Two tanks containing a liquid are placed in series so that the first discharges into the second and the second discharges into a waste outlet. Let \(q_{1}(t)\) and \(q_{2}(t)\) be the flow rates out of the two tanks respectively, and let the height of liquid in each of the tanks be \(h_{1}(t)\) and \(h_{2}(t)\). respectively. The two tanks are identical and each has a constant cross-sectional area \(A\). The outflow from each tank is proportional to the height of liquid in the tank. At \(t=0\) the height of liquid in the first tank is \(h_{0}\) and the second tank is empty. (a) Derive and solve the differential equation for \(h_{1}(t)\). (b) Hence find \(q_{1}(t)\). (c) Derive and solve the differential equation for \(h_{2}(t)\). (d) Hence find \(q_{2}(t)\).

Since the outflow of the first tank is proportional to the height (\(h_1\)) of the liquid in the tank, we can represent the flow rate \(q_1(t)\) as a constant proportionality (\(k\)) times the height of the liquid in the first tank. $$ q_1(t) = k \cdot h_1(t) $$ Taking the derivative of \(h_1(t)\) with respect to time \(t\): $$ \frac{dh_1(t)}{dt} = -\frac{q_1(t)}{A} $$ We replace \(q_1(t)\) with \(k \cdot h_1(t)\). $$ \frac{dh_1(t)}{dt} = -\frac{k}{A} \cdot h_1(t) $$

The differential equation for \(h_1(t)\) is a first-order linear homogeneous differential equation with constant coefficients. To solve it, we can use separation of variables: $$ \frac{dh_1(t)}{h_1(t)} = -\frac{k}{A} dt $$ Integrate both sides: $$ \int \frac{dh_1(t)}{h_1(t)} = -\frac{k}{A} \int dt $$ $$ \ln \left| h_1(t) \right| = -\frac{k}{A}t + C $$ Taking the exponential of both sides: $$ h_1(t) = e^{-\frac{k}{A}t + C} = e^{-\frac{k}{A}t} \cdot e^C $$ Let \(C = \ln{h_0}\), then we get: $$ h_1(t) = h_0 \cdot e^{-\frac{k}{A}t} $$

By substituting the equation for \(h_1(t)\) back into the expression for \(q_1(t)\), we find: $$ q_1(t) = k \cdot h_1(t) = k \cdot h_0 \cdot e^{-\frac{k}{A}t} $$

The inflow to the second tank is equal to the outflow from the first tank, \(q_1(t)\). The outflow from the second tank is given by: $$ q_2(t) = k \cdot h_2(t) $$ Taking the derivative of \(h_2(t)\) with respect to time \(t\): $$ \frac{dh_2(t)}{dt} = \frac{q_1(t)}{A} - \frac{q_2(t)}{A} $$ Replacing \(q_1(t)\) and \(q_2(t)\) with expressions in terms of \(h_1(t)\) and \(h_2(t)\): $$ \frac{dh_2(t)}{dt} = \frac{k \cdot h_1(t)}{A} - \frac{k \cdot h_2(t)}{A} $$

We substitute the expression for \(h_1(t)\) into the differential equation for \(h_2(t)\) and solve it using an integrating factor: $$ \frac{dh_2(t)}{dt} = \frac{k \cdot h_0 \cdot e^{-\frac{k}{A}t}}{A} - \frac{k \cdot h_2(t)}{A} $$ Multiply both sides by the integrating factor \(e^{\frac{k}{A}t}\): $$ e^{\frac{k}{A}t} \frac{dh_2(t)}{dt} + \frac{k}{A} e^{\frac{k}{A}t} h_2(t) = \frac{k \cdot h_0}{A} e^{\frac{k}{A}t} $$ The left side is now the derivative of the product \(e^{\frac{k}{A}t} h_2(t)\). $$ \frac{d}{dt} \left( e^{\frac{k}{A}t} h_2(t) \right) = \frac{k \cdot h_0}{A} e^{\frac{k}{A}t} $$ Integrate both sides: $$ e^{\frac{k}{A}t} h_2(t) = \frac{h_0 \cdot A}{k} e^{\frac{k}{A}t} + C $$ $$ h_2(t) = \frac{h_0 \cdot A}{k} + Ce^{-\frac{k}{A}t} $$ Since the second tank is initially empty, \(h_2(0) = 0\): $$ 0 = \frac{h_0 \cdot A}{k} + C $$ $$ C = - \frac{h_0 \cdot A}{k} $$ So the final expression for \(h_2(t)\) is: $$ h_2(t) = \frac{h_0 \cdot A}{k} \left(1 - e^{-\frac{k}{A}t}\right) $$

By substituting the equation for \(h_2(t)\) back into the expression for \(q_2(t)\), we find: $$ q_2(t) = k \cdot h_2(t) = k \cdot \frac{h_0 \cdot A}{k} \left(1 - e^{-\frac{k}{A}t}\right) $$ $$ q_2(t) = h_0 \cdot A \left(1 - e^{-\frac{k}{A}t}\right) $$