To find the exact length of the curve given by the equation

$x = \frac{y^4}{8} + \frac{1}{4y^2}$ over the interval $1 \leq y \leq 3$, we will use the formula for the arc length of a curve:

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

Step 1: Find $\frac{dx}{dy}$

The given function is: $x = \frac{y^4}{8} + \frac{1}{4y^2}$

Taking the derivative with respect to $y$: $\frac{dx}{dy} = \frac{4y^3}{8} - \frac{1}{2y^3}$ $\frac{dx}{dy} = \frac{y^3}{2} - \frac{1}{2y^3}$

Step 2: Find $\left(\frac{dx}{dy}\right)^2$

$\left(\frac{dx}{dy}\right)^2 = \left(\frac{y^3}{2} - \frac{1}{2y^3}\right)^2$ Expanding this expression: $\left(\frac{dx}{dy}\right)^2 = \left(\frac{y^3}{2}\right)^2 - 2\left(\frac{y^3}{2}\right)\left(\frac{1}{2y^3}\right) + \left(\frac{1}{2y^3}\right)^2$ $= \frac{y^6}{4} - \frac{1}{2} + \frac{1}{4y^6}$

Step 3: Substitute into the arc length formula

The arc length formula becomes: $L = \int_{1}^{3} \sqrt{1 + \frac{y^6}{4} - \frac{1}{2} + \frac{1}{4y^6}} \, dy$ Simplifying the expression under the square root: $L = \int_{1}^{3} \sqrt{\frac{y^6}{4} + \frac{1}{2} + \frac{1}{4y^6}} \, dy$

Step 4: Solve the integral

The exact evaluation of this integral might require more advanced methods or substitution, so let's analyze and express the integral as accurately as possible: $L = \int_{1}^{3} \sqrt{\frac{y^6}{4} + \frac{1}{2} + \frac{1}{4y^6}} \, dy$

This expression represents the exact length of the curve but requires numerical methods or further simplification for a precise numerical result.

Would you like me to proceed with numerical evaluation or provide more detailed steps for solving this integral?

Follow-up Questions

1. What other methods exist for finding arc lengths of more complex curves?
2. How do we handle parametric equations when finding arc lengths?
3. Can the integral expression for arc length be simplified using trigonometric identities?
4. Why do some curves yield easier-to-compute arc lengths compared to others?
5. How do changes in $y$ or $x$ affect the arc length's complexity?

Tip

When finding arc lengths, always start by carefully finding $\frac{dx}{dy}$ or $\frac{dy}{dx}$ to avoid errors in the integral calculation.