(c) The flow of solution into the vat is stopped, and the solution is drained. The depth of solution in the vat is modeled by the function \( h \) that satisfies the differential equation \( \frac{d h}{d t}=-k \sqrt{h} \), where \( h(t) \) is measured in meters, \( t \) is the number of minutes since draining began, and \( k \) is a constant. If the depth of the solution is 16 meters at time \( t=0 \) minutes and 4 meters at time \( t=30 \) minutes, what is \( h(t) \) in terms of \( t \) ?

To solve the differential equation \(\frac{d h}{d t} = -k \sqrt{h}\), we can separate the variables and integrate. Rewriting it gives: \[ \frac{1}{\sqrt{h}} dh = -k dt. \] Integrating both sides yields: \[ 2\sqrt{h} = -kt + C, \] where \(C\) is the constant of integration. Squaring both sides leads us to: \[ h = \left(-\frac{kt}{2} + \frac{C}{2}\right)^2. \] At \(t=0\), it's given that \(h(0) = 16\), so: \[ 16 = \left(\frac{C}{2}\right)^2 \implies C = 8. \] Thus, we have: \[ h = \left(-\frac{kt}{2} + 4\right)^2. \] Next, we can use the second condition \(h(30)=4\): \[ 4 = \left(-\frac{30k}{2} + 4\right)^2 \implies 2 = -15k + 4. \] Solving for \(k\): \[ -15k = -2 \implies k = \frac{2}{15}. \] Now substituting \(k\) back in gives: \[ h(t) = \left(-\frac{t}{15} + 4\right)^2. \] Ultimately, the function describing the depth of solution in the vat is: \[ h(t) = \left(4 - \frac{t}{15}\right)^2. \] This is the depth of the solution over time!