Objectives
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Describe velocity and acceleration as derivatives of a vector-valued position function.
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Express arc length as the integral of the magnitude of a vector-valued function.
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Re-parametrize curves in terms of their arc length.
Section 13.2 Motion in space
Introduction goes here.
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We’ve started with 2D, but point out how the vector notation makes it work for any dimension
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Equation of a line in 3D:\begin{equation*} \bv{r}(t)=\bv{r}_{0}+\bv{v}_{0}t=(x_{0}+v_{0x}t,y_{0}+v_{0y}t,z_{0}+v_{0z}t) \end{equation*}
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Length / norm of a vector is related to distance and Pythagorean theorem:\begin{equation*} \norm{(a,b,c)}=\sqrt{a^{2}+b^{2}+c^{2}} \end{equation*}
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Arc length:\begin{align*} s \amp = \int_a^b \norm{\bv{r}'(t)}\dd{t} \\ \amp = \int_a^b \sqrt{\qty(\dv{r_x}{t})^{2}+\qty(\dv{r_y}{t})^{2}+\qty(\dv{r_z}{t})^{2}}\dd{t} \end{align*}
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If we define \(s(t)=\int_{a}^{t}\norm{\bv{r}'(u)}\dd{u}\text{,}\) then \(s(t)\) is a one-to-one increasing function so it has an inverse. Hence if we solve for \(t\) as a function of \(s\text{,}\) i.e. \(t(s)\text{,}\) then \(\bar{\bv{r}}(s)=\bv{r}(t(s))\) is parametrized by arc length. This is called the natural parametrization of the curve.
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If we parametrize by arc length, then \(\norm{\bar{\bv{r}}'(s)}=1\text{.}\)
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Difficult to do in practice, but often useful for theoretical arguments, and can be turned back into parametrization by \(t\text{.}\)
