计算曲线x=3t,y=3t^2,z=2t^3从(0,0,0)到(3,3,2)一段弧长

Question

计算曲线x=3t,y=3t^2,z=2t^3从(0,0,0)到(3,3,2)一段弧长

Answer 1

计算曲线x=3t,y=3t²,z=2t³,从(0,0,0)到(3,3,2)一段弧长
dx/dt=3,dy/dt=6t,dz/dt=6t²,当t=0时,x=0,y=0,z=0；当t=1时,x=3,y=3,z=2.
故弧A(0,0,0)B(3,3,2)对应于参数t=0与t=1,所以A⌒B弧长S:
S=(0,1)∫[√(9+36t²+36t⁴)]dt=(0,1) 3∫[√(1+4t²+4t⁴)]dt=(0,1) 3∫[√(1+2t²)²]dt
=(0,1) 3∫(1+2t²)dt=3[t+(2/3)t³](0,1)=3[1+(2/3)]=5.