There is a derivation of a formula in my textbook which I don't fully understand. Most of the formulas I encounter I understand without difficulty, but if someone can help me understand one step of the following derivation, I would be very grateful:
Consider a flow of fluid through an inclined core sample of length $\Delta l$, with a constant flow rate, $q$, maintained at a pressure differential $\Delta p$. The flow at an angle $\theta$ above the horizontal can be described by the following version of the Darcy equation:
$q = -A\frac{k}{\mu}\frac{d(p + \rho g z)}{dl}$
where $z$ is the elevation in the gravitational field. Since $z = l \sin \theta$, with $l$ as the direction of flow, the equation written for the pressure gradient, becomes:
$\frac{dp}{dl} = - \left(\frac{q \mu}{Ak} + \rho g \sin \theta\right)$
OK, so it is this last step here I don't quite understand. I see that from the given equation we have:
$q = -A\frac{k}{\mu}\frac{d(p + \rho g z)}{dl}$
$\frac{d(p + \rho g z)}{dl} = - \frac{q \mu}{Ak}$
$\frac{d(p + \rho g l \sin \theta)}{dl} = -\frac{q \mu}{Ak}$
But I don't see this last step from here on. How do you "extract" the $\sin \theta$ part from the differential expression on the left side of this equation?
If anyone can explain this to me, I would appreciate it a lot!