = (0,-10)$, and displacement
vector $\vec (2,-1)$. Compute the projection of $\vec F$ onto
$\vec d$, and then write $\vec F$ as the sum of a vector
parallel to $\vec d$ and a vector orthogonal to $\vec d$. See
\footnote{ The projection is $\proj_{\vec d}\vec F =
\frac{10}{5}(2,-1) = (4,-2)$. This is the parallel component
$\vec F_{|| \text{ to }\vec d} =(4,-2)$. To get the orthogonal
component, we know that $\vec F = \vec F_{|| \text{ to }\vec d}
+ \vec F_{\perp \text{ to }\vec d}$. Vector subtraction gives
$\vec F_{\perp \text{ to }\vec d} = \vec F -\vec F_{|| \text{
to }\vec d} = (0,-10)-(4,-2) = (-4,-8)$. We now write $$\vec F
= (4,-2) + (-4,-8).$$ }.

%

If we throw a pebble from a 64 ft tall cliff, then we could
parameterize the path after $t$ seconds using $\vec r(t) =
(3t,64-16t^2)$. The numbers below get rather large in a hurry, so
let's use a simpler parameterization to gain understanding about
the connection between $\vec v$, $\vec a$, $\vec T$, and $\vec
N$. Sometimes the key to understanding is to simplify the
problem.
% Suppose that a projectile is fired from the ground on some
distance planet, and that the only acceleration is due to
gravity. We can parametrized the path of the particle with
$r(t)=(40t,-5t^2+40t)$. It takes 8 seconds to hit the ground.
%

- %
- % % Find $\vec v$ and $\vec a$. We're on a distant planet because you should find the acceleration is always pointing straight down with magnitude 10. %
- Find $\vec T(t)$, and then at $t=1$ please compute $\text{proj}_{\vec T(1)}\vec a(1)$. %
- State $\vec N(t)$ (remember you can flip the order and change a sign), and then at $t=1$ compute $\text{proj}_{\vec N(1)}\vec a(1)$. %
- If we write $\vec a = a_T\vec T +a_N\vec N$, then what are $a_T$ and $a_N$? %

Consider the parameterization $\vec r(t) = (t,9-t^2)$ (it's a
simplified version of tossing a pebble off a building). Our
goal, at time $t=1$, is to write $\vec a$ in the form $\vec a =
a_T\vec T+a_N\vec N.$ Follow the steps below.

- Compute $\vec v$, $\vec a$, and $\vec T$ at time $t$, and then at time $t=1$. Then at $t=1$, compute the projection of $\vec a$ onto $\vec T$, i.e. compute $\proj_{\vec T(1)}\vec a(1)$.
- State $\vec N(t)$ (remember you can flip the order and change a sign), and then at $t=1$ compute $\text{proj}_{\vec N(1)}\vec a(1)$.
- If we write $\vec a = a_T\vec T +a_N\vec N$, then what are $a_T$ and $a_N$?

Suppose that $\vec r(t)$ is a smooth parametrization of a
moving object. Let $\vec T$ be the unit tangent vector. The
tangential component of acceleration and the normal component
of acceleration are the scalars $a_T$ and $a_N$ that we obtain
by writing the acceleration as the sum of a vector parallel to
$T$ and a vector orthogonal to $\vec T$, i.e. the scalars that
satisfy $$\vec a = a_T\vec T+a_N\vec N.$$

Let's return to the example of Sammy on a merry-go-round.
From this example, we'll see one of the key ideas in this
section.
Suppose that Sammy sits $\rho$ feet away from the center of the
merry-go-ground. His sister decides to spin him around at
different speeds. Let $\vec r(t) = (\rho \cos \omega t, \rho
\sin \omega t)$ be a parametrization of Sammy's postion.

In the problem above, all of the acceleration is in the
normal direction. The interesting thing to note is that the
normal acceleration is ALWAY $a_T = \frac{|\vec v|^2}{\rho}$.
That's what we'll now show. We'll show that the acceleration of
an object moving along a curve $\vec r(t)$ with velocity $\vec
v(t)$ is the sum $$\vec a(t) = a_T\vec T+a_N\vec
N=\frac{d}{dt}|\vec v(t)| \vec T + \kappa |\vec v|^2 \vec N.$$
The scalars $a_T=\dfrac{d}{dt}|\vec v(t)|$ and $a_N=\kappa |\vec
v|^2$
- Show that Sammy's speed is $|\vec v|=\rho \omega$.
- Find the curvature of Sammy's path at any time $t$ (it happens to be constant - and we did this already when we computed the curvature along a circle).
- Find the acceleration vector, and show that $\ds |\vec a| = \kappa |\vec v|^2 = \frac{|\vec v|^2}{\rho}$.

