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Section 9.4 : Arc Length with Parametric Equations

5. Set up, but do not evaluate, an integral that gives the length of the parametric curve given by the following set of parametric equations. You may assume that the curve traces out exactly once for the given range of \(t\)’s.

\[x = {\cos ^3}\left( {2t} \right)\hspace{0.25in}y = \sin \left( {1 - {t^2}} \right)\hspace{0.25in} - \frac{3}{2} \le t \le 0\]

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Start Solution

The first thing we’ll need here are the following two derivatives.

\[\frac{{dx}}{{dt}} = - 6{\cos ^2}\left( {2t} \right)\sin \left( {2t} \right)\hspace{0.5in}\frac{{dy}}{{dt}} = - 2t\cos \left( {1 - {t^2}} \right)\] Show Step 2

We’ll need the \(ds\) for this problem.

\[\begin{align*}ds & = \sqrt {{{\left[ { - 6{{\cos }^2}\left( {2t} \right)\sin \left( {2t} \right)} \right]}^2} + {{\left[ { - 2t\cos \left( {1 - {t^2}} \right)} \right]}^2}} \,dt\\ & = \sqrt {36{{\cos }^4}\left( {2t} \right){{\sin }^2}\left( {2t} \right) + 4{t^2}{{\cos }^2}\left( {1 - {t^2}} \right)} \,dt\end{align*}\] Show Step 3

The integral for the arc length is then,

\[L = \int_{{}}^{{}}{{ds}} = \int_{{ - \,\frac{3}{2}}}^{0}{{\sqrt {36{{\cos }^4}\left( {2t} \right){{\sin }^2}\left( {2t} \right) + 4{t^2}{{\cos }^2}\left( {1 - {t^2}} \right)} \,dt}}\]