the potential energy of a function in a region of space is given as :

U=2x^2 + 3y^3 + 2z [ 2x*x + 3y*y*y + 2z]

here, x,y,z are in metres. find the force acting on the particle at point p ( 1m, 2m, 3m)

Question

the potential energy of a function in a region of space is given as :

U=2x^2 + 3y^3 + 2z     [ 2x*x + 3y*y*y + 2z]

here, x,y,z are in metres. find the force acting on the particle at point p ( 1m, 2m, 3m)

anonymous · Answer

Assuming that the origin of the force is at the origin of the co-ordinate system, $$F=\frac{U}{d}$$$$=\frac{2(m)^2 + 3 (2m)^3 + 2 (3m)}{\sqrt{m^2 + (2m)^2 + (3m)^2}}$$

anonymous · Answer

insert numbers

anonymous · Answer

its not the correct answer 

the correct answer is here.

f = -dU/Dx

f= d/dx (2x^2 + 3y^3 + 2z )

f= 4x + 9y^2 + 2

substituting the coordinates of the point u get

f= 4i + 36j + 2k

anonymous · Answer

Surely it should be $$F=-\frac{\partial}{\partial x}\frac{\partial}{\partial y}\frac{\partial}{\partial z} U \text{   ?}$$Unless the force doesn't vary over y, and z (but I suspect it does).