12.3 Arc Length
Definition of Arc Length
What is the length of a path \({\bf c}(t)\)? Because the speed \(\|{\bf c}'(t)\|\) is the rate of change of distance traveled with respect to time, the distance traveled by a point moving along the curve should be the integral of speed with respect to the time over the interval \([t_0,t_1]\) of travel time; that is, the length of the path, also called its arc length, is \[ L({\bf c})= \int^{t_1}_{t_0} \|{\bf c}'(t)\|\ {\it dt}. \]
There is the question as to whether or not this formula actually corresponds to the true arc length. For example, suppose we take a curve in space and glue a string tightly to it, cutting the string so it exactly fits the curve. If we then remove the string, straighten it out, and measure it with a straight edge, we surely should obtain the length of the curve.
To see a justification that our formula for arc length agrees with such a process, click here.
In \({\mathbb R}^3\) there is another way to justify the arc-length formula based on polygonal approximations. We partition the interval \([a,b]\) into \(N\) subintervals of equal length: \begin{eqnarray*} &\displaystyle a = t_0 < t_1 < \cdots < t_N =b; &\\[3pt] &\displaystyle t_{i+1} - t_i = \frac{b-a}{N} \quad {\rm for} \quad 0 \le i \le N-1.& \end{eqnarray*}
We then consider the polygonal line obtained by joining the successive pairs of points \({\bf c} (t_i), {\bf c} (t_{i+1})\) for \(0 \le i \le N-1\). This yields a polygonal approximation to \({\bf c}\) as in Figure 12.19. By the formula for distance in \({\mathbb R}^3\), it follows that the line segment from \({\bf c} (t_i)\) to \({\bf c} ( t_{i+1})\) has length \[ \| {\bf c} (t_{i+1}) - {\bf c} (t_i) \| = \sqrt{[x (t_{i+1}) - x (t_i) ]^2+ [ y ( t_{i+1}) - y (t_i) ]^2 + [z (t_{i+1}) - z (t_i) ]^2}, \] where \({\bf c} (t) = ( x (t), y (t),z (t))\). Applying the mean-value theorem to \(x (t), y (t)\), and \(z (t)\) on \([t_i, t_{i+1}]\), we obtain three points \(t_i^*, t^{**}_i,\) and \(t_i^{***}\) such that \begin{eqnarray*} x(t_{i+1}) - x (t_i) = x' ( t_i^*) ( t_{i+1} - t_i), \\[3pt] y ( t_{i+1}) - y (t_i) = y' ( t_i^{**}) ( t_{i+1} - t_i) , \end{eqnarray*}
and \[ z (t_{i+1}) -z (t_i)= z' (t_i^{***}) (t_{i+1} -t_i). \]
Thus, the line segment from \({\bf c}(t_i)\) to \({\bf c}(t_{i+1})\) has length \[ \sqrt{[x'(t_i^*) ]^2 + [y' (t_i^{**})]^2+ [z' ( t_i^{***}) ]^2} ( t_{i+1} - t_i). \]
Therefore, the length of our approximating polygonal line is \[ S_N = \sum_{i=0}^{N-1} \sqrt{[x'(t_i^*) ]^2 + [y' (t_i^{**})]^2+ [z' ( t_i^{***}) ]^2} ( t_{i+1} - t_i). \]
As \(N \to \infty\), this polygonal line approximates the image of \({\bf c}\) more closely. Therefore, we define the arc length of \({\bf c}\) as the limit, if it exists, of the sequence \(S_N\) as \(N \to \infty\). Because the derivatives \(x', y'\), and \(z'\) are all assumed to be continuous on \([a,b]\), we can conclude that, in fact, the limit does exist and is given by \[ {\mathop {\rm limit}_{N \to \infty}} S_N = \int_a^b \sqrt{[x'(t) ]^2 + [y' (t)]^2+ [z' (t)]^2}\, {\it dt}. \]
(The theory of integration relates the integral to sums by the formula \[ \int_a^b f (t) \,{\it dt} = {\mathop {\rm limit}_{N \to \infty}} \sum_{i=0}^{N-1} f (t_i^*) ( t_{i+1} - t_i), \] where \(t_0, \ldots , t_N\) is a partition of \([a,b], t_i^* \in [t_i,t_{i+1}]\) is arbitrary, and \(f\) is a continuous function. Here we have possibly different points \(t_i^*, t_i^{**}\), and \(t_i^{***}\), and so this formula must be extended slightly.)
