1. **State the problem:** Find the arc length function for the curve $$y = x^2 - \frac{1}{8} \log x$$ starting from the point $$P_0(1,1)$$. 2. **Formula for arc length function:** The arc length function from $$x=a$$ to $$x$$ is given by $$ S(x) = \int_a^x \sqrt{1 + \left(\frac{dy}{dt}\right)^2} \, dt $$ where $$\frac{dy}{dt}$$ is the derivative of $$y$$ with respect to $$t$$. 3. **Find the derivative $$\frac{dy}{dx}$$:** $$ y = x^2 - \frac{1}{8} \log x $$ Using derivative rules, $$ \frac{dy}{dx} = 2x - \frac{1}{8} \cdot \frac{1}{x} = 2x - \frac{1}{8x} $$ 4. **Square the derivative:** $$ \left(\frac{dy}{dx}\right)^2 = \left(2x - \frac{1}{8x}\right)^2 = 4x^2 - 2 \cdot 2x \cdot \frac{1}{8x} + \frac{1}{64x^2} = 4x^2 - \frac{1}{2} + \frac{1}{64x^2} $$ 5. **Set up the integrand:** $$ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + 4x^2 - \frac{1}{2} + \frac{1}{64x^2} = \frac{1}{2} + 4x^2 + \frac{1}{64x^2} $$ 6. **Simplify the expression under the square root:** $$ \frac{1}{2} + 4x^2 + \frac{1}{64x^2} = \left(2x - \frac{1}{8x}\right)^2 + 1 $$ But from step 4, we see it equals $$\left(2x - \frac{1}{8x}\right)^2 + 1$$, so the integrand is $$ \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{\frac{1}{2} + 4x^2 + \frac{1}{64x^2}} $$ 7. **Rewrite the integrand as a perfect square:** Try to write $$ \frac{1}{2} + 4x^2 + \frac{1}{64x^2} = \left(2x + \frac{1}{8x}\right)^2 $$ Calculate: $$ \left(2x + \frac{1}{8x}\right)^2 = 4x^2 + 2 \cdot 2x \cdot \frac{1}{8x} + \frac{1}{64x^2} = 4x^2 + \frac{1}{2} + \frac{1}{64x^2} $$ This matches exactly. 8. **Therefore, the integrand simplifies to:** $$ \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = 2x + \frac{1}{8x} $$ 9. **Set up the arc length function:** $$ S(x) = \int_1^x \left(2t + \frac{1}{8t}\right) dt $$ 10. **Integrate:** $$ \int 2t \, dt = t^2 $$ $$ \int \frac{1}{8t} \, dt = \frac{1}{8} \log t $$ 11. **Evaluate the definite integral:** $$ S(x) = \left[t^2 + \frac{1}{8} \log t\right]_1^x = \left(x^2 + \frac{1}{8} \log x\right) - \left(1 + \frac{1}{8} \log 1\right) $$ Since $$\log 1 = 0$$, $$ S(x) = x^2 + \frac{1}{8} \log x - 1 $$ **Final answer:** $$ \boxed{S(x) = x^2 + \frac{1}{8} \log x - 1} $$