13.3 Arc Length and Speed

In Section 11.2, we derived a formula for the arc length of a plane curve given in parametric form. This discussion applies to paths in three-space with only minor changes.

Recall that arc length is defined as the limit of the lengths of polygonal approximations. To produce a polygonal approximation to a path \[ {\bf r}(t) = \langle x(t),y(t),z(t)\rangle,\qquad\quad a\le t \le b \] we choose a partition \(a = t_0<t_1<t_2<\cdots <t_N=b\) and join the terminal points of thev vectors \({\bf r}(t_j)\) by segments, as in Figure 13.15. As in Section 11.2, we find that if \({\bf r}'(t)\) exists and is continuous on \([a,b]\), then the lengths of the polygonal approximations approach a limit \(L\) as the maximum of the widths \(|t_j-t_{j-1}|\) tends to zero. This limit is the length \(s\) of the path which is computed by the integral in the next theorem.

REMINDER

The length of a path or curve is referred to as the arc length.

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Figure 13.15: Polygonal approximation to the arc r(t) for \(a≤t≤b\).

THEOREM 1 Length of a Path

Assume that \({\bf r}(t)\) is differentiable and that \({\bf r}'(t)\) is continuous on \([a,b]\). Then the length \(s\) of the path \({\bf r}(t)\) for \(a\le t\le b\) is equal to \[ \boxed{\bbox[#FAF8ED,5pt]{{s = \int_a^b\|{{\bf r}'(t)}\|\, dt = \int_a^b \sqrt{x'(t)^2+y'(t)^2+z'(t)^2}\, dt}}}\tag{1} \]

Note

Keep in mind that the length \(s\) in Eq. (1) is the distance traveled by a particle following the path \({\bf r}(t)\). The path length \(s\) is not equal to the length of the underlying curve unless \({\bf r}(t)\) traverses the curve only once without reversing direction.

EXAMPLE 1

Find the arc length \(s\) of \(\displaystyle{{\bf r}(t) = \langle\cos 3t, \sin 3t, 3t\rangle}\) for \(0\le t \le 2\pi\).

Solution  The derivative is \({\bf r}'(t) = \langle-3\sin 3t, 3\cos 3t, 3\rangle\), and \[ \|{{\bf r}'(t)}\|^2 = 9\sin^23t+9\cos^23t+9=9(\sin^23t+\cos^23t)+9=18 \] Therefore, \(\displaystyle{s = \int_0^{2\pi}\|{{\bf r}'(t)}\|\,dt = \int_0^{2\pi}\sqrt{18}\,dt = 6\sqrt 2\pi}\).

Speed, by definition, is the rate of change of distance traveled with respect to time \(t\). To calculate the speed, we define the arc length function: \[ \boxed{\bbox[#FAF8ED,5pt]{ s(t) = \int_a^t\|{{\bf r}'(u)}\|\, du}} \]

Thus \(s(t)\) is the distance traveled during the time interval \([a,t]\). By the Fundamental Theorem of Calculus, \[ \boxed{\bbox[#FAF8ED,5pt]{ \textrm{Speed at time \(t\)} = \frac{s}{t} = \|{\bf r}'(t)\|}} \]

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Now we can see why \({\bf r}'(t)\) is known as the velocity vector (and also as the tangent vector). It points in the direction of motion, and its magnitude is the speed (Figure 13.16). We often denote the velocity vector by \({\bf v}(t)\) and the speed by \(v(t)\): \[ {\bf v}(t) = {\bf r}'(t),\qquad v(t) = \|{\bf v}(t)\| \]

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Figure 13.16: The velocity vector is longer at \(t_0\) than at \(t_1\), indicating that the particle is moving faster at \(t_0\).

EXAMPLE 2

Find the speed at time \(t=2\) s of a particle whose position vector is \[ {\bf r}(t) = t^3{\bf i}-e^t{\bf j}+4t{\bf k} \]

Solution  The velocity vector is \(\displaystyle{{\bf v}(t) = {\bf r}'(t)= 3t^2{\bf i}-e^t{\bf j}+4{\bf k}}\), and at \(t = 2\), \[ {\bf v}(2) = 12{\bf i}-e^2{\bf j}+4{\bf k} \]

The particle’s speed is \(v(2) = \|{{\bf v}(2)}\| = \sqrt{12^2+(-e^2)^2+4^2}\approx 14.65 \text{ft/s}\).

Arc Length Parametrization

We have seen that parametrizations are not unique. For example, \({\bf r}_1(t)=\langle t,t^2\rangle\) and \({\bf r}_2(s)=\langle s^3,s^6\rangle\) both parametrize the parabola \(y=x^2\). Notice in this case that \({\bf r}_2(s)\) is obtained by substituting \(t=s^3\) in \({\bf r}_1(t)\).

Note

Keep in mind that a parametrization \({\bf r}(t)\) describes not just a curve, but also how a particle traverses the curve, possibly speeding up, slowing down, or reversing direction along the way. Changing the parametrization amounts to describing a different way of traversing the sameunderlying curve.

