Question

1. a rectangular well is 6 feet long, 4 feet wide, and 8 feet deep. if water is running into the well at the rate of 3 ft³/sec, find how fast the water is rising (keep in mind which variables are constant and which are changing).

Explanation:

Step1: Write the volume formula for the water in the well

The volume \(V\) of water in the rectangular well is \(V = l\times w\times h\), where \(l = 6\) (constant), \(w = 4\) (constant), and \(h\) is the height (depth) of the water. So \(V=24h\).

Step2: Differentiate both sides with respect to time \(t\)

Using the chain - rule, \(\frac{dV}{dt}=24\frac{dh}{dt}\).

Step3: Substitute the given value of \(\frac{dV}{dt}\)

We know that \(\frac{dV}{dt} = 3\) \(ft^{3}/sec\). Then \(3 = 24\frac{dh}{dt}\).

Step4: Solve for \(\frac{dh}{dt}\)

\(\frac{dh}{dt}=\frac{3}{24}=\frac{1}{8}\) \(ft/sec\).

Answer:

The water is rising at a rate of \(\frac{1}{8}\) \(ft/sec\).