Drew Youngren dcy2@columbia.edu
Suppose a particle starts at the origin with initial velocity $\left\langle 1, 0 \right\rangle$ and acceleration \[ \vec a(t) = \begin{cases} \left\langle -t, 1 \right\rangle, \text{ if } t \leq 1 \\ \left\langle 1, -t-1 \right\rangle, \text{ if } t > 1 \\ \end{cases}. \] Find its position at $t = 2$.
\[ \vec v(t) = \begin{cases} \left\langle -\frac12 t^2 + 1, t \right\rangle, \text{ if } t \leq 1 \\ \left\langle t - \frac12, -t - \frac12 t^2 + \frac52 \right\rangle, \text{ if } t > 1 \\ \end{cases}. \]
\[ \vec x(t) = \begin{cases} \left\langle -\frac16 t^3 + t, \frac12 t^2 \right\rangle, \text{ if } t \leq 1 \\ \left\langle \frac12 t^2 - \frac12 t + \frac56, \frac12 t^2 - \frac16 t^3 + \frac52 t - \frac43 \right\rangle, \text{ if } t > 1 \\ \end{cases}. \]
A parametrized curve $\vec r(t)$ is called smooth provided \[\vec r'(t) \neq \vec 0\] for all $t$. A smooth curve has a well-defined unit tangent vector \[\vec T(t) = \frac{\vec r'(t)}{|\vec r'(t)|}\]
A counterexample might express this more easily. \[ \vec r(t) = \left\langle t^2, t^3 \right\rangle \]
To measure the arc length of a path in space, we would like to just "straighten it out" (without stretching) and use the distance formula, but this is difficult, mathematically.
Instead, measure distances between sample points along the curve.
For $\vec r(t)$ with $a \leq t \leq b$, we define $t_i = a + i \frac{b - a}{N}$ and then arc length of the curve is
\[ s = \lim\limits_{N\to\infty}\sum_{i = 1}^N |\vec r(t_i) - \vec r(t_{i - 1})|\]
For $\vec r(t)$ with $a \leq t \leq b$, we define $t_i = a + i \frac{b - a}{N}$ and then arc length of the curve is
\[ s = \lim\limits_{N\to\infty}\sum_{i = 1}^N \frac{|\vec r(t_i) - \vec r(t_{i - 1})|}{\Delta t} \Delta t \]
Find the arc length of one coil of a helix. \[\vec r(t) =\left\langle \cos t, \sin t, t \right\rangle \]
\[s = \int_0^{2\pi} |\left\langle -\sin t, \cos t, 1 \right\rangle|\,dt \] \[ = \int_0^{2\pi} \sqrt{\sin^2 t + \cos^2 t + 1}\,dt = \int_0^{2\pi} \sqrt2\,dt = 2\sqrt 2 \pi \]
$\langle \cos t, \sin t\rangle$ traces out the unit circle over the interval $[0,2\pi]$.
$\langle \cos (6\pi t), \sin (6\pi t)\rangle$ traces out the unit circle much "faster", over the interval $[0,\frac13]$. Take the derivatives of each of the above to see this.
$\langle \cos (\pi \sin t), \sin (\pi \sin t)\rangle$ traces out the unit circle "back and forth."
Consider a smooth curve $\vec r(t)$ for $a \leq t \leq b$. A smooth reparametrization is a choice of scalar function $f(s)$ with $f'(s) > 0$, $f(c) = a$, and $f(d) = b$ such that \[\vec q(s) = \vec r(f(s))\] traces out the same path.
Arc length is independent of parametrization. Using the notation of previous slide,
\[ \int_c^d \left| \vec q '(s) \right|\,ds = \int_c^d \left| \frac{d}{ds} \vec r (f(s)) \right|\,ds \] \[ = \int_c^d \left| \vec r '(f(s)) \right| f'(s)\,ds \qquad \begin{cases} w =f(s) \\ dw = f'(s)\,ds \\ \end{cases} \] \[ = \int_a^b |\vec r'(t) |\,dt \]
\[ s = f(t) = \int_a^t |\vec r'(w)|\,dw. \]
\[ s = f(t) = \int_a^t |\vec r'(w)|\,dw. \]
\[\frac{d}{ds} \vec r(f^{-1}(s)) = \vec r '(f^{-1}(s)) (f^{-1})'(s) = \vec r'(t) \frac{1}{f'(f^{-1}(s))}\]
\[ = \frac{\vec r'(t)}{|\vec r'(t)|} = \vec T \]
Using the definition from the previous slide, a curve is paramtrized by arc length $ \vec q(s) $ for $0 \leq s \leq L$ if $|\vec q'(s)| = 1$.
Parametrize the curve \[ \langle e^{-t}, 1-2e^{-t}, 2e^{-t} \rangle \] for $t \geq 0$ by arc length.
We aim to quantify how curvy a curve can be.
Which of these paths is "curviest" at the origin?
Recall that $\vec T$ is the unit tangent vector to a curve $\vec r(t)$. The curvature $\kappa$ of $\vec r$ at a given point is given by \[ \kappa = \left|\frac{d\vec T}{ds} \right|. \]
That is, we measure how much the direction changes per unit arc length. Alternatively, \[\kappa = \frac{|\vec T'(t)|}{|\vec r'(t)|}\]
\[\kappa = \frac{|\vec T'(t)|}{|\vec r'(t)|}\]
Because \[\left| \vec T'(t) \right| = \left|\frac{d\vec T}{dt}\right| = \left|\frac{d\vec T}{ds}\frac{ds}{dt} \right|= \kappa |\vec r'(t)|. \]
Find the relation between the radius $R$ of a circle and its curvature.
$\vec r(t) = \left\langle R\cos t, R\sin t \right\rangle$ has arclength $s = Rt$. Thus we parametrize by arc length \[\vec q(s) = \left\langle R\cos \frac{s}{R}, R\sin \frac{s}{R} \right\rangle \]
\[\kappa = |\vec q''(s) |= \frac{1}{R} \]
Use a curve plot in 3Demos to explore a curve, its reparametrization by arc length, and the so-called osculating circle, which best approximates the curve at a point with a circle with a matching $\kappa$ value.