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I hope my question is not too ridiculous, but why can differentiate a physical equation in the following manner:

$$\rho VA=\textrm{constant}$$

$$\frac{d\rho}{\rho}+\frac{dA}{A}+\frac{dV}{V}=0$$

It is the equation of continuity at steady state, and I understand that all the members are then divided by $\rho VA$ but why can we cancel $dt$ as if it was a constant?

Przemysław Scherwentke
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    Take the derivative of both sides (what is the derivative of a constant?) and apply the product rule to the left side. Where is this $dt$ you mention? – nluigi Dec 20 '17 at 14:46
  • Well if you want to take the derivative you have to do something like $\frac{d}{dt}$ right? I know that the derivative of $\frac{d( \rho AV)}{dt}=0$ but then in the expression there will be this $dt$ term, how can I cancel a differential element? –  Dec 20 '17 at 14:53
  • The derivative is not with respect to t. It should be with respect to x along the duct. –  Dec 20 '17 at 15:11
  • Why should the properties be constant along the duct? And also how would i be able to cancel the $dx$? –  Dec 20 '17 at 15:19
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    LHS: $d(\rho V A) = d(\rho)VA + \rho d(V)A + \rho Vd(A)$. RHS: $d(constant) = 0$. Now divide by $\rho VA$. See more here. – JM1 Dec 20 '17 at 15:35
  • But can you take a derivative without making it with respect to something, i.e, $d/d()$? –  Dec 20 '17 at 15:36
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    You're not finding the derivative. You're finding the differential. – JM1 Dec 20 '17 at 15:42
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    $\rho V A$ represents the mass flow rate along the duct, and, at steady state, by conservation of mass, it is certainly going to be constant along the duct. Each individual parameter is not necessarily going to be constant along the duct, but the product of the three will be. So, if you take the derivative of the mass flow rate with respect to position along the duct, you get: $$\frac{d\rho}{dx}VA+\rho\frac{dV}{dx}A+\rho V\frac{dA}{dx}=0$$, or, equivalently $$\frac{1}{\rho}\frac{d\rho}{dx}+\frac{1}{V}\frac{dV}{dx}+\frac{1}{A}\frac{dA}{dx}=0$$ –  Dec 20 '17 at 15:44

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