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}

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}

• 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 $$(4,8)$$.

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 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 $$x=8$$.

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