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,

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