Finding the Arc Length of a Graph
Find the arc length of the graph of the function \(f(x) =x^{2/3}\) from \(x=1\) to \(x=8\).
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The derivative of \(f(x)=x^{2/3}\) is \(f' (x)=\dfrac{2}{3}x^{-1/3}=\dfrac{2}{3x^{1/3}}\). Notice that \(f' \) is not continuous at \(0\). However, since \(f'\) is continuous on an interval containing \(1\) and \( 8\) (use \(\left[ \dfrac{1}{2},9\right]\), for example, which avoids \(0)\), we use the arc length formula (1). The arc length \(s\) from \(x=1\) to \(x=8\) is \begin{eqnarray*} s&=&\int_{1}^{8}\sqrt{1+\left[ f' (x) \right] ^{2}} ~{\it dx}=\int_{1}^{8}\sqrt{1+\left( \frac{2}{3x^{1/3}}\right) ^{2}} ~{\it dx}=\int_{1}^{8}\sqrt{1+\frac{4}{9x^{2/3}}}~{\it dx} \\[5pt] &=&\int_{1}^{8}\sqrt{\frac{9x^{2/3}+4}{9x^{2/3}}}~{\it dx} \underset{\underset{\underset{\color{#0066A7}{\hbox{so \(\sqrt{x^{2/3}} =x^{1/3}\)}}}{\color{#0066A7}{\hbox{\(x>0\) on \([1,8]\),}}}} {\color{#0066A7}{\uparrow }}} {=} {\frac{1}{3}} \int_{1}^{8}\Big( x^{-1/3}\sqrt{9x^{2/3}+4}\Big) ~{\it dx}\\[-9pt] \end{eqnarray*}
We use the substitution \(u=9x^{2/3}+4.\) Then \({\it du}=6x^{-1/3}{\it dx}\) and, \( x^{-1/3}{\it dx}=\dfrac{{\it du}}{6}\). The limits of integration change to \(u=13\) when \(x=1\), and to \(u=40\) when \(x=8\). Then \begin{eqnarray*} s&=&\dfrac{1}{3} \displaystyle \int_{1}^{8}\Big[ x^{-1/3}\sqrt{~9x^{2/3}+4}\Big] ~{\it dx}= \dfrac{1}{3}\displaystyle \int_{13}^{40}\sqrt{~u}~\dfrac{{\it du}}{6}= \dfrac{1}{18} \left[\dfrac{u^{3/2}}{\dfrac{3}{2}}\right] _{13}^{40}\\[5pt] &=&\dfrac{1}{27}\big( 80\sqrt{10}-13\sqrt{13}\big) \end{eqnarray*}