Length of a Curve

In this section, we want to find the length of the curve \(y=f(x)\) from \(x=a\) to \(x=b\). A piece of a curve that lies between two specific points is called an arc. Like the previous sections of this chapter, we construct a formula for the length of an arc by considering infinitesimals.

Let \(s\) be the length of the arc from the fixed point \((a,f(a))\) to a variable point \((x,f(x))\) as shown in Figure 1. Assume \(s\) increases by an infinitesimally small amount \(ds\), and \(x\) and \(y\) by \(dx\) and \(dy\), respectively. Because \(ds\) is so small, this part of the curve is virtually straight and by the Pythagorean theorem we can write \[ \bbox[#F2F2F2,5px,border:2px solid black]{ds=\sqrt{(dx)^{2}+(dy)^{2}}}\] or \[ds=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx.\] But \(dy/dx=f'(x).\) Therefore, \[ds=\sqrt{1+[f'(x)]^{2}}dx.\] As \(ds\) sweeps along the curve from \((a,f(a))\) to \((b,f(b))\), we add up all the infinitesimal lengths: \[\begin{aligned} \text{length of arc} & =\int_{a}^{b}ds=\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx\\ & =\int_{a}^{b}\sqrt{1+[f'(x)]^{2}}dx.\end{aligned}\]

Figure 1
Figure 1

 

If we want to compute the length of the curve \(x=h(y)\) between \((h(c),c)\) and \((h(d),d)\) (Figure 2), then we can still use \[ds=\sqrt{(dx)^{2}+(dy)^{2}}.\] Therefore, \[\begin{aligned} ds & =\sqrt{\left(\frac{dx}{dy}\right)^{2}+1}\ dy\\ & =\sqrt{[h'(y)]^{2}+1}\ dy\end{aligned}\] and \[\begin{aligned} \text{length of arc} & =\int_{c}^{d}ds=\int_{c}^{d}\sqrt{\left(\frac{dx}{dy}\right)^{2}+1}\ dy\\ & =\int_{c}^{d}\sqrt{[h'(y)]^{2}+1}\ dy\end{aligned}\]

Figure 2
  • In the above discussion we assumed that the curve is rectifiable; that is, the curve has a finite arc length. It is possible to provide an example of a continuous function \(y=f(x)\) (\(a\leq x\leq b\)) whose graph is not rectifiable.1 A sufficient condition for the curve \(y=f(x)\) to be rectifiable is that \(f’\) be bounded. That is, if there is some number \(K\) such that \(|f'(x)|<K\) for \(x\) between \(a\) and \(b\).
Example 1
Find the length of the curve \(y^{2}=x^{3}\) between the points \((0,0)\) and \((4,8)\).

Figure 3
Solution 1

Solving \(y^{2}=x^{3}\) for \(y\), we obtain \[y=x^{\frac{3}{2}}\ (\text{for }y>0)\Rightarrow\frac{dy}{dx}=\frac{3}{2}x^{1/2}\] The arc length formula yields \[\begin{aligned} L= & \int_{0}^{4}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx\\ = & \int_{0}^{4}\sqrt{1+\frac{9}{4}x}dx\end{aligned}\] Let \(u=1+\frac{9}{4}x\) \(du=\frac{9}{4}dx\) \(\Rightarrow dx=\frac{4}{9}du\)
\[x=0\quad\Leftrightarrow\quad u=1\] \[x=4\quad\Leftrightarrow\quad u=10\] Therefore, \[L=\int_{1}^{10}\frac{4}{9}\sqrt{u}\ du=\frac{4}{9}\left(\frac{2}{3}\right)u^{\frac{3}{2}}\Bigg|_{u=1}^{u=10}=\frac{8}{27}(\sqrt{1000}-1)\approx9.07342.\]

Example 2
Find the length of the curve \(y=\ln\sec x\) on \([0,\frac{\pi}{4}]\).

Figure 3
Figure 4: Graph of \(y=\ln\sec x\).

