Consider the curve \( f(x)=\frac{1}{16} x^{4}+\frac{1}{2 x^{2}} \) for \( 1 \leq x \leq 2 \). Find the arc length of the curve.

To find the arc length of the curve \( f(x) = \frac{1}{16} x^4 + \frac{1}{2 x^2} \) from \( x = 1 \) to \( x = 2 \), we'll use the formula for the arc length of a function \( y = f(x) \): \[ \text{Arc Length} = \int_{a}^{b} \sqrt{1 + \left(f'(x)\right)^2} \, dx \] **Step 1: Compute the derivative \( f'(x) \)** \[ f(x) = \frac{1}{16} x^4 + \frac{1}{2 x^2} \] Differentiate with respect to \( x \): \[ f'(x) = \frac{4}{16} x^3 - \frac{2}{2 x^3} = \frac{1}{4} x^3 - \frac{1}{x^3} \] **Step 2: Compute \( 1 + \left(f'(x)\right)^2 \)** \[ \left(f'(x)\right)^2 = \left(\frac{1}{4} x^3 - \frac{1}{x^3}\right)^2 = \frac{1}{16} x^6 - \frac{2}{4} + \frac{1}{x^6} = \frac{1}{16} x^6 - \frac{1}{2} + \frac{1}{x^6} \] Adding 1: \[ 1 + \left(f'(x)\right)^2 = \frac{1}{2} + \frac{1}{16} x^6 + \frac{1}{x^6} \] Notice that this expression can be written as a perfect square: \[ 1 + \left(f'(x)\right)^2 = \left(\frac{1}{4} x^3 + \frac{1}{x^3}\right)^2 \] **Step 3: Compute the integral for arc length** \[ \text{Arc Length} = \int_{1}^{2} \left(\frac{1}{4} x^3 + \frac{1}{x^3}\right) dx \] Integrate term by term: \[ \int \frac{1}{4} x^3 \, dx = \frac{1}{4} \cdot \frac{x^4}{4} = \frac{x^4}{16} \] \[ \int \frac{1}{x^3} \, dx = \int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2} \] Combine the results: \[ \text{Arc Length} = \left[\frac{x^4}{16} - \frac{1}{2x^2}\right]_{1}^{2} \] Evaluate from \( x = 1 \) to \( x = 2 \): At \( x = 2 \): \[ \frac{2^4}{16} - \frac{1}{2 \cdot 2^2} = \frac{16}{16} - \frac{1}{8} = 1 - \frac{1}{8} = \frac{7}{8} \] At \( x = 1 \): \[ \frac{1^4}{16} - \frac{1}{2 \cdot 1^2} = \frac{1}{16} - \frac{1}{2} = \frac{1}{16} - \frac{8}{16} = -\frac{7}{16} \] Subtract the two results: \[ \frac{7}{8} - \left(-\frac{7}{16}\right) = \frac{7}{8} + \frac{7}{16} = \frac{14}{16} + \frac{7}{16} = \frac{21}{16} \] **Final Answer:** The arc length of the curve is \( \frac{21}{16} \) units.