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The problem provided involves a trough with isosceles triangle ends, where we are asked to solve two parts related to rates of change as water is being pumped into it.

### Problem Setup:
- The trough is **12 feet long**, and its triangular ends have a base of **3 feet** and an altitude (height) of **3 feet**.
- The rate of water being pumped into the trough is given in part (a) as **2 cubic feet per minute** and in part (b) we are given a different rate in inches per minute.

The two questions ask us to calculate:
1. (a) **How fast the water level is rising** when the height \( h = 1.6 \) feet and water is pumped at 2 cubic feet per minute.
2. (b) **The rate at which water is being pumped** when the height \( h = 2.2 \) feet and the water is rising at 3/8 inch per minute.

### Solution Approach:

#### Geometry of the Trough:
The cross-section of the trough is a triangle, and the volume \( V \) of water in the trough depends on the height \( h \) of the water. The formula for the area \( A \) of the triangular cross-section as a function of \( h \) can be found from similar triangles:
- The full height of the triangular end is 3 feet, and the base is also 3 feet.
- So, for a given water height \( h \), the corresponding width of the water at that height is proportional to the height, i.e. the base of the water-filled portion at height \( h \) is \( \frac{h}{3} \times 3 = h \) feet.

Thus, the area of the triangular cross-section when the water height is \( h \) is:
\[
A(h) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times h \times h = \frac{h^2}{2}
\]

The volume of water in the trough, which is 12 feet long, is:
\[
V(h) = A(h) \times 12 = \frac{h^2}{2} \times 12 = 6h^2
\]

#### Differentiation for Rates of Change:
Now, to find the rate of change of the volume with respect to the height, we differentiate \( V(h) \) with respect to \( h \):
\[
\frac{dV}{dh} = 12h
\]

We use this in both parts of the problem.

---

#### Part (a):
Given:
- Water is being pumped into the trough at a rate of \( \frac{dV}{dt} = 2 \) cubic feet per minute.
- We need to find the rate of change of height \( \frac{dh}{dt} \) when \( h = 1.6 \) feet.

Using the chain rule:
\[
\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}
\]
Substituting known values:
\[
2 = 12(1.6) \cdot \frac{dh}{dt}
\]
\[
2 = 19.2 \cdot \frac{dh}{dt}
\]
Solving for \( \frac{dh}{dt} \):
\[
\frac{dh}{dt} = \frac{2}{19.2} \approx 0.1042 \, \text{ft/min}
\]

This is the correct answer for part (a).

---

#### Part (b):
Given:
- The water level is rising at \( \frac{dh}{dt} = \frac{3}{8} \) inches per minute \( = \frac{3}{8} \div 12 = \frac{1}{32} \, \text{ft/min} \) when \( h = 2.2 \) feet.
- We need to determine \( \frac{dV}{dt} \).

Again, using:
\[
\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}
\]
Substituting known values:
\[
\frac{dV}{dt} = 12(2.2) \cdot \frac{1}{32}
\]
\[
\frac{dV}{dt} = 26.4 \cdot \frac{1}{32} = 0.825 \, \text{cubic feet per minute}
\]

This is the correct answer for part (b).

---

Let me know if you'd like more details or if you have any further questions.

### Related Questions:
1. How does the shape of the trough affect the rate of water level rise?
2. How would the results change if the trough were only half the length?
3. What if the triangular cross-section were not isosceles but scalene?
4. How does the relationship between volume and height differ for other shapes (like cylindrical troughs)?
5. What would happen if the rate of water flow into the trough were variable instead of constant?

### Tip:
Always use the chain rule when dealing with related rates problems; it helps link the changing quantities with respect to time.

A trough is 12 feet long and 3 feet across the top. Its ends are isosceles triangles with altitudes of 3 feet. If water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when h = 1.6 feet? Also, if the water is rising at a rate of 3/8 inch per minute when h = 2.2 feet, determine the rate at which water is being pumped into the trough.

Solve related rates problems involving a trough with triangular ends. Learn how to calculate the rate at which water level rises and the rate at which water is pumped. This solution covers differentiation and the chain rule.

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