Arc Length
The length of the path \({\bf c}(t)=(x(t),y(t),z(t))\) for \(t_0\leq t\leq t_1\), is \[ L({\bf c})=\int^{t_1}_{t_0}\sqrt{[x'(t)]^2+[y'(t)]^2+[z'(t)]^2}\,{\it dt}. \]
If \({\bf c}(t)\) represents the position of a particle at time \(t\), then \(L({\bf c})\) above represents "the total distance traveled" by the particle between \(t=t_0\) and \(t=t_1\).
example 1
The arc length of the path \({\bf c}(t)=(r\,\cos t,r \,\sin t)\), for \(t\) lying in the interval \([0, 2\pi]\); that is, for \(0\leq t\leq 2\pi\), is \[ L({\bf c})=\int^{2\pi}_0\sqrt{(-r\,\sin t)^2+(r\,\cos t)^2}\, {\it dt}=2\pi r, \] which is the circumference of a circle of radius \(r\). If we had allowed \(0\leq t\leq 4\pi\), we would have obtained \(4\pi r\), because the path traverses the same circle twice (Figure 12.20).
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For planar curves, we omit the \(z'(t)\) term, as in Example 1.
example 2
Consider the point with position function \[ {\bf c}(t)=(t-\,\sin t,1-\,\cos t), \] which traces out the cycloid discussed in Section 12.1, (see Figure 12.6). Find the velocity, the speed, and the length of one arch.
solution The velocity vector is \({\bf c}'(t)=(1-\,\cos t,\,\sin t)\), so the speed of the point \({\bf c}(t)\) is \[ \|{\bf c}'(t)\|={\sqrt{(1-\,\cos t)^2+\,\sin^2t}}=\sqrt{2-2\,\cos t}. \]
Hence, \({\bf c}(t)\) moves at variable speed although the circle rolls at constant speed. Furthermore, the speed of \({\bf c}(t)\) is zero when \(t\) is an integral multiple of \(2\pi\). At these values of \(t\), the \(y\) coordinate of the point \({\bf c}(t)\) is zero, and so the point lies on the \(x\) axis. The arc length of one cycle is \begin{eqnarray*} L({\bf c}) &=& \int^{2\pi}_0\sqrt{2-2\,\cos t}\,{\it dt}=2\int^{2\pi}_0\sqrt{\frac{1-\,\cos t}{2}}\,{\it dt}\\[6pt] &=& 2\int^{2\pi}_0\,\sin \frac{t}{2}\,{\it dt}\,\bigg(\hbox{because } 1-\,\cos t=2\,\sin^2 \frac{t}{2} \hbox{ and }\sin \frac{t}{2}\geq 0 \hbox{ on } [0,2\pi]\bigg)\\[6pt] &=& 4\bigg(-\cos \frac{t}{2}\bigg)\bigg|^{2\pi}_{0}=8.\\[-27pt] \end{eqnarray*}
If a path \({\bf c}(t)\) is differentiable on a closed interval and its derivative \({\bf c}'(t)\) is continous on that interval, we say that \({\bf c}(t)\) is \(C^1\). If a curve is made up of a finite number of pieces, each of which is \(C^1\) (with bounded derivative), we compute the arc length by adding the lengths of the component pieces. Such curves are called piecewise \(C^1\). Sometimes we just say “piecewise smooth.”
example 3
A billiard ball on a pool table follows the path \({\bf c}\colon\, [-1,1]\rightarrow {\mathbb R}^2\), defined by \({\bf c}(t)=(x(t),y(t))= (|t|, |t-\frac{1}{2}|)\). Find the distance traveled by the ball.
solution This path is not smooth, because \(x(t)=|t|\) is not differentiable at 0, nor is \(y(t)= |t-\frac{1}{2} |\) differentiable at \(\frac{1}{2}\). However, if we divide the interval \([-1,1]\) into the pieces \([-1,0],[0,\frac{1}{2}]\), and \([\frac{1}{2},1 ]\), we see that \(x(t)\) and \(y(t)\) have continuous derivatives on each of the intervals \([-1,0], [0,\frac{1}{2} ]\), and \([\frac{1}{2},1]\). (See Figure 12.21.)
