Section 9.4 : 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.
\[x = 8{t^{\frac{3}{2}}}\hspace{0.25in}y = 3 + {\left( {8 - t} \right)^{\frac{3}{2}}}\hspace{0.25in}0 \le t \le 4\]Show All Steps Hide All Steps
Start SolutionThe 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 2We’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 3The integral for the arc length is then,
\[L = \int_{{}}^{{}}{{ds}} = \int_{0}^{4}{{\sqrt {\frac{{567}}{4}t + 18} \,dt}}\] Show Step 4This is a simple integral to compute with a quick substitution. Here is the integral work,
\[L = \int_{0}^{4}{{\sqrt {\frac{{567}}{4}t + 18} \,dt}} = \left. {\frac{4}{{567}}\left( {\frac{2}{3}} \right){{\left( {\frac{{567}}{4}t + 18} \right)}^{\frac{3}{2}}}} \right|_0^4 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{8}{{1701}}\left( {{{585}^{\,\frac{3}{2}}} - {{18}^{\,\frac{3}{2}}}} \right) = 66.1865}}\]