Calculate the arc length of the curve x = (t² - 2) sin t

Calculate the arc length of the curve $x = (t^2 - 2)\sin t + 2t\cos t$, $y = (2 - t^2)\cos t + 2t\sin t$ from $t = 0$ to $t = \pi$.

We differentiate both components carefully; a substantial cancellation occurs in each.

Step 1: Differentiate $x(t)$

\[\frac{dx}{dt} = 2t\sin t+(t^2-2)\cos t + 2\cos t - 2t\sin t = (t^2-2)\cos t+2\cos t = t^2\cos t\]

Step 2: Differentiate $y(t)$

\[\frac{dy}{dt} = -2t\cos t-(2-t^2)\sin t+2\sin t+2t\cos t = -(2-t^2)\sin t+2\sin t = t^2\sin t\]

Step 3: Simplify the arc-length integrand

\[\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2 = t^4\cos^2t+t^4\sin^2t = t^4 \implies \sqrt{\cdots} = t^2\]

Step 4: Integrate

\[L = \int_0^{\pi}t^2\,dt = \left[\frac{t^3}{3}\right]_0^{\pi} = \frac{\pi^3}{3}\]

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

Step 5: Numerical value

\[L = \frac{\pi^3}{3} \approx 10.335 \text{ units}\]

Summary

\[\boxed{L = \frac{\pi^3}{3} \approx 10.335 \text{ units}}\]

Key insight: This curve is deliberately constructed (an involute-style construction) so that the messy product-rule terms cancel almost entirely, leaving the strikingly clean derivative pair $(t^2\cos t,\ t^2\sin t)$ and an arc-length integrand of bare $t^2$.