Find the arc length of the curve x=13t3x = \frac{1}{3}t^3x=31t3, y=12t2y = \frac{1}{2}t^2y=21t2 for 0≤t≤10 \le t \le 10≤t≤1.
13(23/2−1)\frac{1}{3}(2^{3/2} - 1)31(23/2−1)
23(23/2−1)\frac{2}{3}(2^{3/2} - 1)32(23/2−1)
13(21/2−1)\frac{1}{3}(2^{1/2} - 1)31(21/2−1)
23(21/2−1)\frac{2}{3}(2^{1/2} - 1)32(21/2−1)