(1 point)
Find the length of the parametric curve
\[
x=1+27 t^{2}, \quad y=6+18 t^{3}, \quad 0 \leq t \leq 5 .
\]

Step 1 ：First, we need to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

Step 2 ：We have \(\frac{dx}{dt} = \frac{d(1+27t^2)}{dt} = 54t\) and \(\frac{dy}{dt} = \frac{d(6+18t^3)}{dt} = 54t^2\).

Step 3 ：Then, we substitute these into the formula for the length of a parametric curve: \(L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt\).

Step 4 ：This gives us \(L = \int_{0}^{5} \sqrt{(54t)^2 + (54t^2)^2} dt = \int_{0}^{5} \sqrt{2916t^2 + 2916t^4} dt = \int_{0}^{5} \sqrt{2916(t^2 + t^4)} dt = \int_{0}^{5} 54t \sqrt{1 + t^2} dt\).

Step 5 ：This integral is not easy to solve directly, so we use a substitution: let \(u = t^2 + 1\), then \(du = 2t dt\), and \(dt = \frac{du}{2t}\).

Step 6 ：Substituting these into the integral, we get \(L = \int_{1}^{26} 27 \sqrt{u} du = [18u^{3/2}]_{1}^{26} = 18*26^{3/2} - 18*1^{3/2} = 18*169 - 18 = 3024\).

Step 7 ：So, the length of the parametric curve is \(\boxed{3024}\).