Question

求问高数题目

题目如图

Answer 1

曲线L：(x-y)^2=a(x+y)……① x^2-y^2=(9z^2)/8……②

令z=t，代入②
x^2-y^2=(9/8)*t^2
由①得，(x-y)^3=a(x^2-y^2)=(9a/8)*t^2
x-y=(9a/8)^(1/3)*t^(2/3)
由①得，x+y=[(x-y)^2]/a=(81/64a)^(1/3)*t^(4/3)
x=(81/512a)^(1/3)*t^(4/3)+(9a/64)^(1/3)*t^(2/3)
y=(81/512a)^(1/3)*t^(4/3)-(9a/64)^(1/3)*t^(2/3)
综上，曲线L的参数方程为：
x=(81/512a)^(1/3)*t^(4/3)+(9a/64)^(1/3)*t^(2/3)
y=(81/512a)^(1/3)*t^(4/3)-(9a/64)^(1/3)*t^(2/3)
z=t
其中t定义域介于0和z0之间
x'(t)=(3/8a)^(1/3)*t^(1/3)+(a/24)^(1/3)*t^(-1/3)
y'(t)=(3/8a)^(1/3)*t^(1/3)-(a/24)^(1/3)*t^(-1/3)
z'(t)=1
则(0,0,0)到(x0,y0,z0)的弧长

=∫(0,z0) √[x'(t)^2+y'(t)^2+1]dt
=∫(0,z0) √[2*(3/8a)^(2/3)*t^(2/3)+2*(a/24)^(2/3)*t^(-2/3)+1]dt
=√2*∫(0,z0) [(3/8a)^(1/3)*t^(1/3)+(a/24)^(1/3)*t^(-1/3)]dt
=√2*[(3/4)*(3/8a)^(1/3)*t^(4/3)+(3/2)*(a/24)^(1/3)*t^(2/3)]|(0,z0)
=(3√2/4)*[(3/8a)^(1/3)*z0^(4/3)+(a/3)^(1/3)*z0^(2/3)]