Solution 2
\(y=\ln\sec x\). \[\sec x=u\Rightarrow y’=u’\frac{1}{u}=(\sec x\tan x)\left(\frac{1}{\sec x}\right)=\tan x\] So the arc length formula \[\begin{aligned} L & =\int_{0}^{\frac{\pi}{4}}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx\\ & =\int_{0}^{\frac{\pi}{4}}\sqrt{1+\tan^{2}x}dx\\ & =\int_{0}^{\frac{\pi}{4}}\sqrt{\sec^{2}x}dx\qquad\qquad{\small{(1+\tan^{2}x=\sec^{2}x)}}\\ & =\int_{0}^{\frac{\pi}{4}}|\sec x|dx\\ & =\int_{0}^{\frac{\pi}{4}}\sec xdx\qquad\qquad{\small{(\sec x>0\text{ for }0\leq x\leq\frac{\pi}{4})}}\\ & =\Bigg[\ln|\sec x+\tan x|\Bigg]_{x=0}^{x=\frac{\pi}{4}}\\ & =\ln\left|\sec\frac{\pi}{4}+\tan\frac{\pi}{4}\right|-\ln|\sec0+\tan0|\\ & =\ln\left|\frac{1}{\cos\frac{\pi}{4}}+\tan\frac{\pi}{4}\right|-\ln\left|\frac{1}{\cos0}+\tan0\right|\\ & =\ln\left|\frac{2}{\sqrt{2}}+1\right|-\ln|1+0|\\ & =\ln\left(\frac{2}{\sqrt{2}}+1\right)-0\\ & =\ln(\sqrt{2}+1).\end{aligned}\]

Example 3
  Find the length of the curve \(y=a\cosh\dfrac{x}{a}\) on \([0,a]\).

Solution 3
Recall that \(\frac{d}{dx}\cosh x=\sinh x\). Therefore \[\begin{aligned} y=a\cosh\underbrace{\frac{x}{a}}_{u}\quad\Rightarrow\quad y’ & =au’\sinh u\\ & =a\frac{1}{a}\sinh\frac{x}{a}\\ & =\sinh\frac{x}{a}\end{aligned}\] The arc length formula gives \[\begin{aligned} L & =\int_{0}^{a}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx\\ & =\int_{0}^{a}\sqrt{1+\sinh^{2}\frac{x}{a}}dx\\ & =\int_{0}^{a}\cosh\frac{x}{a}dx\qquad\qquad{\small{(\cosh^{2}x-\sinh^{2}x=1 \text{ and } \cosh x>0)}}\\ & =a\left.\sinh\frac{x}{a}\right|_{x=0}^{x=a}\\ & =a(\sinh1-\sinh0)\\ & =a\sinh1\end{aligned}\] [Recall that \(\int\cosh xdx=\sinh x+C\) and hence \[\begin{aligned} \int\cosh\frac{x}{a}dx= & \int\cosh u\frac{du}{u}\qquad\qquad{\small{(x=au\Rightarrow dx=a\ du)}}\\ = & \frac{1}{a}\sinh u+C\\ = & \frac{1}{a}\sinh\frac{x}{a}+C]\end{aligned}\]

Discontinuities in dy/dx

Even when \(dy/dx\) is not defined at some points on the curve \(y=f(x)\), it is possible that \(dx/dy\) is defined everywhere on this curve. An example of such a situation is when the curve \(y=f(x)\) has a vertical tangent. In such cases, to compute the arc length of the curve \(y=f(x)\), we can express \(x\) in terms of \(y\) and write \[ds=\sqrt{dx^{2}+dy^{2}}\] \[\Rightarrow\frac{ds}{dy}=\sqrt{\left(\frac{dx}{dy}\right)^{2}+1}.\]

Example 4
Find the length of the curve \(y=x^{2/3}\) from \(x=0\) to \(x=8\).

