Question
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: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.

Ask by Gough Edwards. in the United States
Jan 25,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

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

Solution

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

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The Deep Dive

To find the arc length of the curve given by \( f(y) = \frac{1}{8} y^{4} + \frac{1}{4 y^{2}} \) from \( y = 1 \) to \( y = 2 \), we can use the formula for the arc length of a curve defined as \( f(y) \): \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{df}{dy} \right)^{2}} \, dy \] 1. First, we need to compute the derivative \( \frac{df}{dy} \): \[ \frac{df}{dy} = \frac{1}{2} y^{3} - \frac{1}{2 y^{3}} \] 2. Next, square the derivative: \[ \left( \frac{df}{dy} \right)^{2} = \left( \frac{1}{2} y^{3} - \frac{1}{2 y^{3}} \right)^{2} = \frac{1}{4} \left( y^{6} - 2 + \frac{1}{y^{6}} \right) = \frac{1}{4} \left( y^{6} + \frac{1}{y^{6}} - 2 \right) \] 3. Now, we will use the arc length formula: \[ L = \int_{1}^{2} \sqrt{1 + \left( \frac{df}{dy} \right)^{2}} \, dy \] Which simplifies to: \[ L = \int_{1}^{2} \sqrt{1 + \frac{1}{4} \left( y^{6} + \frac{1}{y^{6}} - 2 \right)} \, dy \] This further simplifies to: \[ L = \int_{1}^{2} \sqrt{\frac{1}{4} \left( y^{6} + 4 + \frac{1}{y^{6}} - 2 \right)} \, dy = \int_{1}^{2} \sqrt{\frac{1}{4}(y^{6} + 2 + \frac{1}{y^{6}})} \, dy \] 4. We can write: \[ L = \frac{1}{2} \int_{1}^{2} \sqrt{y^{6} + 2 + \frac{1}{y^{6}}} \, dy \] Using the identity \( y^{6} + 2 + \frac{1}{y^{6}} = \left( y^{2} + \frac{1}{y^{2}} \right)^{3} - 3\left( y^{2} + \frac{1}{y^{2}} \right) + 2 \), this way we can explore a simpler integral if possible. 5. The final integral can be calculated either analytically or numerically over the evaluated limits from \( y = 1 \) to \( y = 2 \). After solving, the arc length \( L \) evaluates to: \[ L = \frac{1}{2} \int_{1}^{2} (y^{3} + \frac{1}{y^{3}}) \, dy = \frac{1}{2} \left[ \frac{y^{4}}{4} - \frac{1}{2y^{2}} \right]_{1}^{2} \] Eventually, computing all the terms will provide the arc length of the curve from \( y = 1 \) to \( y = 2 \). Therefore, the arc length, when calculated completely, is found to be: \[ L = \frac{9}{4} \] after simplification and evaluation of constants.

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