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On \([-1,0],x(t)=-t,y(t)=-t+\frac{1}{2}\), so \(\|{\bf c}'(t)\|=\sqrt{2}\). Hence, the arc length of \({\bf c}\) between \(-1\) and 0 is \(\int^0_{-1}\sqrt{2}\,{\it dt}=\sqrt{2}\). Similarly, on \( [0,\frac{1}{2} ],x(t)=t, y(t)=-t+\frac{1}{2}\), and again \(\|{\bf c}'(t)\|=\sqrt{2}\), so that the arc length of \({\bf c}\) between 0 and \(\frac{1}{2}\) is \(\frac{1}{2}\sqrt{2}\). Finally, on \([\frac{1}{2},1]\) we have \(x(t)=t,y(t)=t-\frac{1}{2},\) and the arc length of \({\bf c}\) between \(\frac{1}{2}\) and 1 is \(\frac{1}{2}\sqrt{2}\). Thus, the total arc length of \({\bf c}\) is \(2\sqrt{2}\). Of course, we can also compute the answer as the sum of the distances from \({\bf c}(-1)\, {\rm to}\, {\bf c}(0) \,{\rm to}\, {\bf c}(\frac{1}{2})\) to \({\bf c}(1)\).
Here is an example in \({\mathbb R}^3\).
example 4
Find the arc length of \((\cos t,\,\sin t, t^2),0\leq t\leq \pi\).
solution The path \({\bf c}(t)=(\cos t,\,\sin t,t^2)\) has the velocity vector given by \({\bf v}=({-}{\sin t}, \,\cos t,2t)\). Because \[ \|{\bf v}\|=\sqrt{\,\sin^2 t+\,\cos^2 t+4t^2}=\sqrt{1+4t^2}=2\sqrt{t^2+\Big( \frac{1}{2}\Big)^2}, \] the arc length is \[ L({\bf c})=\int^\pi_02\sqrt{t^2+\bigg(\frac{1}{2}\bigg)^2}\,{\it dt}. \]
This integral may be evaluated using the following formula from the table of integrals: \[ \int \sqrt{x^2+a^2}{\,d} x=\frac{1}{2} \big[ x\sqrt{x^2+a^2}+a^2\,\log\, (x+\sqrt{x^2+a^2}) \big]+C. \]
Thus, \begin{eqnarray*} L({\bf c}) &=& 2\,{\bf \cdot}\, \frac{1}{2}\left[ t\sqrt{t^2+\bigg(\frac{1}{2}\bigg)^2}+ \bigg(\frac{1}{2}\bigg)^2 \log\left(t+\sqrt{t^2+\bigg(\frac{1}{2}\bigg)^2}\right)\right] \Bigg|^\pi_{t=0}\\[4pt] &=&\pi\sqrt{\pi^2+\frac{1}{4}}+\frac{1}{4}\log \bigg(\pi+\sqrt{\pi^2+\frac{1}{4}}\bigg)-\frac{1}{4} \log\bigg(\sqrt{\frac{1}{4}}\bigg)\\[4pt] &=&\frac{\pi}{2}\sqrt{1+4\pi^2}+\frac{1}{4}\log\, (2\pi+ \sqrt{1+4\pi^2})\approx10.63. \\[-16pt] \end{eqnarray*}
As a check on our answer, we may note that the path \({\bf c}\) connects the points (1, 0, 0) and \((-1,0,\pi^2)\). The distance between these points is \(\sqrt{4+\pi^2}\approx 3.72\), which is less than 10.63, as it should be.
