Discussion
A reservoir discharges water through holes with shutters at depth h and surface area A(h). The values are given in the following table. If t denotes the time in minutes and the rate of fall of the water level is given by the equation \(\frac{dh}{dt} = -\frac{48h}{A}\), what is the time taken in minutes for the water level to fall from 14 ft. to 10 ft. above the holes ?
| h (in ft.) | A (in sq. ft.) |
|---|---|
| 10 | 940 |
| 11 | 1078 |
| 12 | 1200 |
| 13 | 1352 |
| 14 | 1540 |
Answer: A
The time taken is about 8.41 minutes.
Separate the variables: \(dt = -\frac{A}{48h}\,dh\), so the time to fall from 14 ft to 10 ft is \(t = \int_{10}^{14}\frac{A}{48h}\,dh\). Tabulate \(f = \frac{A}{48h}\):
\(f(10) = 1.9583\), \(f(11) = 2.0417\), \(f(12) = 2.0833\), \(f(13) = 2.1667\), \(f(14) = 2.2917\).
Simpson's 1/3 rule with step 1:
\[ \begin{aligned} t &= \tfrac{1}{3}[f_{0} + 4(f_{1} + f_{3}) + 2f_{2} + f_{4}] \\ &= \tfrac{1}{3}[1.9583 + 16.8333 + 4.1667 + 2.2917] \\ &= \tfrac{1}{3}(25.25) \approx 8.42 \end{aligned} \]
which is the keyed 8.41. The other options are far off: the integrand lies between about 1.96 and 2.29 over a 4 ft fall, so the time must lie between about 7.8 and 9.2 minutes.
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