A rectangle trough is 8 meters long, 2 meters across at the top, and 4 meters deep. If water flows in at the rate of 2 cubic meters per minute, how fast is the surface rising when the water is 1 meter deep?

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"rectangle trough". What does the cross section look like? is the cross section a triangle, or a truncated triangle?

I assume a triangular cross-section. So, the width of the water surface when it has a depth of y is y/2. Thus the volume of water of depth y is

v = (1/2)(y)(y/2)(8) = 2y^2
dv/dt = 4y dy/dt

Now just plug in your numbers

To solve this problem, we need to apply the principles of related rates. Let's break it down step by step:

Step 1: Identify the given information:
- Length of the rectangle trough (l) = 8 meters
- Width across the top of the trough (w) = 2 meters
- Depth of the trough (h) = 4 meters
- Rate at which water flows in (dV/dt, the change in volume with respect to time) = 2 cubic meters per minute
- We need to find the rate at which the surface is rising, which is given by dh/dt (the change in depth with respect to time).

Step 2: Visualize the problem:
Imagine the trough as a rectangular prism with a length of 8 meters, a width of 2 meters, and a variable depth (h) that is initially 4 meters. As water flows in at a rate of 2 cubic meters per minute, the depth of the water will increase over time.

Step 3: Set up the formula:
The formula for the volume of a rectangular prism is V = lwh. In this case, the formula should be rearranged to express the depth (h) in terms of the volume (V). The formula is h = V / (lw).

Step 4: Find the volume in terms of the depth:
Given that the depth (h) is 4 meters, we can substitute this value into the formula to find the corresponding volume (V). Plugging in the values, we have V = 8 * 2 * 4 = 64 cubic meters.

Step 5: Differentiate the volume equation with respect to time:
We need to differentiate the volume equation (V = lwh) with respect to time (t) to determine how the volume is changing. This is because the rate at which the surface rises is directly related to the rate at which the volume is changing.

dV/dt = d/dt (lwh) = lw (dh/dt)

Step 6: Substitute known values into the differentiation equation:
We already know that the width (w) is 2 meters and the rate of change in volume (dV/dt) is 2 cubic meters per minute. We can substitute these values into the equation:

2 = 2 * 2 * (dh/dt)

Step 7: Solve for dh/dt to find the rate at which the surface is rising:
To find dh/dt (the rate at which the surface is rising), we can rearrange the equation:

2 = 4 * (dh/dt)

Dividing both sides by 4:

1/2 = (dh/dt)

Therefore, when the water is 1 meter deep, the surface is rising at a rate of 1/2 meters per minute.