Derivation of Acceleration from Velocity with Partial derivatives

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 7K views
fluidmech
Messages
17
Reaction score
1

Homework Statement


I'm taking a fluid mechanics class and I'm having an issue with acceleration and background knowledge. I know this is ridiculous, but I was hoping someone might be able to explain it for me.

Homework Equations


I definitely understand:
##a=\frac{d\vec{V}}{dt}##

And I know that u, v, and w are components of the velocity, ##\vec{V}=<u,v,w>##

But how do I use the chain rule of differentiation to get to:

##\vec{a}=\frac{d\vec{V}}{dt}=\frac{\partial \vec{V}}{\partial t} +\frac{\partial \vec{V}}{\partial x}\frac{dx}{dt} +\frac{\partial \vec{V}}{\partial y}\frac{dy}{dt} +\frac{\partial \vec{V}}{\partial z}\frac{dz}{dt}##

Thanks in advance!

- Matt
 
Physics news on Phys.org
fluidmech said:

Homework Statement


I'm taking a fluid mechanics class and I'm having an issue with acceleration and background knowledge. I know this is ridiculous, but I was hoping someone might be able to explain it for me.

Homework Equations


I definitely understand:
##a=\frac{d\vec{V}}{dt}##

And I know that u, v, and w are components of the velocity, ##\vec{V}=<u,v,w>##

But how do I use the chain rule of differentiation to get to:

##\vec{a}=\frac{d\vec{V}}{dt}=\frac{\partial \vec{V}}{\partial t} +\frac{\partial \vec{V}}{\partial x}\frac{dx}{dt} +\frac{\partial \vec{V}}{\partial y}\frac{dy}{dt} +\frac{\partial \vec{V}}{\partial z}\frac{dz}{dt}##

Thanks in advance!

- Matt

You want to think of V as a function of four variables V(t,x,y,z).
 
I see, I'm still a bit hazy on the mathematics of the partials, would you mind elaborating on that?
 
Last edited:
That helped me tremendously. Now I understand it, thank you!