Speed and arc length

Now to distance traveled. Intuitively, since the speed $v(t)$ of a moving object is the length of its velocity vector, the distance the object travels from time $t=a$ to time $t=b$ should be the integral of $\|{\bf r}'(t)\|$ over the time interval $[a,\, b]$. Equivalently, this will be the arc length of the curve parametrized by ${\bf r}(t), \, a \le t \le b\,.$ This is the same formula that we derived for plane curves, only now $\| {\bf r}'(t)\| = \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}$ instead of $\sqrt{(x'(t))^2 + (y'(t))^2}$

It's important to remember that an integral is more than an anti-derivative. An integral is a limit of a sum, and our strategy for computing bulk quantities is always the same. (1) Break the sum into pieces. (2) Estimate the size of each piece. (3) Add up the pieces. (4) Take a limit to get an integral. So let $C$ be the curve shown in blue to the right below and parametrized by ${\bf r} : [a,\,b] \to {\mathbb R}^3$.

Partition $[a,\,b]$ by $$a=t_0 < t_1 < t_2 < \ldots t_{k-1} < t_k \, < \, \ldots \, t_n=b$$ witn $\Delta t = (b-a)/n$ and $t_k = a+k\Delta t$, and use this to decompose $C$ into consecutive arcs $$C_1,\, C_2,\, \ldots \,C_k,\, \ldots,\, C_n$$ parametrized by $$C_k \,=\, \{ {\bf r} (t) : t_{k-1} \le t \le t_k\}\,.$$ The total length of $C$ is the sum of the lengths of the $C_k$'s. Each $C_k$ can be approximated by the red vector ${\bf r}(t_k) - {\bf r}(t_{k-1})$ so that $$\hbox{length} (C_k) \approx \|{\bf r}(t_k) - {\bf r}(t_{k-1})\|\,.$$

Now choose an (orange) sample point $\ {\bf r}(t_k^*)$ in $C_k$, so that ${\bf r}(t_k) - {\bf r}(t_{k-1}) \approx {\bf r}'(t_k^*) \Delta t$.

Adding up the pieces gives the Riemann Sum $$\sum_{k\,=\,1}^n\, \|{\bf r}'(t_k^*)\| \Delta k$$ as an approximation to to the length of $C$ between ${\bf r}(a)$ and ${\bf r}(b)$. Taking a limit as $n \to \infty$ gives the integral $$\lim_{n\,\to\, \infty}\sum_{k\,=\,1}^n\, \|{\bf r}'(t_k^*)\| \Delta t \,=\,\int_a^b\, \|{\bf r}'(t)\|\, dt\,.$$ The arc length up to time $t$ is usually denoted $s$, so we have $s = \int \|{\bf r}'(t) \| dt$ and the total length is often written $\displaystyle \int_C\, ds\,$. Formally,

For a space curve $C$ parametrized by ${\bf r} (t),\, a \le t \le b,$ its $$\hbox{arc length}\ \int_C\, ds \ = \ \int_a^b\,\big \|{\bf r}'(t)\big\|\, dt\,.$$ When ${\bf r} (t)$ is the position of an object moving in 3-space, then this integral expression also gives the distance traveled by the object from time $t=a$ to time $t = b$.

Example 2: Find the integral that represents the length of the graph shown in of the vector function $${\bf r}(t) \ = \ \langle\,4 \cos^{3}t,\ 4 \sin^{3}t\, \rangle\,.$$

Solution: when $${\bf r}(t) \ = \ \langle\,4 \cos^{3}t,\ 4 \sin^{3}t\, \rangle\,,$$ then $${\bf r}'(t) \ = \ \langle\,-12\sin t \cos^{2}t,\ 12 \cos t \sin^{2}t\, \rangle\,,$$ and so $$\|{\bf r}'(t)\|\ = \ 12 \sqrt{\cos^2 t \sin^2 t (\cos^2 t + \sin^2 t)}\,.$$ $$\qquad \qquad = \ 12|\cos t \sin t |\,.$$ The graph of ${\bf r}(t)$ starts at $(4,\,0)$ and returns to $(4,\,0)$ when $t = 2\pi$. Thus $$\hbox{length}\ = \ 12 \int_0^{2\pi}\,|\cos t \sin t |\,dt \,.$$

Example 3: find the distance traveled over the time interval $[0, \ln 2]$ for a particle whose position function is $${\bf r}(t) \ = \ \big\langle \sqrt{2}\,t,\, e^t\,,e^{-t}\big\rangle\,.$$ Solution: the distance traveled by the particle between $t=a$ and $t= b$ is given by $$I \ = \ \int_a^b\,\| {\bf r}'(t)\|\,dt\,.$$

Now ${\bf r}'(t) = \big\langle \sqrt{2},\, e^t\,,-e^{-t}\big\rangle\,,$ so $$\|{\bf r}'(t)\| = \sqrt{2 + e^{2t}+e^{-2t}}=\sqrt{((e^t+e^{-t})^2} \,.$$ Thus $$I\ = \ \int_0^{\ln 2}\, (e^t + e^{-t})\, dt\hskip0.5in$$ $$\hskip0.5in = \ \Bigl[ e^t - e^{-t }\Bigl]_0^{\ln 2} \ = \ \frac{3}{2}\,.$$