Question: Find the arc length of the following curve on the given interval. \[ y = \frac{1}{3} x^{3/2}…

Find the arc length of the following curve on the given interval.

\[ y = \frac{1}{3} x^{3/2} \text{ on } [0, 54] \]

Solution

To find the arc length of the curve \( y = \frac{1}{3} x^{3/2} \) on the interval \([0, 54]\), we use the arc length formula: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] First, find the derivative \(\frac{dy}{dx}\). Given: \[ y = \frac{1}{3} x^{3/2} \] Differentiate with respect to \(x\): \[ \frac{dy}{dx} = \frac{1}{3} \cdot \frac{3}{2} x^{1/2} = \frac{1}{2} x^{1/2} \] Next, find \(\left(\frac{dy}{dx}\right)^2\): \[ \left( \frac{dy}{dx} \right)^2 = \left( \frac{1}{2} x^{1/2} \right)^2 = \frac{1}{4} x \] Substitute into the arc length formula: \[ L = \int_{0}^{54} \sqrt{1 + \frac{1}{4} x} \, dx \] Simplify the expression under the square root: \[ L = \int_{0}^{54} \sqrt{\frac{4 + x}{4}} \, dx = \int_{0}^{54} \frac{\sqrt{4 + x}}{2} \, dx \] Now solve the integral: \[ L = \frac{1}{2} \int_{0}^{54} \sqrt{4 + x} \, dx \] Use the substitution \(u = 4 + x\), then \(du = dx\) and when \(x = 0\), \(u = 4\), when \(x = 54\), \(u = 58\): \[ L = \frac{1}{2} \int_{4}^{58} \sqrt{u} \, du \] Integrate: \[ L = \frac{1}{2} \cdot \left[ \frac{2}{3} u^{3/2} \right]{4}^{58} \] \[ L = \frac{1}{2} \cdot \frac{2}{3} \left[ u^{3/2} \right]{4}^{58} \] \[ L = \frac{1}{3} \left[ 58^{3/2} - 4^{3/2} \right] \] Calculate the values: \[ 58^{3/2} = \sqrt{58^3} = \sqrt{195112} \approx 441.6729 \] \[ 4^{3/2} = \sqrt{4^3} = \sqrt{64} = 8 \] \[ L = \frac{1}{3} (441.6729 - 8) \] \[ L = \frac{1}{3} \times 433.6729 \approx 144.5576 \] The arc length of the curve on the interval \([0, 54]\) is approximately \(144.558\).