Find the arc length of the curve x = (t^6)/6, y = 2

Find the arc length of the curve $x = \frac{t^6}{6}$, $y = 2 - \frac{t^4}{4}$ between the intersection points with the coordinate axes.

We locate the axis intercepts, then integrate using a substitution that lands on nice round numbers.

Step 1: Intersection with the $y$-axis ($x=0$)

$\frac{t^6}{6}=0 \implies t=0$, giving $y=2$, i.e. the point $(0,2)$.

Step 2: Intersection with the $x$-axis ($y=0$)

$2-\frac{t^4}{4}=0 \implies t^4=8 \implies t=\pm2^{3/4}$ (giving the same point $x=\frac{8\sqrt2}{3}$ either way, by the even powers involved).

Step 3: Compute derivatives

$\frac{dx}{dt}=t^5$, $\frac{dy}{dt}=-t^3$.

Step 4: Set up the arc length (from $t=0$ to $t=2^{3/4}$)

\[L = \int_0^{2^{3/4}}\sqrt{t^{10}+t^6}\,dt = \int_0^{2^{3/4}}t^3\sqrt{t^4+1}\,dt\]

Step 5: Substitute $u=t^4+1$

$du=4t^3\,dt$. When $t=0$: $u=1$; when $t=2^{3/4}$: $u=(2^{3/4})^4+1=2^3+1=9$.

\[L = \frac14\int_1^9\sqrt u\,du = \frac14\left[\frac{2u^{3/2}}{3}\right]_1^9 = \frac16\left(9^{3/2}-1\right) = \frac16(27-1) = \frac{26}{6} = \frac{13}{3}\]

\[\boxed{L = \frac{13}{3} \text{ units}}\]

Summary

\[\boxed{L = \frac{13}{3} \text{ units}}\]

Key insight: Although the curve's own coordinates involve sixth and fourth powers of $t$, the arc-length integrand simplifies to $t^3\sqrt{t^4+1}$, exactly matched to the substitution $u=t^4+1$, which lands on the clean bound $u=9$ — producing the tidy rational answer $\frac{13}{3}$.