The equation shown is:

$36y^2 = (x^2 - 4)^3$

with the domain $5 \leq x \leq 6$ and the condition $y \geq 0$.

To find the exact length of the curve, we will use the arc length formula:

$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Step 1: Express $y$ in terms of $x$

Starting with the given equation:

$36y^2 = (x^2 - 4)^3$

Solve for $y$:

$y^2 = \frac{(x^2 - 4)^3}{36}$

$y = \frac{(x^2 - 4)^{3/2}}{6}, \quad y \geq 0$

Step 2: Differentiate $y$ with respect to $x$

Now, find $\frac{dy}{dx}$.

$y = \frac{(x^2 - 4)^{3/2}}{6}$

Using the chain rule:

$\frac{dy}{dx} = \frac{1}{6} \cdot \frac{d}{dx} \left( (x^2 - 4)^{3/2} \right)$

$\frac{dy}{dx} = \frac{1}{6} \cdot \frac{3}{2} (x^2 - 4)^{1/2} \cdot 2x$

$\frac{dy}{dx} = \frac{x (x^2 - 4)^{1/2}}{2}$

Step 3: Plug into the arc length formula

Now substitute $\frac{dy}{dx}$ into the arc length formula:

$L = \int_5^6 \sqrt{1 + \left( \frac{x (x^2 - 4)^{1/2}}{2} \right)^2} \, dx$

Simplify the integrand:

$L = \int_5^6 \sqrt{1 + \frac{x^2 (x^2 - 4)}{4}} \, dx$

This integral is quite complex, so solving it requires advanced techniques, such as substitution or numerical methods. I'll evaluate this integral next to find the exact length of the curve.

Let me calculate that for you!The result of the arc length integral simplifies to a piecewise expression, which can be further evaluated numerically. The integral is:

$L = \frac{1}{2} \int_5^6 (x^2 - 2) \, dx$

Now, I'll calculate the numerical value of this definite integral.The exact length of the curve is approximately:

$L \approx 14.17 \text{ units}$

Let me know if you would like a detailed explanation of any of the steps.

Here are 5 related questions to further your understanding:

1. How do we apply the chain rule in the process of differentiation?
2. What are the steps to simplify an integral involving radicals?
3. Why does the arc length formula include the derivative of $y$ with respect to $x$?
4. How do we solve more complex arc length integrals that require numerical methods?
5. What would change in the process if the limits of integration were different?

Tip: When solving arc length problems, simplifying the derivative before substituting into the formula helps avoid unnecessary complexity in the integral.