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To solve this problem, we need to calculate the rate at which the water level is rising when the water is 3 inches deep. We’ll go through it in steps.

### Step 1: Understanding the Trough's Geometry
The trough is 4 feet long, with cross-sectional ends in the shape of isosceles triangles:
- Each triangle has a base of 3 feet and a height of 3 feet.
  
### Step 2: Volume of Water in the Trough
The volume of water in the trough can be calculated using the formula for the volume of a triangular prism:
\[
V = \text{Area of Cross Section} \times \text{Length}
\]
where the area of the triangular cross-section \( A \) depends on the water depth \( h \).

### Step 3: Cross-sectional Area as a Function of Depth \( h \)
As the water rises, the width of the water surface changes proportionally with the depth. Since the full triangle has a base of 3 feet and height of 3 feet, the width \( w \) at any depth \( h \) is given by similar triangles:
\[
\frac{w}{3} = \frac{h}{3} \Rightarrow w = h
\]
So, at a given depth \( h \), the area of the triangular cross-section is:
\[
A = \frac{1}{2} \times w \times h = \frac{1}{2} \times h \times h = \frac{h^2}{2}
\]

### Step 4: Volume in Terms of \( h \)
Substitute this area into the volume formula:
\[
V = \frac{h^2}{2} \times 4 = 2h^2
\]

### Step 5: Differentiate with Respect to Time
To find how fast the water level is rising, we need to differentiate the volume \( V \) with respect to time \( t \):
\[
\frac{dV}{dt} = 4h \frac{dh}{dt}
\]
We are given:
\[
\frac{dV}{dt} = 12 \text{ ft}^3/\text{min}
\]
We also need to convert 3 inches to feet for \( h \):
\[
3 \text{ inches} = \frac{3}{12} = 0.25 \text{ feet}
\]

### Step 6: Solve for \( \frac{dh}{dt} \)
Substitute \( h = 0.25 \) feet and \( \frac{dV}{dt} = 12 \) ft\(^3\)/min:
\[
12 = 4 \times 0.25 \times \frac{dh}{dt}
\]
\[
12 = 1 \times \frac{dh}{dt}
\]
\[
\frac{dh}{dt} = 12 \text{ ft/min}
\]

Thus, the water level is rising at a rate of **12 ft/min** when the water is 3 inches deep.

Would you like a detailed breakdown of any part of this solution? 

Here are some related questions that build on this problem:
1. How would the rate change if the trough's length was different?
2. How would the rate of water level rising change if the water was being added at a slower rate?
3. What if the cross-sectional triangle had different dimensions?
4. How would the problem change if the trough’s cross-section was a different shape, like a rectangle?
5. How would you calculate the rate if the trough was not uniform in shape along its length?

**Tip:** Always convert units to a consistent system before starting calculations to avoid mistakes.

A trough is 4 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 3 ft. If the trough is being filled with water at a rate of 12 ft³/min, how fast is the water level rising when the water is 3 in deep?

Learn how to calculate the rate at which the water level rises in a trough with isosceles triangle ends, given the trough dimensions and water fill rate. This problem applies related rates and geometry, using similar triangles to solve a calculus-based real-world problem.

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