Partial Derivative

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  • #1
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Main Question or Discussion Point

I have the equation

[itex]\frac{d\rho}{dt}[/itex]=-[itex]\nabla[/itex][itex]\cdot[/itex][itex]\rho v[/itex]

where the vector v depends only x and t.

I want to take the partial derivative of this whole equation with respect to t.

Just not sure how to take the partial of the divergence. Thanks!
 

Answers and Replies

  • #2
SteamKing
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Why don't you write out the equivalent equation, replacing the del operator with the partial derivatives incorporated in its meaning?
 
  • #3
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[itex]\frac{\partial}{\partial t} \left(\nabla\cdot\rho v\right)[/itex]

v a vector again x component only

[itex]\frac{\partial\rho}{\partial t}[/itex][itex]\cdot v[/itex]+[itex]\rho[/itex][itex]\frac{dv}{dx}[/itex]

knowing rho is with respect to time and v with x.

Yay or nay?
 
  • #4
HallsofIvy
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Well, that's not the same problem now! In your first post you stated that v was a function of both x and t and did not say anything about [itex]\rho[/itex]. Assuming the most general case, that both v and [itex]\rho[/itex] are functions of both x and t, and writing v= f(x,t)i+ g(x,t)j+ h(x,t)k then we have [itex]\rho v= \rho(x,t)f(x,t)i+ \rho(x,t)g(x,t)j+ \rho(x,t)h(x,t)k[/itex] and
[tex]\nabla\cdot \rho v= \frac{\partial \rho f}{\partial x}+ \frac{\partial\rho g}{\partial y}+ \frac{\partial \rho h}{\partial z}[/tex]
[tex]= \frac{\partial\rho}{\partial x}f+ \rho \frac{\partial f}{\partial x}+ \frac{\partial\rho}{\partial y}g+ \rho \frac{\partial g}{\partial y}+ \frac{\partial\rho}{\partial z}h+ \rho\frac{\partial h}{\partial z}[/tex]
[tex]= \nabla\rho\cdot v+ \rho \nabla\cdot v[/tex]
 

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