Engineers often use the equivalent formula $a_N = \frac{|\vec
v|^2}{\rho}$, as $\rho$ is a physical distance that they can
measure.

are the tangential and normal components of acceleration.
All we have to do is write the vector $\vec a(t)$ as the sum of a
vector parallel to $\vec T$ and a vector orthogonal to $\vec T$.
Before we decompose the acceleration into its tangential and
normal components, let's look at two examples to see what these
facts physically represent.
%

% Consider the force vector $\vec F = (0,-10)$ and
displacement vector $\vec (2,-1)$. Draw $\vec F$, $\vec d$,
and the projection of $\vec F$ onto $\vec d$. See
\footnote{The solution is %

% draw[->] (0,0)->(0,-10); % draw[->]
(0,0)->(2,-1); %

% }. %
% %

% \marginparbmw{See 13.5: 17-20 for more practice.}% %
Consider the path of an object in projectile motion that has
been fired from the origin. Draw a typical path followed by a
projectile. The acceleration $\vec a(t)=(0,-g)$ acts straight
down for any time $t$. %

- %
- Pick a point on your path before the max height occurs. At that point, draw both $\vec T$, $\vec a$, and the projection of $\vec a$ onto $\vec T$. Is $a_T$ positive or negative? %
- At the point you chose above, is the speed of the projectile increasing or decreasing as it climbs higher? %
- Now pick a point after the projectile passes the peak. Repeat the last two parts at this new point. %

Imagine that you are riding as a passenger on a road and
encounter a series of switchbacks (so the road starts to zigzag
up the mountain). Right before each bend in the road, you see a
yellow sign that tells you a U-turn is coming up, and that you
should reduce your speed from 45 mi/hr to 15 mi/hr. Assume the
largest curvature along the turn is $\kappa$. Recall that
$a_N=\kappa |\vec v|^2$. The engineers of the road designed the
road so that if you are moving at 15 mi/hr, then the normal
acceleration will be at most $A$ units.

- Suppose that your driver (Ben) ignores the suggestion to slow down to 15 mi/hr. He keeps going 45 mi/hr through the turn. Had he slowed down, the max acceleration would be $A$. You're traveling 3 times faster than suggested. What will your maximum normal acceleration be? [It's more than $3A$.]
- You yell at Ben to slow down (you don't want to die). So Ben decides to only slow to 30 mi/hr. He figures this means you'll only feel twice as much acceleration as $A$. Explain why this line of reasoning is flawed.
- Ben gets frustrated by the fact that he has to slow down. He complains about the engineers who designed the road, and says, “they should have just built a larger corner so I could keep going 45.” How much larger should the radius of the circle be so that you can travel 45 mi/hr instead of 15 mi/hr, and still feel the same acceleration $A$?
- Which will cause the normal acceleration to decrease more, halving your speed or halving the curvature (doubling the radius)?

%

% We defined the principle unit normal vector as $\vec N =
\dfrac{d\vec T/dt}{|d\vec T/dt|}$. Explain why we can write
$\vec N = \dfrac{d\vec T/ds}{|d\vec T/ds|}$ as well. Then use
this fact to explain why , which means we can write
$\ds\kappa|\vec v|\vec N=\frac{d\vec T}{dt}$. %

Watch a
YouTube Video.

% Prove that $\ds \vec a(t) = a_T\vec T+a_N\vec
N=\frac{d}{dt}|\vec v| \vec T + \kappa |\vec v|^2 \vec N.$
Here's some hints.
- Rewrite the velocity $\vec v$ as a magnitude $|\vec v|$ times a direction $\vec T$, so $\vec v = |v|\vec T$.
- We know that $\vec a(t) = \frac{d}{dt}\vec v(t)$. Take the derivative of $\vec v = |\vec v|\vec T$ by using the product rule (on the scalar product $|\vec v|\vec T$).
- You should encounter the quantity $d\vec T/dt$. We know that $\frac{d\vec T}{dt} = |\vec v|\frac{d\vec T}{ds}$. Why does $\ds d\vec T/dt=|\vec v|\kappa\vec N$?
- Conclude to explain why $a_N =\kappa |\vec v|^2$.

Show that $$\kappa = \frac{|\vec v\times \vec a|}{|\vec v|^3} =
\frac{|\vec r'\times \vec r”|}{|\vec r'|^3}.$$ [Hint: We
know that $\ds \vec a = a_T\vec T+a_N\vec N=\frac{d}{dt}|\vec
v| \vec T + \kappa |\vec v|^2 \vec N.$ Cross both sides with
$\vec v$. You should be able to cancel $\vec v\times \vec v$
(why). Then take the magnitude of each side and solve for
$\kappa$. You'll have to explain why $|\vec v\times \vec N| =
|\vec v|$.]