In general, we obtain a new parametrization by making a substitution \(t = g(s)\)—that is, by replacing \({\bf r}(t)\) with \({\bf r}_1(s) = {\bf r}(g(s))\) Figure 13.17. If \(t=g(s)\) increases from \(a\) to \(b\) as \(s\) varies from \(c\) to \(d\), then the path \({\bf r}(t)\) for \(a\le t \le b\) is also parametrized by \({\bf r}_1(s)\) for \(c\le s \le d\).

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Figure 13.17: The path is parametrized by \({\bf r}(t)\) and by \({\bf r}_1(s)={\bf r}(g(s))\).

EXAMPLE 3

Parametrize the path \({\bf r}(t)=(t^2, \sin t, t)\) for \(3\le t \le 9\) using the parameter \(s\), where \(t = g(s)=e^s\).

Solution Substituting \(t=e^s\) in \({\bf r}(t)\), we obtain the parametrization \[ {\bf r}_1(s) = {\bf r}(g(s)) = \langle e^{2s}, \sin e^s, e^s\rangle \]

Because \(s = \ln t\), the parameter \(t\) varies from \(3\) to \(9\) as \(s\) varies from \(\ln 3\) to \(\ln 9\). Therefore, the path is parametrized by \({\bf r}_1(s)\) for \(\ln 3 \le s\le \ln 9\).

One way of parametrizing a path is to choose a starting point and “walk along the path” at unit speed (say, \(1\) m/s). A parametrization of this type is called an arc length parametrization [Figure 13.18(A)]. It is defined by the property that the speed has constant value \(1\): \[ \|{{\bf r}'(t)}\|=1\qquad\textrm{for all \(t\) \]

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In an arc length parametrization,the distance traveled over any time interval \([a,b]\)is equal to the length of the interval: \[ \textrm{Distance traveled over \([a,b]\) } = \int_a^b\|{{\bf r}'(t)}\|\, dt =\int_a^b \,1\, dt = b-a \]

Note

Arc length parametrizations are also called unit speed parametrizations. We will use arc length parametrizations to define curvature in Section 13.4.

To find an arc length parametrization, start with any parametrization \({\bf r}(t)\) such that \({\bf r}'(t)\ne {\bf 0}\) for all \(t\), and form the arc length integral \[ s(t) = \int_0^t\|{{\bf r}'(u)}\|\,du \]

Note

The letter \(s\) is often used as the parameter in an arc length parametrization.

Because \(\|{{\bf r}'(t)}\|\ne 0\), \(s(t)\) is an increasing function and therefore has an inverse \(t = g(s)\). By the formula for the derivative of an inverse (and since \(s'(t) = \|{{\bf r}'(t)}\|\)), \[ g'(s) = \frac1{s'(g(s))} = \frac1{\|{{\bf r}'(g(s))}}\| \]

REMINDER

By Theorem 1 in Section 3.8, if \(g(x)\) is the inverse of \(f(x)\), then \[ g'(x) = \frac1{f'(g(x))} \]

Now we can show that the parametrization \[ {\bf r}_1(s) = {\bf r}(g(s)) \] is an arc length parametrization. Indeed, by the Chain Rule, \[ \|{{\bf r}_1'(s)}\|=\|{{\bf r}'(g(s))g'(s)}\| =\|{{\bf r}'(g(s))}\|\, \frac1{\|{{\bf r}'(g(s))}}\| = 1 \]

In most cases we cannot evaluate the arc length integral \(s(t)\) explicitly, and we cannot find a formula for its inverse \(g(s)\) either. So althougharc length parametrizations exist in general, we can find them explicitly only in special cases.

EXAMPLE 4 Finding an Arc Length Parametrization

Find the arc length parametrization of the helix \({\bf r}(t) = \langle \cos 4t, \sin 4t, 3t\rangle\).

Solution  First, we evaluate the arc length function \[ \begin{align*} \|{{\bf r}'(t)}\|&= \|{\langle -4\sin 4t, 4\cos t, 3\rangle}\| = \sqrt{16\sin^2 4t+16\cos^2 4t+3^2}= 5 \\ s(t) &= \int_0^t\|{{\bf r}'(t)}\|\, dt = \int_0^t 5\, dt= 5t \end{align*} \]

Then we observe that the inverse of \(s(t)=5t\) is \(\displaystyle{t = {s}/5}\); that is, \(\displaystyle{g(s) = {s}/5}\). As shown above, an arc length parametrization is \[ {\bf r}_1(s) = {\bf r}(g(s)) = {\bf r}\left(\frac{s}5\right) = \langle \cos \frac{4s}5, \sin \frac{4s}5, \frac{3s}5\rangle \]

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As a check, let’s verify that \({\bf r}_1(s)\) has unit speed: \[ \|{{\bf r}_1'(s)}\| = \left\Vert \langle -\frac45\sin \frac{4s}5, \frac45\cos \frac{4s}5, \frac{3}5\rangle \right\Vert = \sqrt{\frac{16}{25}\sin^2\frac{4s}{5} +\frac{16}{25}\cos^2\frac{4s}{5}+\frac9{25}}=1 \]

13.3.1 Summary