From Problem 3 above we know that the arc length function for this vector function is,

$$s\left( t \right) = \frac{1}{{135}}\left[ {{{\left( {4 + 45{t^2}} \right)}^{\frac{3}{2}}} - 8} \right]$$

We need to solve this for $t$. Doing this gives,

$$\begin{align*}{\left( {4 + 45{t^2}} \right)^{\frac{3}{2}}} - 8 & = 135s\\ {\left( {4 + 45{t^2}} \right)^{\frac{3}{2}}} & = 135s + 8\\ 4 + 45{t^2} & = {\left( {135s + 8} \right)^{\frac{2}{3}}}\\ {t^2} & = \frac{1}{{45}}\left[ {{{\left( {135s + 8} \right)}^{\frac{2}{3}}} - 4} \right]\hspace{0.25in} \to \hspace{0.25in}t = \sqrt {\frac{1}{{45}}\left[ {{{\left( {135s + 8} \right)}^{\frac{2}{3}}} - 4} \right]} \end{align*}$$

Note that we only used the positive $t$ after taking the root because the implicit assumption from the arc length function is that $t$ is positive.

Show Step 2