Vector form of $vdv=adx$ [duplicate]

Question

I have studied that $vdv=adx$ in motion in 1D, it is derived as

$$a= \frac{dv}{dt}=\frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$$

I want to know if its vector form, $v \cdot dv= a \cdot dx$, is proper and how we would prove it.

Answer 1

If you want to see its vector form, you can repeat the previous derivation, but in 2D:

$$\vec{a} = \frac{d \vec{v}}{dt} = \frac{\partial \vec{v}}{\partial x_1} \frac{dx_1}{dt} + \frac{\partial \vec{v}}{\partial x_2} \frac{dx_2}{dt} = v_1 \frac{\partial \vec{v}}{\partial x_1} + v_2 \frac{\partial \vec{v}}{\partial x_2} = \vec{v} \cdot \vec{\nabla} \vec{v}$$

Be aware that the dot product happens between $\vec{v}$ and $\vec{\nabla}$. You need to know something about differential calculus. The same reasoning is valid also in 3D or any number of space dimensions.

Edit: The comments below refer to a previous version of the answer. Also, it is worth checking this and this related posts.