To solve this problem, we'll use the concept of related rates. We need to find how fast the surface of the water rises, which is the rate of change of the height of the water in the trough over time, \(dh/dt\).
Here's a step-by-step method to solve the problem:
Step 1: Identify what you know.
- The length (\(L\)) of the trough is 10 feet.
- The width (\(W\)) of the trough is 3 feet.
- The rate at which water flows into the trough, which is the rate of change of volume (\(dV/dt\)), is 12 cubic feet per minute.
Step 2: Set up the relationship between volume and dimensions.
The volume (\(V\)) of water in the trough at any time is given by the formula for the volume of a rectangular prism:
\[ V = L \times W \times h\]
where \(h\) is the height of the water in the trough.
Step 3: Differentiate both sides of the volume equation with respect to time (\(t\)).
Taking the derivative gives us the relationship between the rate of change of the volume and the rate of change of the height:
\[ dV/dt = L \times W \times dh/dt\]
Step 4: Solve for \(dh/dt\).
We want to find \(dh/dt\), so we rearrange the equation from Step 3 to solve for \(dh/dt\):
\[ dh/dt = dV/dt / (L \times W)\]
Step 5: Plug in the known values and solve.
\(L = 10\) feet, \(W = 3\) feet, and \(dV/dt = 12\) cubic feet per minute. Now substitute these values into the equation:
\[ dh/dt = 12 \, \text{ft}^3/\text{min} / (10 \, \text{ft} \times 3 \, \text{ft})\]
\[ dh/dt = 12 \, \text{ft}^3/\text{min} / 30 \, \text{ft}^2\]
\[ dh/dt = 0.4 \, \text{ft/min} \]
So the surface of the water rises at a rate of 0.4 feet per minute.
