F(x,y,z)=< y2z3,2xyz3,3xy2z2> Is this conservative or not. I got it to be not conservative. Please check?
回答 (5)
2015-12-05 9:06 pm
F(x,y,z)= < y²z³, 2xyz³, 3xy²z² > If conservative then curl F = 0
I ....i........j............k...I
I ..∂/∂x..∂/∂y....∂/∂z...I
I..y²z³..2xyz³,..3xy²z²I = i [(∂/∂y(3xy²z²)-∂/∂z(2xyz³)] - j [∂/∂x(3xy²z²)-∂/∂z(y²z³)] + k[∂/∂x(2xyz³)-∂/∂y(y²z³)] =
......................................i (6xyz²-6xyz²) - j (3y²z²-3y²z²) + k (2xyz³-2xyz³) = 0 => conservative
I ....i........j............k...I
I ..∂/∂x..∂/∂y....∂/∂z...I
I..y²z³..2xyz³,..3xy²z²I = i [(∂/∂y(3xy²z²)-∂/∂z(2xyz³)] - j [∂/∂x(3xy²z²)-∂/∂z(y²z³)] + k[∂/∂x(2xyz³)-∂/∂y(y²z³)] =
......................................i (6xyz²-6xyz²) - j (3y²z²-3y²z²) + k (2xyz³-2xyz³) = 0 => conservative
2015-12-05 8:06 pm
For any conservative force field, partial d(f(x,y,z))/DX= partial d(f(x,y,z))/Dy= partial df(x,y,z))/dz, which is not true here, so it is non conservative
2015-12-05 7:19 pm
You get that right.
2015-12-05 7:02 pm
No
收錄日期: 2021-04-21 15:38:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151205110120AAU9dMs