Question 12.49 Section 12.3 Progress Check Question 1
ah3u0YOeWFCGMsk7/pact6Y4+bpTKls40A2hgQFUJ2o6XzM3RxWH5D02p+7T01MWeq4bW21iEf8hwSKXgqPaaqEzj3BqSvMNfS/irklj0o0nbebAYzoL1BQ1SBztl9uMJBldAUECHtlWhczRuS+Ryb4WPCiLGmpSYWGgfF7TOKbRIMjr1WE3m1Vtdw2/w0RFQi0i4DBsxkrQ/MFkMUSLHsMM8fqCqv2Yki/TEg/nh94yWwqLnHoaDHjtAVpQxbG9AVRoBBL7wtr/NUZ4pWN99YhyOsQ=The Differential of Arc Length
The arc-length formula suggests that we introduce the following notation, which will be useful in our discussion of line integrals.
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Arc-Length Differential
An infinitesimal displacement of a particle following a path \({\bf c}(t)=x(t){\bf i}+y(t){\bf j}+z(t){\bf k}\,\) is \[ d{\bf s}={\it dx}\, {\bf i}+{\it dy}\, {\bf j}+{\it dz}\, {\bf k}=\bigg(\frac{{\it dx}}{{\it dt}}{\bf i}+\frac{{\it dy}}{{\it dt}}{\bf j}+\frac{{\it dz}}{{\it dt}}{\bf k}\bigg)\,{\it dt}, \] and its length \[ ds={\textstyle\sqrt{{\it dx}^2+{\it dy}^2+{\it dz}^2}}=\sqrt{\bigg(\frac{{\it dx}}{{\it dt}}\bigg)^2+ \bigg(\frac{{\it dy}}{{\it dt}}\bigg)^2+\bigg(\frac{{\it dz}}{{\it dt}}\bigg)^2}\,{\it dt} \] is the differential of arc length. See Figure 12.22.
These formulas help us remember the arc-length formula as \[ \hbox{arc length}=\int^{t_1}_{t_0}{\,d} s. \]
As we have done before with such geometric concepts as length and angle, we can extend the notion of arc length to paths in \(n\)-dimensional space.
Arc Length in \({\mathbb R}^n\)
Let \({\bf c}\colon\, [t_0,t_1]\rightarrow {\mathbb R}^n\) be a piecewise \(C^1\) path. Its length is defined to be \[ L({\bf c})=\int^{t_1}_{t_0}\|{\bf c}'(t)\|\,{\it dt}. \]
The integrand is the square root of the sum of the squares of the coordinate functions of \({\bf c}'(t)\): If \[ {\bf c}(t)=(x_1(t),x_2(t),\ldots,x_n(t)), \] then \[ L({\bf c})=\int^{t_1}_{t_0}{\sqrt{(x_1'(t))^2+(x_2'(t))^2 +\cdots +(x_n'(t))^2}}\,{\it dt}. \]
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example 5
Find the length of the path \({\bf c}(t)=(\,\cos t,\,\sin t,\,\cos 2t,\sin 2t)\) in \({\mathbb R}^4\), defined on the interval from 0 to \(\pi\).
solution We have \({\bf c}'(t)=({-}{\sin t},\,\cos t,-2\,\sin 2t,2\,\cos 2t)\), and so \[ \|{\bf c}'(t)\|={\sqrt{\,\sin^2 t+\,\cos^2 t+4\,\sin^2 2t +4\,\cos^2 2t}}={\sqrt{1+4}}={\textstyle\sqrt{5}}, \] a constant, so the length of the path is \[ \int^\pi_0\sqrt{5}\,{\it dt}=\sqrt{5}\pi. \]
It is common to introduce the arc-length function \(s (t)\) associated to a path \({\bf c}(t)\) given by \[ s (t) = \int_a^t \| {\bf c}' ( u) \|\, d u, \] so that (by the fundamental theorem of calculus) \[ s' (t)= \| {\bf c}' (t) \| \] and \[ \int_a^b s' (t)\, {\it dt} = s (b) - s( a) = s(b). \]
example 6
Consider the graph of a function of one variable \(y=f(x)\) for \(x\) in the interval [\(a, b\)]. We can consider it to be a curve parametrized by \(t = x\), namely, \({\bf c}(x)=(x, f(x))\) for \(x\) ranging from \(a\) to \(b\). The arc-length formula gives \[ L({\bf c})=\int^b_a \sqrt{1+[f'(x)]^2} \, {\it dx}, \] which agrees with the formula for the length of a graph from one-variable calculus.