onsider the curve \( f(y)=\frac{1}{20} y^{4}+\frac{5}{8 y^{2}} \) for \( 1 \leq y \leq 3 \). Find the arc length of the curve.

To find the arc length of the curve \( f(y) = \frac{1}{20} y^4 + \frac{5}{8 y^2} \) for \( 1 \leq y \leq 3 \), we use the formula for arc length when the function is expressed as \( x = f(y) \): \[ \text{Arc Length} = \int_{a}^{b} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy \] ### Step 1: Compute the Derivative \(\frac{dx}{dy}\) Given: \[ f(y) = \frac{1}{20} y^4 + \frac{5}{8 y^2} \] Differentiate with respect to \( y \): \[ \frac{dx}{dy} = f'(y) = \frac{4}{20} y^3 - \frac{5}{4} y^{-3} = \frac{1}{5} y^3 - \frac{5}{4 y^3} \] ### Step 2: Compute \(1 + \left(\frac{dx}{dy}\right)^2\) \[ \left(\frac{dx}{dy}\right)^2 = \left(\frac{1}{5} y^3 - \frac{5}{4 y^3}\right)^2 = \frac{1}{25} y^6 - \frac{1}{2} + \frac{25}{16 y^6} \] Thus: \[ 1 + \left(\frac{dx}{dy}\right)^2 = \frac{1}{25} y^6 + \frac{1}{2} + \frac{25}{16 y^6} \] Notice that this expression is a perfect square: \[ \frac{1}{25} y^6 + \frac{1}{2} + \frac{25}{16 y^6} = \left(\frac{1}{5} y^3 + \frac{5}{4 y^3}\right)^2 \] So, \[ \sqrt{1 + \left(\frac{dx}{dy}\right)^2} = \frac{1}{5} y^3 + \frac{5}{4 y^3} \] ### Step 3: Integrate to Find the Arc Length \[ \text{Arc Length} = \int_{1}^{3} \left(\frac{1}{5} y^3 + \frac{5}{4 y^3}\right) dy \] Integrate term by term: \[ \int \frac{1}{5} y^3 \, dy = \frac{1}{20} y^4 \] \[ \int \frac{5}{4 y^3} \, dy = \frac{5}{4} \left(-\frac{1}{2 y^2}\right) = -\frac{5}{8 y^2} \] So the definite integral is: \[ \left[\frac{1}{20} y^4 - \frac{5}{8 y^2}\right]_{1}^{3} \] Evaluate from \( y = 1 \) to \( y = 3 \): \[ \left(\frac{81}{20} - \frac{5}{72}\right) - \left(\frac{1}{20} - \frac{5}{8}\right) = \frac{41}{9} \] ### Final Answer The arc length of the curve from \( y = 1 \) to \( y = 3 \) is: \[ \boxed{\dfrac{41}{9}} \]