We can use the above formula for curvature to get a quick
way to compute the curvature of a function $y=f(x)$. If you use
the previous problem, this formula falls out almost instantly.
You'd see this formula in dynamics, and it shows up on the
Fundamentals of Engineering exam (where you just have to use the
formula, not prove where it comes from). This is the culminating
idea from this chapter that you'll use again and again in
engineering courses.
\label{formula for curvature} \marginparbmw{See 13.4: 5.}% The
function $y=f(x)$ can be given the parametrization $\vec r(x) =
(x,f(x))$. Use this parametrization (and the previous problem)
to show that the curvature is $$\kappa(x) =
\frac{|f”(x)|}{(1+(f')^2)^{3/2}},$$ and that the radius
of curvature is $$\rho(x) =
\frac{(1+(f')^2)^{3/2}}{|f”(x)|}.$$

Let $\vec r(t)$ be a parametrization of a smooth curve $C$ with
unit tangent vector $\vec T(t)$. The derivative of $\vec B$
with respect to $s$ tells us how rapidly the plane containing
$\vec T$ and $\vec N$ rotates. We'll define the torsion vector
to be

The computations involved in getting $\tau$ require a lot
of work. Let's use the computer to help us. You can do all of
this with the aid of Sage. I'll let you decide from the code what
$\tau$ is. That will be your decision to make.
Watch a
YouTube Video.

% $$\vec \tau = \dfrac{d\vec B}{ds} = \dfrac{d\vec
B/dt}{ds/dt}=\dfrac{d\vec B/dt}{|d\vec r/dt|}.$$ The torsion
$\tau$, up to a sign, is the length of this vector. We say
there is positive torsion if $\vec \tau$ causes a
counterclockwise rotation about $\vec T$ (as you look down
$\vec T$), which occurs precisely when $\vec tau$ and $\vec N$
point in opposite directions. We can summarize this is
$$\tau=\left|\dfrac{d\vec B}{ds}\right|\quad \text{or}\quad
\tau=-\left|\dfrac{d\vec B}{ds}\right|,$$ where you choose
“$+$” if $\vec N$ and $\vec \tau$ point in opposite
directions.
\marginparbmw{See 13.4: 9-16 and 13.5: 9-16 (the relevant
parts) for more practice.}% Consider the helix $r(t)=(3\cos t,
3\sin t, 4t)$. In problem TODO we found
\begin{align*} \vec T &= (-\frac{3}{5}\sin
t,\frac{3}{5}\cos t,\frac{4}{5})\\ \vec N &= (-\cos t,-\sin
t,0)\\ \vec B &= (\frac{4}{5}\sin t,-\frac{4}{5}\cos
t,\frac{3}{5}) \end{align*} Compute the torsion vector $\vec
\tau=\dfrac{d\vec B}{ds}$, and then give the torsion $\tau$
(you'll need to determine the speed). Is the torsion positive
or negative. Ask me in class to show you how you would be able
to determine this physically (without any computations).

Consider the helix $r(t)=(4\sin t, 4\cos t, 3t)$. Use a
computer to find $\vec T$, $\vec N$, $\vec B$, $\vec \kappa$,
and $\vec \tau$. State your answers. (This sage link will help.) Use your
answers to then give $\kappa$ and $\tau$. (When you present on
the board, just write down the 5 vectors, and then explain how
you obtained $\kappa$ and $\tau$ from these vectors. If you
follow the link, this is mostly already done for you. )

In the examples above, you should have noticed that $\vec
\tau$ was either parallel to $\vec N$ or anti-parallel to $\vec
N$. Let's now show this is always the case. The key is to use the
product rule on the cross product, together with some key fact
about the cross product.
What is the cross product of $(1,2,3)$ and $(2,4,6)$? If two
vectors are parallel, then what is their cross product? In
particular, what is the cross product of $\vec N$ and $\vec
\kappa$? See \footnote{The cross product of parallel vectors is
always the zero vector $(0,0,0)$. This is because the area of
the parallelogram formed using the parallel vectors is always
zero. So all three answers are $(0,0,0)$.} for an answer.

Watch a
YouTube Video.

% Suppose a curve $\vec r(t)$ has the frame $\vec T(t)$,
$\vec N(t)$, and $\vec B(t)$. Prove that $\dfrac{d\vec B}{ds}$
is either parallel to $\vec N$, or points opposite $\vec N$.
Here are some steps.
- Why is $\dfrac{d\vec B}{ds}$ orthogonal to $\vec B$? [Hint: How long is $\vec B$? See Theorem TODO.]
- We know $\vec B=\vec T\times \vec N$. Compute the derivative of both sides using the product rule and explain why $\frac{d\vec T}{ds}\times \vec N$ cancels out. Then explain why $\dfrac{d\vec B}{ds}$ is orthogonal to $\vec T$.
- If $\dfrac{d\vec B}{ds}$ is orthogonal to both $\vec B$ and $\vec T$ why must it be either parallel or anti-parallel to $\vec N$?