Finding the Arc Length of a Function
Find the arc length of \(f(x) =x^{2/3}\) from \(x=-1\) to \(x=8.\)
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From Figure 52, we observe that the length of the graph of \(f\) from \(x=-1\) to \(x=8\) is the same as the length of the graph of \(x=g_{1}( y) =-y^{3/2}\) from \(y=0\) to \(y=1\) plus the length of the graph of \(x=g_{2}( y) =y^{3/2}\) from \(y=0\) to \(y=4.\)
We investigate \(x=g_{1}(y) =-y^{3/2}\) first. Since \(g_{1}' (y) =-\dfrac{3}{2}y^{1/2}\) is continuous for all \(y\geq 0.\) we can use arc length formula (2) to find the arc length \(s_{1}\) of \(g_{1}\) from \(y=0\) to \(y=1\). \begin{eqnarray*} s_{1}&=&\int_{0}^{1}\sqrt{1+[g_{1}'(y)]^{2}}{\it dy} \underset{\underset{\color{#0066A7}{\hbox{\(g_{1}'(y)\)} =-\frac{3}{2}y^{1/2}}}{\color{#0066A7}{\uparrow }}}{=}\int_{0}^{1} \sqrt{1+\left(\frac{3}{2}y^{1/2}\right) ^{2}}~{\it dy}\\[-11pt] &=&\int_{0}^{1}\sqrt{1+ \frac{9}{4}y}~{\it dy}=\frac{1}{2}\int_{0}^{1}\sqrt{4+9y}~{\it dy} \end{eqnarray*}
We use the substitution \(u=4+9y\). Then \({\it du}=9~{\it dy}\) or, equivalently, \({\it dy}= \dfrac{{\it du}}{9}\). The limits of integration are \(u=4\) when \(y=0\), and \(u=13\) when \(y=1.\) \begin{equation*} s_{1}=\frac{1}{2}\displaystyle \int_{4}^{13}\sqrt{u}~\frac{{\it du}}{9}=\frac{1}{18}\left[ \frac{u^{3/2}}{\dfrac{3}{2}}\right] _{4}^{13}=\frac{1}{27}\big( 13\sqrt{13}-8\big) \end{equation*}
Now we investigate \(x=g_{2}(y)\). Since \(g_{2}' (y) =\dfrac{3}{2}y^{1/2}\) is continuous for all \(y\geq 0,\) we can use arc length formula (2) to find the arc length \(s_{2}\) of \(g_{2}\) from \(y=0\) to \(y=4\). \[ \begin{array}{rcl@{\qquad}l} s_{2} &=&\int_{0}^{4}\sqrt{1+[g_{2}'(y)] ^{2}}~{\it dy}=\int_{0}^{4}\sqrt{1+\left(\dfrac{3}{2}y^{1/2}\right)^{~2}} ~{\it dy}\\ &=&\int_{0}^{4}\sqrt{1+\dfrac{9}{4}y}~{\it dy} =\frac{1}{2}\int_{0}^{4}\sqrt{4+9y}~{\it dy} &\color{#0066A7}{\hbox{Let \(u=4+9y \); }}\\ &&&\color{#0066A7}{\hbox{\(du=9dy \).}}\\ &=&\frac{1}{2}\int_{4}^{40}\sqrt{u}~\frac{{\it du}}{9}=\frac{1}{18}\left[ \frac{u^{3/2}}{\frac{3}{2}}\right] _{4}^{40}=\frac{1}{27}\big( 80\sqrt{10}-8\big) \end{array}\]
The arc length \(s\) of \(y=f(x) =x^{2/3}\) from \(x=-1\) to \(x=8\) is the sum \begin{eqnarray*} s&=&s_{1}+s_{2}=\frac{1}{27}\big( 13\sqrt{13}-8\big) +\dfrac{1}{27}\big( 80\sqrt{10}-8\big)\\[4pt] &=&\frac{1}{27} \big( 80\sqrt{10}+13\sqrt{13}-16\big) \end{eqnarray*}