The problem requires us to find the arc length of the curve defined by the parametric equations $x = t - 3$ and $y = \sqrt{2t}$, where $0 \le t \le 8$.

1. Problem Description

The problem requires us to find the arc length of the curve defined by the parametric equations $x = t - 3$ and $y = \sqrt{2t}$, where $0 \le t \le 8$.

2. Solution Steps

The formula for the arc length of a parametric curve defined by $x = f(t)$ and $y = g(t)$ from $t = a$ to $t = b$ is given by:

$L = \int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$

First, we need to find the derivatives of $x$ and $y$ with respect to $t$.

$\frac{dx}{dt} = \frac{d(t-3)}{dt} = 1$

$\frac{dy}{dt} = \frac{d(\sqrt{2t})}{dt} = \frac{d((2t)^{1/2})}{dt} = \frac{1}{2}(2t)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2t}}$

Now, we substitute these derivatives into the arc length formula:

$L = \int_{0}^{8} \sqrt{(1)^2 + (\frac{1}{\sqrt{2t}})^2} dt$

$L = \int_{0}^{8} \sqrt{1 + \frac{1}{2t}} dt$

$L = \int_{0}^{8} \sqrt{\frac{2t+1}{2t}} dt$

Let $u = 2t+1$. Then $du = 2 dt$, so $dt = \frac{1}{2} du$. Also, $t = \frac{u-1}{2}$.

Then $\sqrt{\frac{2t+1}{2t}} = \sqrt{\frac{u}{u-1}}$. This substitution does not seem to simplify the integral.

Instead, let's try to manipulate the integrand directly.

$L = \int_{0}^{8} \sqrt{1 + \frac{1}{2t}} dt = \int_{0}^{8} \sqrt{\frac{2t+1}{2t}} dt$

This integral does not have a simple closed-form solution using elementary functions. The question may have an error. However, assuming the given function is right, the arc length integral is

$L = \int_0^8 \sqrt{1 + \frac{1}{2t}} \, dt$.

Let $u = \sqrt{2t+1}$, then $u^2 = 2t+1$, $2t = u^2 - 1$, $t = \frac{u^2-1}{2}$ and $dt = u du$.

The limits of integration are $\sqrt{2(0)+1} = 1$ and $\sqrt{2(8)+1} = \sqrt{17}$.

Then $\sqrt{1 + \frac{1}{2t}} = \sqrt{1 + \frac{1}{u^2-1}} = \sqrt{\frac{u^2-1+1}{u^2-1}} = \sqrt{\frac{u^2}{u^2-1}} = \frac{u}{\sqrt{u^2-1}}$.

The integral becomes $L = \int_1^{\sqrt{17}} \frac{u}{\sqrt{u^2-1}} u \, du = \int_1^{\sqrt{17}} \frac{u^2}{\sqrt{u^2-1}} \, du$.

Let $u = \sec \theta$. Then $du = \sec \theta \tan \theta \, d\theta$ and $\sqrt{u^2 - 1} = \tan \theta$.

Then $\int \frac{u^2}{\sqrt{u^2 - 1}} \, du = \int \frac{\sec^2 \theta}{\tan \theta} \sec \theta \tan \theta \, d\theta = \int \sec^3 \theta \, d\theta$.

$\int \sec^3 \theta \, d\theta = \frac{1}{2} (\sec \theta \tan \theta + \ln |\sec \theta + \tan \theta|)$.

For $u=1$, $\sec \theta = 1$, $\theta = 0$.

For $u = \sqrt{17}$, $\sec \theta = \sqrt{17}$, $\tan \theta = \sqrt{16} = 4$.

Then $L = \frac{1}{2} (\sqrt{17} \cdot 4 + \ln|\sqrt{17} + 4|) - \frac{1}{2}(1 \cdot 0 + \ln |1+0|) = 2\sqrt{17} + \frac{1}{2} \ln (4 + \sqrt{17}) - 0$.

Therefore, $L = 2\sqrt{17} + \frac{1}{2} \ln (4 + \sqrt{17})$.

3. Final Answer

$2\sqrt{17} + \frac{1}{2}\ln(4 + \sqrt{17})$