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### Section 3-4 : Arc Length with Parametric Equations

2. 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.

$x = 3t + 1\hspace{0.25in}y = 4 - {t^2}\hspace{0.25in} - 2 \le t \le 0$

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Start Solution

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

$\frac{{dx}}{{dt}} = 3\hspace{0.5in}\frac{{dy}}{{dt}} = - 2t$ Show Step 2

We’ll need the $$ds$$ for this problem.

$ds = \sqrt {{{\left[ 3 \right]}^2} + {{\left[ { - 2t} \right]}^2}} \,dt = \sqrt {9 + 4{t^2}} \,dt$ Show Step 3

The integral for the arc length is then,

$L = \int_{{}}^{{}}{{ds}} = \int_{{ - 2}}^{0}{{\sqrt {9 + 4{t^2}} \,dt}}$ Show Step 4

This integral will require a trig substitution (as will quite a few arc length integrals!).

Here is the trig substitution we’ll need for this integral.

$t = \frac{3}{2}\tan \theta \hspace{0.25in}dt = \frac{3}{2}{\sec ^2}\theta \,d\theta \,$ $\sqrt {9 + 4{t^2}} = \sqrt {9 + 9{{\tan }^2}\theta } = 3\sqrt {1 + {{\tan }^2}\theta } = 3\sqrt {{{\sec }^2}\theta } = 3\left| {\sec \theta } \right|$

To get rid of the absolute value on the secant will need to convert the limits into $$\theta$$ limits.

\begin{align*}t & = - 2: & \hspace{0.25in} - 2 & = \frac{3}{2}\tan \theta \hspace{0.25in} \to \hspace{0.25in}\tan \theta = - \frac{4}{3}\hspace{0.25in} \to \hspace{0.25in}\theta = {\tan ^{ - 1}}\left( { - \frac{4}{3}} \right) = - 0.9273\\ t & = 0:& \hspace{0.25in}0 & = \frac{3}{2}\tan \theta \hspace{0.25in} \to \hspace{0.25in}\tan \theta = 0\hspace{0.25in}\, \to \hspace{0.25in}\theta = 0\end{align*}

Okay, the corresponding range of$$\theta$$ for this problem is $$- 0.9273 \le \theta \le 0$$ (fourth quadrant) and in this range we know that secant is positive. Therefore the root becomes,

$\sqrt {9 + 4{t^2}} = 3\sec \theta$

The integral is then,

\begin{align*}L & = \int_{{ - 2}}^{0}{{\sqrt {9 + 4{t^2}} \,dt}} = \int_{{ - 0.9273}}^{0}{{\left( {3\sec \theta } \right)\left( {\frac{3}{2}{{\sec }^2}\theta } \right)\,d\theta }}\\ & = \int_{{ - 0.9273}}^{0}{{\frac{9}{2}{{\sec }^3}\theta \,d\theta }} = \left. {\frac{9}{4}\left[ {\sec \theta \tan \theta + ln\left| {\sec \theta + \tan \theta } \right|} \right]} \right|_{ - 0.9273}^0 = \require{bbox} \bbox[2pt,border:1px solid black]{{7.4719}}\end{align*}