Find the arc length of the curve y² = x³ cut off

Find the arc length of the curve $y^2 = x^3$ cut off by the line $x = \frac{4}{3}$.

We use the upper branch of Neil's parabola and a linear substitution.

Step 1: Set up using the upper branch

$y=x^{3/2} \implies \dfrac{dy}{dx}=\dfrac{3\sqrt x}{2}$. By symmetry (doubling for the lower branch):

\[L = 2\int_0^{4/3}\sqrt{1+\frac{9x}{4}}\,dx = \int_0^{4/3}\sqrt{4+9x}\,dx\]

Step 2: Substitute $u=4+9x$

$du=9\,dx$. When $x=0$: $u=4$; when $x=\frac43$: $u=4+12=16$.

\[L = \frac19\int_4^{16}\sqrt u\,du = \frac{2}{27}\left[u^{3/2}\right]_4^{16} = \frac{2}{27}(64-8) = \frac{112}{27}\]

\[\boxed{L = \frac{112}{27} \text{ units}}\]

Step 3: Numerical value

\[L = \frac{112}{27} \approx 4.148 \text{ units}\]

Summary

\[\boxed{L = \frac{112}{27} \approx 4.148 \text{ units}}\]

Key insight: Neil's semicubical parabola was historically the first algebraic curve (other than the line) to have its arc length computed in closed form (William Neile, 1657) — the linear substitution $u=4+9x$ makes the computation entirely elementary.