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Section 3-4 : Arc Length with Parametric Equations

1. Determine 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.

\[x = 8{t^{\frac{3}{2}}}\hspace{0.25in}y = 3 + {\left( {8 - t} \right)^{\frac{3}{2}}}\hspace{0.25in}0 \le t \le 4\]

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Start Solution

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

\[\frac{{dx}}{{dt}} = 12{t^{\,\frac{1}{2}}}\hspace{0.5in}\frac{{dy}}{{dt}} = - \frac{3}{2}{\left( {8 - t} \right)^{\frac{1}{2}}}\] Show Step 2

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

\[ds = \sqrt {{{\left[ {12{t^{\,\frac{1}{2}}}} \right]}^2} + {{\left[ { - \frac{3}{2}{{\left( {8 - t} \right)}^{\frac{1}{2}}}} \right]}^2}} \,dt = \sqrt {144t + \frac{9}{4}\left( {8 - t} \right)} \,dt = \sqrt {\frac{{567}}{4}t + 18} \,dt\] Show Step 3

The integral for the arc length is then,

\[L = \int_{{}}^{{}}{{ds}} = \int_{0}^{4}{{\sqrt {\frac{{567}}{4}t + 18} \,dt}}\] Show Step 4

This is a simple integral to compute with a quick substitution. Here is the integral work,

\[L = \int_{0}^{4}{{\sqrt {\frac{{567}}{4}t + 18} \,dt}} = \left. {\frac{4}{{567}}\left( {\frac{2}{3}} \right){{\left( {\frac{{567}}{4}t + 18} \right)}^{\frac{3}{2}}}} \right|_0^4 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{8}{{1701}}\left( {{{585}^{\,\frac{3}{2}}} - {{18}^{\,\frac{3}{2}}}} \right) = 66.1865}}\]