Find the arc length of the curve y² = 4/9(2 - x)³ cut off

Find the arc length of the curve $y^2 = \dfrac{4}{9}(2 - x)^3$ cut off by the line $x = -1$.

The curve $y=\pm\frac23(2-x)^{3/2}$ has a cusp at $(2,0)$ and is symmetric about the $x$-axis.

Step 1: Endpoints

At $x=-1$: $y^2=\frac49\cdot27=12$, so $y=\pm2\sqrt3$. At $x=2$: $y=0$ (the cusp).

Step 2: Differentiate the upper branch

\[y=\frac23(2-x)^{3/2} \implies y'=-\sqrt{2-x}\]

Step 3: Simplify $1+(y')^2$

\[1+(y')^2 = 1+(2-x) = 3-x\]

Step 4: Integrate

\[L_{\text{upper}}=\int_{-1}^{2}\sqrt{3-x}\,dx\]

With $u=3-x$: $L_{\text{upper}}=\int_1^4\sqrt u\,du=\left[\frac23u^{3/2}\right]_1^4=\frac23(8-1)=\frac{14}3$

Step 5: Double for both branches

By symmetry the lower branch contributes the same length, so

\[L=2\cdot\frac{14}3=\frac{28}3\]

\[\boxed{L=\frac{28}3\approx9.333}\]