Compute the arc length of the parametric curve x = t^2, y = t^3 from t = 0 to t = 1.

• A. (4√13 - 2)/9
• B. (8√13 - 4)/27
• C. (13√13 - 8)/27
• D. (9√13 - 5)/27

Answer: (C)

The arc length for a parametric curve uses the derivatives of the components: L = ∫ from t=a to t=b of sqrt[(dx/dt)^2 + (dy/dt)^2] dt. For x = t^2 and y = t^3, dx/dt = 2t and dy/dt = 3t^2, so the integrand is sqrt(4t^2 + 9t^4) = t sqrt(4 + 9t^2) on [0,1] (t is nonnegative there).

Thus L = ∫_0^1 t sqrt(4 + 9t^2) dt. Let u = 4 + 9t^2, so du = 18t dt. This gives L = (1/18) ∫4^{13} sqrt(u) du = (1/18) * (2/3) u^{3/2} |{4}^{13} = (1/27) [13^{3/2} − 4^{3/2}] = (1/27)(13√13 − 8).

Therefore, the arc length is (13√13 − 8)/27.