題目 3. (10 Points) Find the length of the parametric curve x=3t2,y=2t3,0≤t≤1.x=3t^2,\quad y=2t^3,\quad 0\le t\le 1.x=3t2,y=2t3,0≤t≤1. 解答 解法一 dxdt=6t,dydt=6t2.\frac{\mathrm{d}x}{\mathrm{d}t}=6t,\qquad \frac{\mathrm{d}y}{\mathrm{d}t}=6t^2.dtdx=6t,dtdy=6t2. 故弧長為 ∫0136t2+36t4 dt=∫016t1+t2 dt=2(22−1).\int_0^1\sqrt{36t^2+36t^4}\,\mathrm{d}t =\int_0^1 6t\sqrt{1+t^2}\,\mathrm{d}t =\boxed{2(2\sqrt2-1)}.∫0136t2+36t4dt=∫016t1+t2dt=2(22−1).
