Process Systems Analysis and Control (3rd Edition) chap.5 problem 2

I want to know what happens to the differential parts.

Problem A liquid-level system, such as the one shown in Fig. 5-1, has a cross-sectional area of $3.0 \mathrm{ft} 2$. The valve characteristics are \[ q=8 \sqrt{h} \] where $q=$ flow rate, $\mathrm{cfm}$, and $h=$ level above the valve, ft. Calculate the time constant for this system if the average operating level above the valve is (a) $3 \mathrm{ft}$ (b) $9 \mathrm{ft}$

Step 1 of 4 Liquid Level System: Given that, - The cross-sectional area of the tank, $A=3.0 \mathrm{ft}^{2}$. - The valve characteristics are given as the flow rate through it by the equation $q=8 \sqrt{h}$; Here $h$ is the level above the valve. From the characteristic equation it is clear that the $\mathrm{h}$ term is of degree not equal to 1 it is $1 / 2$ so it is of linear structure. Comment Step 2 of 4 By taking the derivative of $q$ with respect to $\mathrm{h}$ will give the reciprocal of resistance of the valve. That is, \[ \begin{aligned} \frac{1}{R} & =\frac{d q}{d h} \\ & =\frac{d(8 \sqrt{h})}{d h} \\ & =8 \frac{1}{2}(h)^{-1 / 2} \\ R & =2 \frac{(h)^{1 / 2}}{8} \end{aligned} \] The liquid level system are first order systems. And the time constant for the liquid level system are calculated from, $\tau=R A$.