1. **State the problem:** We need to find the arc length function $s(x)$ for the curve given by $$y = x^2 - \frac{1}{8} \log x$$ starting from the point $P_0(1,1)$. 2. **Recall the arc length formula:** The arc length $s$ from $x=a$ to $x=b$ for a curve $y=f(x)$ is given by $$s = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$ 3. **Find the derivative $\frac{dy}{dx}$:** Given $$y = x^2 - \frac{1}{8} \log x,$$ we differentiate term-by-term: $$\frac{dy}{dx} = 2x - \frac{1}{8} \cdot \frac{1}{x} = 2x - \frac{1}{8x}$$ 4. **Square the derivative:** $$\left(\frac{dy}{dx}\right)^2 = \left(2x - \frac{1}{8x}\right)^2 = (2x)^2 - 2 \cdot 2x \cdot \frac{1}{8x} + \left(\frac{1}{8x}\right)^2 = 4x^2 - \frac{1}{2} + \frac{1}{64x^2}$$ 5. **Add 1 inside the square root:** $$1 + \left(\frac{dy}{dx}\right)^2 = 1 + 4x^2 - \frac{1}{2} + \frac{1}{64x^2} = \frac{1}{2} + 4x^2 + \frac{1}{64x^2}$$ 6. **Simplify the expression under the root:** Rewrite as $$4x^2 + \frac{1}{2} + \frac{1}{64x^2}$$ Try to express as a perfect square: $$\left(2x - \frac{1}{8x}\right)^2 = 4x^2 - \frac{1}{2} + \frac{1}{64x^2}$$ Our expression is $$4x^2 + \frac{1}{2} + \frac{1}{64x^2} = \left(2x + \frac{1}{8x}\right)^2$$ 7. **Therefore, the integrand simplifies to:** $$\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{\left(2x + \frac{1}{8x}\right)^2} = \left|2x + \frac{1}{8x}\right|$$ Since $x > 0$ (logarithm domain), this is $$2x + \frac{1}{8x}$$ 8. **Set up the arc length function starting at $x=1$:** $$s(x) = \int_1^x \left(2t + \frac{1}{8t}\right) dt$$ 9. **Integrate term-by-term:** $$\int 2t \, dt = t^2$$ $$\int \frac{1}{8t} \, dt = \frac{1}{8} \log t$$ 10. **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 $\log 1 = 0$, this simplifies to $$s(x) = x^2 + \frac{1}{8} \log x - 1$$ **Final answer:** $$\boxed{s(x) = x^2 + \frac{1}{8} \log x - 1}$$ This function gives the arc length of the curve from $x=1$ to any $x > 0$.