One method of delivering water at a slow rate for irrigation

Solution

a. The differential equation for the height of water in the reservoir is written in the following form, using a power (1/2) for the square root,
dh /dt = -sqrt(0.07853h)
This differential equation is solved using the separation of variables technique. The variables are separated with the dependent variable h in the integral on the left below and the independent variable t in the integral on the right below. Thus, we have the two integrals below to solve
Integral (h ^ -1/2) dh = - Integral (0.07835) dt

These two integrals are easily solved, giving the following equation
2h^1/2 = - 0.07835t + C

This equation is solved explicitly for h(t) by dividing by 2 and squaring both sides, resulting in the equation
h(t) = [C/2 – 0.07835t]^2

Next we use the initial condition h(0) = 91 to find the constant C. With the initial condition, it follows that
h(0) = 91 = (C/2)^2

or
C = 19
Thus, the solution is given by the equation
h(t) = (9.53 – 0.07835t)2.



b. The reservoir is empty when h(t) = 0. Thus, we must solve the following:

h(t) = (9.53 - 0.07835t)2 = 0 or 9.53 - 0.07835t= 0.
It follows that
0.07835t =9.53
or
t = 121.75 hr.

The reservoir empties in 121.75 hr.or 5.07 days