Figure 4
Figure 5

Solution 4
\[y=x^{2/3}\Rightarrow y’=\frac{2}{3}x^{-1/3}\] Because \(y’\) is not defined at \(x=0\), we cannot find the length of the curve. However, if we express \(x\) in terms of \(y\), we will find \[x=y^{3/2}\Rightarrow\frac{dx}{dy}=\frac{3}{2}y^{1/2}\] and \[x=0\Leftrightarrow y=0\] \[x=8\Leftrightarrow y=4\] The length of the curve is \[\begin{aligned} s & =\int_{0}^{4}\sqrt{\left(\frac{dx}{dy}\right)^{2}+1}\ dy\\ & =\int_{0}^{4}\sqrt{\frac{9}{4}y+1}\ dy\\ & =\frac{1}{2}\int_{0}^{4}\sqrt{9y+4}\ dy\\ & =\frac{1}{2}\int_{4}^{40}\sqrt{u}\ \frac{du}{9}\qquad\qquad{\small{(u=9y+4 \text{ then } du=9dy)}}\\ & =\frac{1}{18}\left[\frac{2}{3}u^{3/2}\right]_{4}^{40}\\ & =\frac{1}{27}(40^{3/2}-8)\\ & =\frac{1}{27}(8\sqrt{10^{3}}-8)\\ & =\frac{8}{27}(\sqrt{1000}-1)\approx9.07342.\end{aligned}\]

The Arc Length Function

The length of the curve \(y=f(x)\) from \((a,f(a))\) to \((b,f(b))\) is \[\int_{a}^{b}\sqrt{1+[f'(x)]^{2}}dx.\] Let \(s(x)\) be a function that measures the arc length from \((a,f(a))\) to a variable point \((x,f(x))\). To find the formula of \(s(x)\), we just need replace \(b\) in the above formula with \(x\): \[s(x)=\int_{a}^{x}\sqrt{1+[f'(t)]^{2}}\ dt.\] [Because we can denote the variable of integration by any arbitrary letter and because \(x\) appears as the upper limit of integration, we have replaced the variable of integration by \(t\) so that \(x\) does not have two different meanings.] This function is called the arc length function.

  • By the Fundamental Theorem of Calculus : \[\frac{ds}{dx}=\sqrt{1+[f'(x)]^{2}}=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\] or \[ds=\sqrt{dx^{2}+dy^{2}}.\]
  • Note that \(s(x)\) is an increasing function. To mathematically prove this property, we need to show \(s'(x)>0\): \[s'(x)=\sqrt{1+[f'(x)]^{2}}>0.\]
Example 5
Find the arc length function for the curve \(y=\frac{1}{2}x^{2}-\frac{1}{4}\ln x\) taking \((1,0.5)\) as the starting point. Then, using this function, find the arc length along this curve from \((1,0.5)\) to \((e,0.5e^{2}-0.25)\).

Solution 5
\[f(x)=\frac{1}{2}x^{2}-\frac{1}{4}\ln x\Rightarrow f'(x)=x-\frac{1}{4x}\] then \[\begin{aligned} \sqrt{1+[f'(x)]^{2}} & =\sqrt{1+(x-\frac{1}{4x})^{2}}\\ & =\sqrt{1+x^{2}-\frac{1}{2}+\frac{1}{16x^{2}}}\\ & =\sqrt{(x+\frac{1}{4x})^{2}}\\ & =x+\frac{1}{4x}\qquad{\small (\text{because } x>0)}\end{aligned}\] Therefore, the arc length function is \[\begin{aligned} s(x) & =\int_{1}^{x}\sqrt{1+[f'(t)]^{2}}\ dt\\ & =\int_{1}^{x}(t+\frac{1}{4t})dt\\ & =\left[\frac{1}{2}t^{2}+\frac{1}{4}\ln t\right]_{1}^{x}\\ & =\frac{1}{2}x^{2}+\frac{1}{4}\ln x-\frac{1}{2}-\frac{1}{4}\underbrace{\ln1}_{=0}\\ & =\frac{1}{2}x^{2}+\frac{1}{4}\ln x-\frac{1}{2}.\end{aligned}\] The arc length along this curve from \((1,1/2)\) to \((e,0.5e^{2}-0.25)\) is \[s(e)=\frac{1}{2}e^{2}+\frac{1}{4}\underbrace{\ln e}_{=1}-\frac{1}{2}=\frac{e^{2}}{2}-\frac{1}{4}\approx3.44453.\]


  1. Let \[f(x)=\begin{cases} x\sin\frac{\pi}{x} & x\neq0\\ 0 & x=0 \end{cases}\] The function \(f\) is not rectifiable on \([0,1]\).↩︎