Proving the Work-Energy Theorem A lot of calculus, anyone?

[Elysian]

Homework Statement

A particle moves on a curved path from (x₁, y₁,z₁) to (x₂, y₂,z₂). At the start, the particle has a velocity of v = v_1xi+v_1yj+v_1zk. This curved path can be divided into segments infinitesimally, which are, dl = dxi +dyj +dzk. It is acted on by a net force F = Fxj + Fyi + Fzk. The force components Fx, Fy, and Fz, are in general functions of position. Prove the work energy theorem for this general case. That is prove that

W_tot=K₂-K₁

where

W_tot=\intF dl = \int F_xdx + F_ydy + F_zdz

where the limits of integration are from (x₁, y₁,z₁) to (x₂, y₂,z₂) for each.

Homework Equations

W = Fd
∫F=W

a_x= \stackrel{dvx}{dt} = \stackrel{dvx}{dx} *\stackrel{dx}{dt} = v_x\stackrel{dvx}{dx}

The Attempt at a Solution

I don't really have any idea where to start but what I did was I took the velocities and made them into accelerations then changed the Fx, Fy, and Fz, into max, may, and maz values, then I'm confused now because I have an integral which looks like this..

\int m*v_x*dv_x + m*v_y*dv_y/dx dy + m*v_z*dv_z/dx dz

which makes no sense because the dx only canceled out for the X direction, and I need to prove that W_tot = 1/2 mvf^2 - 1/2mvi^2


If anyone could help it'd be appreciated because I've got really no idea what I am doing

 

[Harrisonized]

Bolded means vectors. Note that x = (x₁,x₂,...x_n). The period denotes dot product.

Let W = ∫F.dx
F = m(d²x/dt²)

W = m∫(d²x/dt²).dx
W = m∫(dv/dt).dx

Note that this step is somewhat hand-wavy but it's true. I think it would be a good exercise for you to prove this. To do that, you first need to learn about how line integrals work, and the process of parametrization. Hint: dx=(dx/dt)dt

W = m∫dv.(dx/dt)
W = m∫(dx/dt).dv
W = m∫v.dv
W = m(1/2 v.v) = 1/2 m|v|²

Is your name Ping?