Answer
$s \approx 1.0319$
Work Step by Step
$$\eqalign{
& {\text{Let the function }}y = \tan \pi x{\text{ on the interval }}\left[ {0,\frac{1}{4}} \right] \cr
& {\text{The arc length of the function }}y{\text{ on the interval }}\left[ {a,b} \right]{\text{ is}} \cr
& s = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\tan \pi x} \right] \cr
& \frac{{dy}}{{dx}} = \pi {\sec ^2}\pi x \cr
& {\text{Substituting }}\frac{{dy}}{{dx}}{\text{ into the arc length formula}} \cr
& s = \int_0^{1/4} {\sqrt {1 + {{\left( {\pi {{\sec }^2}\pi x} \right)}^2}} dx} \cr
& s = \int_0^{1/4} {\sqrt {1 + {\pi ^2}{{\sec }^4}\pi x} dx} \cr
& {\text{Integrate using a graphing utility or a computer}}{\text{, we obtain}} \cr
& s \approx 1.0319 \cr} $$
