Find the length $s$ of the curve represented by the parametric equations $$\begin{equation*} x(t) =t^{3}+2\qquad\hbox{}\qquad y(t) =2t^{9/2} \end{equation*}$$

from the point where $t=1$ to the point where $t=3$. Figure 16 shows the graph of the curve.

Figure 16 $x(t) =t^{3}+2,$ $y(t) =2t^{9/2},$ $1\leq t\leq 3$

Solution We begin by finding the derivatives $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}.$ $$\begin{equation*} \dfrac{dx}{dt}=3t^{2}\qquad\hbox{and}\qquad \dfrac{dy}{dt}=9t^{7/2} \end{equation*}$$

The curve is smooth for $1\leq t\leq 3.$ Now using the arc length formula for parametric equations, we have $$\begin{eqnarray*} s&=&\int_{a}^{b}\sqrt{\left( \dfrac{dx}{dt}\right) ^{2}+\left( \dfrac{dy}{dt} \right) ^{2}}~dt=\int_{1}^{3}\sqrt{(3t^{2})^{2}+(9t^{7/2})^{2}} \,dt\\[6pt] &=&\int_{1}^{3}\sqrt{9t^{4}+81\,t^{7}}\,dt =\int_{1}^{3}3t^{2}\sqrt{1+9t^{3} }\,dt \end{eqnarray*}$$

We use the substitution $u=1+9t^{3}$. Then $du=27t^{2}dt,$ or equivalently, $3t^{2}dt=\dfrac{du}{9}.$ Changing the limits of integration, we find that when $t=1,$ then $u=10;$ and when $t=3,$ then $u=1+9\cdot 3^{3}=244.$ The arc length $s$ is $$\begin{eqnarray*} s&=&\int_{1}^{3}3t^{2}\sqrt{1+9t^{3}}\,dt=\int_{10}^{244}\sqrt{u}\left( \frac{ du}{9}\right) =\dfrac{1}{9}\int_{10}^{244}u^{1/2}du=\frac{1}{9}\left[ \frac{ u^{3/2}}{\dfrac{3}{2}}\right] _{10}^{244}\\ &=&\frac{2}{27}[244^{3/2}-10^{3/2}] = \frac{4}{27}\big[244\sqrt{61}- 5\sqrt{10}\big] \end{eqnarray*}$$