Derivation of Acceleration from Velocity with Partial derivatives

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To derive acceleration from velocity using partial derivatives, it's essential to view velocity as a function of time and spatial variables: V(t, x, y, z). The acceleration is expressed as the total derivative of velocity, which incorporates both the time derivative and the spatial derivatives through the chain rule. The formula for acceleration combines these derivatives, resulting in the equation that includes the partial derivatives of velocity with respect to time and space. Understanding the chain rule for partial derivatives is crucial for grasping this concept. This mathematical approach is fundamental in fluid mechanics for analyzing motion.
fluidmech
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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
 
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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:
fluidmech said:
I see, I'm still a bit hazy on the mathematics of the partials, would you mind elaborating on that?

Look up the chain rule for partial derivatives. E.g. http://mathworld.wolfram.com/ChainRule.html
 
That helped me tremendously. Now I understand it, thank you!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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