Given: \[ y = \frac{3}{5}x^{5/3} - \frac{3}{4}x^{1/3} + 6 \] for \( 1 \leq x \leq 27 \). We are to find the **length of the curve**. --- ### **Step 1: Formula for Arc Length** The arc length from \( x=a \) to \( x=b \) is: \[ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] --- ### **Step 2: Find \( \frac{dy}{dx} \)** Given: \( y = \frac{3}{5}x^{5/3} - \frac{3}{4}x^{1/3} + 6 \) Take the derivative: - \( \frac{d}{dx}\left(\frac{3}{5}x^{5/3}\right) = \frac{3}{5} \cdot \frac{5}{3} x^{2/3} = x^{2/3} \) - \( \frac{d}{dx}\left(-\frac{3}{4}x^{1/3}\right) = -\frac{3}{4} \cdot \frac{1}{3} x^{-2/3} = -\frac{1}{4}x^{-2/3} \) - \( \frac{d}{dx}(6) = 0 \) So, \[ \frac{dy}{dx} = x^{2/3} - \frac{1}{4}x^{-2/3} \] --- ### **Step 3: Plug into the Arc Length Formula** \[ L = \int_{1}^{27} \sqrt{1 + \left( x^{2/3} - \frac{1}{4}x^{-2/3} \right)^2} \, dx \] --- #### **Expand the Square** \[ \left( x^{2/3} - \frac{1}{4}x^{-2/3} \right)^2 = \left(x^{2/3}\right)^2 - 2 \cdot x^{2/3} \cdot \frac{1}{4}x^{-2/3} + \left(\frac{1}{4}x^{-2/3}\right)^2 \] \[ = x^{4/3} - \frac{1}{2} + \frac{1}{16}x^{-4/3} \] Now, \[ 1 + \left( x^{2/3} - \frac{1}{4}x^{-2/3} \right)^2 = 1 + x^{4/3} - \frac{1}{2} + \frac{1}{16}x^{-4/3} \] \[ = x^{4/3} + \frac{1}{2} + \frac{1}{16}x^{-4/3} \] --- ### **Step 4: Simplify the Square Root** \[ \sqrt{x^{4/3} + \frac{1}{2} + \frac{1}{16}x^{-4/3}} \] Let's try to factor or rewrite the expression: Let’s try: \[ \left(x^{2/3} + \frac{1}{4}x^{-2/3}\right)^2 = (x^{2/3})^2 + 2x^{2/3} \cdot \frac{1}{4}x^{-2/3} + \left(\frac{1}{4}x^{-2/3}\right)^2 \] \[ = x^{4/3} + \frac{1}{2} + \frac{1}{16}x^{-4/3} \] So, \[ x^{4/3} + \frac{1}{2} + \frac{1}{16}x^{-4/3} = \left(x^{2/3} + \frac{1}{4}x^{-2/3}\right)^2 \] So the integrand is: \[ \sqrt{\left(x^{2/3} + \frac{1}{4}x^{-2/3}\right)^2} = \left|x^{2/3} + \frac{1}{4}x^{-2/3}\right| \] But for \( x > 0 \), this is just \( x^{2/3} + \frac{1}{4}x^{-2/3} \). --- ### **Step 5: Integrate** \[ L = \int_{1}^{27}\left( x^{2/3} + \frac{1}{4}x^{-2/3} \right) dx \] Split: \[ L = \int_1^{27} x^{2/3} dx + \int_1^{27} \frac{1}{4} x^{-2/3} dx \] Integrate each term: **First term:** \[ \int x^{2/3} dx = \frac{3}{5} x^{5/3} \] **Second term:** \[ \int x^{-2/3} dx = \frac{3}{1} x^{1/3} \] So, \[ \int \frac{1}{4} x^{-2/3} dx = \frac{3}{4} x^{1/3} \] So, \[ L = \left[ \frac{3}{5}x^{5/3} + \frac{3}{4}x^{1/3} \right]_{1}^{27} \] --- ### **Step 6: Evaluate from 1 to 27** Plug in \( x = 27 \): - \( 27^{5/3} = (27^{1/3})^5 = 3^5 = 243 \) - \( 27^{1/3} = 3 \) Plug in \( x = 1 \): - \( 1^{5/3} = 1 \) - \( 1^{1/3} = 1 \) So, \[ L = \left[ \frac{3}{5}(243) + \frac{3}{4}(3) \right] - \left[ \frac{3}{5}(1) + \frac{3}{4}(1) \right] \] \[ = \left[ \frac{729}{5} + \frac{9}{4} \right] - \left[ \frac{3}{5} + \frac{3}{4} \right] \] Find a common denominator (20): - \( \frac{729}{5} = \frac{2916}{20} \) - \( \frac{9}{4} = \frac{45}{20} \) - \( \frac{3}{5} = \frac{12}{20} \) - \( \frac{3}{4} = \frac{15}{20} \) So, \[ L = \left( \frac{2916}{20} + \frac{45}{20} \right) - \left( \frac{12}{20} + \frac{15}{20} \right) \] \[ = \frac{2961}{20} - \frac{27}{20} \] \[ = \frac{2961 - 27}{20} \] \[ = \frac{2934}{20} \] \[ = \frac{1467}{10} \] --- ## **Final Answer** \[ \boxed{\frac{1467}{10}} \] The length of the curve is \(\frac{1467}{10}\).

The arc length of a curve is found by integrating the square root of \(1 + (\frac{dy}{dx})^2\) over the given interval, which accounts for the infinitesimal distance along the curve.