To find the arc length of the curve $x = \frac{1}{3} \sqrt{y}(y - 3)$ over the interval $1 \leq y \leq 25$, we use the formula for arc length in terms of $y$:

$L = \int_{y_1}^{y_2} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy$

Step 1: Differentiate $x$ with respect to $y$

The given function is:

$x = \frac{1}{3} \sqrt{y}(y - 3)$

We need to find $\frac{dx}{dy}$.

First, simplify the expression for $x$:

$x = \frac{1}{3} y^{1/2} (y - 3)$

Now apply the product rule to differentiate this:

$\frac{dx}{dy} = \frac{1}{3} \left( \frac{d}{dy}(y^{1/2}) (y - 3) + y^{1/2} \frac{d}{dy}(y - 3) \right)$

$= \frac{1}{3} \left( \frac{1}{2} y^{-1/2} (y - 3) + y^{1/2} \right)$

Simplify the expression:

$\frac{dx}{dy} = \frac{1}{3} \left( \frac{1}{2} \frac{y - 3}{\sqrt{y}} + \sqrt{y} \right)$

$= \frac{1}{3} \left( \frac{y - 3}{2 \sqrt{y}} + \sqrt{y} \right)$

This is the expression for $\frac{dx}{dy}$.

Step 2: Plug into the arc length formula

Now, the arc length formula becomes:

$L = \int_1^{25} \sqrt{1 + \left( \frac{1}{3} \left( \frac{y - 3}{2 \sqrt{y}} + \sqrt{y} \right) \right)^2} \, dy$

This integral is challenging to compute by hand, so it requires numerical methods. I can evaluate it numerically to find the approximate arc length.

Let me calculate that for you.The arc length of the graph of the function $x = \frac{1}{3} \sqrt{y}(y - 3)$ over the interval $1 \leq y \leq 25$ is approximately 45.33 units.

Would you like further clarification or a detailed breakdown of any step? Here are five related questions that could expand your understanding:

1. How do we derive the formula for arc length in terms of $y$?
2. Can we compute arc length for functions expressed as $x = f(y)$ and $y = f(x)$ in the same way?
3. What are the numerical methods for solving integrals like this one?
4. How would the arc length change if we modified the function slightly?
5. Can arc length be applied to three-dimensional curves as well?

Tip: Always simplify expressions before differentiation or integration to avoid complex terms later.