Short and Sweet Calculus

7.5 Arc length

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 7.15. 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 \[\boxed{ds=\sqrt{(dx)^{2}+(dy)^{2}}}\] 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 \((h(d),d)\), then we can still use \[ds=\sqrt{(dx)^{2}+(dy)^{2}}.\] Therefore, \[ds=\sqrt{\left(\frac{dx}{dy}\right)^{2}+1}\ dy=\sqrt{[h'(y)]^{2}+1}\ dy\] and \[\text{length of arc}=\int_{c}^{d}ds=\int_{c}^{d}\sqrt{(dx)^{2}+(dy)^{2}}=\int_{c}^{d}\sqrt{\left(\frac{dx}{dy}\right)^{2}+1}\ dy=\int_{c}^{d}\sqrt{[h'(y)]^{2}+1}\ dy.\]

Example 7.7. Find the length of the curve \(y^{2}=x^{3}\) between the points \((0,0)\) and \((4,8)\).


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 \[L=\int_{0}^{4}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx=\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx.\] Let \(u=1+\frac{9}{4}x\). Then \(du=\frac{9}{4}dx\) or \(dx=\frac{4}{9}du\) and \[x=0\quad\Leftrightarrow\quad u=1\] \[x=4\quad\Leftrightarrow\quad u=10\] Therefore, \[L=\int_{1}^{10}\frac{4}{9}\sqrt{u}\ du=\frac{4}{9}\left(\frac{2}{3}\right)u^{\frac{3}{2}}\Bigg|_{u=1}^{u=10}=\frac{8}{27}(\sqrt{1000}-1)\approx9.07342.\]

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