:onsider the curve $$ f(y)=\frac{1}{8} y^{4}+\frac{1}{4 y^{2}} $$ for $$ 1 \leq y \leq 2 $$. Find the arc length of the curve.

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To find the arc length of the curve $$ f(y) = \frac{1}{8} y^{4} + \frac{1}{4 y^{2}} $$ for $$ 1 \leq y \leq 2 $$, we'll use the arc length formula for a function expressed in terms of $$ y $$:

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

### Step 1: Compute $$ \frac{dx}{dy} $$

Given:
$$
f(y) = \frac{1}{8} y^{4} + \frac{1}{4 y^{2}}
$$

Differentiating with respect to $$ y $$:
$$
\frac{dx}{dy} = f'(y) = \frac{1}{2} y^{3} - \frac{1}{2} y^{-3} = \frac{y^3 - y^{-3}}{2}
$$

### Step 2: Compute $$ 1 + \left(\frac{dx}{dy}\right)^2 $$

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

### Step 3: Integrate to Find the Arc Length

$$
\text{Arc Length} = \int_{1}^{2} \frac{y^3 + y^{-3}}{2} \, dy = \frac{1}{2} \int_{1}^{2} y^3 \, dy + \frac{1}{2} \int_{1}^{2} y^{-3} \, dy
$$

Calculating the integrals:
$$
\frac{1}{2} \left[ \frac{y^4}{4} \right]_1^2 = \frac{1}{8} (16 - 1) = \frac{15}{8}
$$
$$
\frac{1}{2} \left[ -\frac{1}{2 y^2} \right]_1^2 = \frac{1}{2} \left( -\frac{1}{8} + \frac{1}{2} \right) = \frac{3}{16}
$$

Adding them together:
$$
\frac{15}{8} + \frac{3}{16} = \frac{30}{16} + \frac{3}{16} = \frac{33}{16}
$$

### Final Answer

The arc length of the curve from $$ y = 1 $$ to $$ y = 2 $$ is $$ \frac{33}{16} $$ units.

**Answer:**  
After simplifying, the arc length is 33⁄16 units
The arc length of the curve from $$ y = 1 $$ to $$ y = 2 $$ is $$ \frac{33}{16} $$ units.

:onsider the curve for . Find the arc length of the curve.

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