Find the arc length on the parabola y² = 2x + 1 cut off

Find the arc length on the parabola $y^2 = 2x + 1$ cut off by the line $x - y = 1$.

We find the intersections, then integrate with respect to $y$ using the standard $\sqrt{1+y^2}$ antiderivative.

Step 1: Find intersection points

From the line, $x=y+1$; substituting into the parabola: $y^2=2(y+1)+1 \implies y^2-2y-3=0 \implies (y-3)(y+1)=0$, so $y=3$ or $y=-1$, giving points $(4,3)$ and $(0,-1)$.

Step 2: Express $x$ as a function of $y$

\[x = \frac{y^2-1}{2} \implies \frac{dx}{dy}=y\]

Step 3: Set up the arc length

\[L = \int_{-1}^{3}\sqrt{1+y^2}\,dy\]

Step 4: Evaluate using the standard antiderivative

\[\int\sqrt{1+y^2}\,dy = \frac{y\sqrt{1+y^2}}{2}+\frac12\ln\left(y+\sqrt{1+y^2}\right)+C\]

At $y=3$: $\dfrac{3\sqrt{10}}{2}+\dfrac12\ln(3+\sqrt{10})$. At $y=-1$: $-\dfrac{\sqrt2}{2}+\dfrac12\ln(\sqrt2-1)$.

Step 5: Combine

\[L = \frac{3\sqrt{10}+\sqrt2}{2} + \frac12\left[\ln(3+\sqrt{10})-\ln(\sqrt2-1)\right]\]

Since $\dfrac{1}{\sqrt2-1}=\sqrt2+1$, this simplifies to:

\[\boxed{L = \frac{3\sqrt{10}+\sqrt2}{2} + \frac12\ln\left[(3+\sqrt{10})(\sqrt2+1)\right] \text{ units}}\]

Step 6: Numerical value

\[L \approx 5.449 + 1.352 \approx 6.800 \text{ units}\]

Summary

\[\boxed{L = \frac{3\sqrt{10}+\sqrt2}{2} + \frac12\ln\left[(3+\sqrt{10})(\sqrt2+1)\right] \approx 6.800 \text{ units}}\]

Key insight: Integrating with respect to $y$ instead of $x$ avoids the awkward two-branch square root $y=\pm\sqrt{2x+1}$ entirely, since $x$ is single-valued in $y$ along this arc — a routine but essential simplification for parabolas opening sideways.