Find the length of the curve. \[ x=\frac{t^{2}}{2}, y=\frac{(2 t+1)^{3 / 2}}{3}, 0 \leq t \leq 6 \]

Step 1 ：We are given the parametric equations \(x=\frac{t^{2}}{2}\) and \(y=\frac{(2 t+1)^{3 / 2}}{3}\) for \(0 \leq t \leq 6\).

Step 2 ：The length of a curve defined by parametric equations can be found using the formula: \[L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt\]

Step 3 ：First, we need to find the derivatives of x and y with respect to t. The derivative of \(x\) with respect to \(t\) is \(t\), and the derivative of \(y\) with respect to \(t\) is \(1.0*(2*t + 1)^{0.5}\).

Step 4 ：Substitute these derivatives into the formula for the length of the curve, we get: \[L = \int_{0}^{6} \sqrt{1.0*t^2 + 1.0*(2*t + 1)^1.0} dt\]

Step 5 ：Evaluating this integral from t=0 to t=6, we find the numerical value of the length to be 24.

Step 6 ：Final Answer: The length of the curve is \(\boxed{24}\).