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Let's solve each part step-by-step.

### Given Information
- The swimming pool dimensions are **16 ft wide** and **20 ft long**.
- Let:
  - \( h \): height of water in the pool at any time \( t \).
  - \( V \): volume of water in the pool at any time \( t \).
  - \(\frac{dV}{dt}\): rate at which the volume of water is increasing.
  - \(\frac{dh}{dt}\): rate at which the height of the water is rising.

### Part (a): Finding the Relationship Between \( V \), \( h \), \( \frac{dV}{dt} \), and \( \frac{dh}{dt} \)

Since the pool is rectangular with a uniform cross-section, the volume of water in the pool can be calculated as:
\[
V = \text{Area of base} \times h = (16 \, \text{ft} \times 20 \, \text{ft}) \times h = 320h
\]

So, the relationship between \( V \) and \( h \) is:
\[
V = 320h
\]

Differentiating both sides with respect to time \( t \):
\[
\frac{dV}{dt} = 320 \frac{dh}{dt}
\]

This equation relates \( \frac{dV}{dt} \) to \( \frac{dh}{dt} \).

---

### Part (b): Rate of Increase in Volume When Water Level is Rising at \( \frac{1}{4} \) ft/min

We are given:
\[
\frac{dh}{dt} = \frac{1}{4} \, \text{ft/min}
\]

Using the equation from part (a):
\[
\frac{dV}{dt} = 320 \cdot \frac{1}{4} = 80 \, \text{ft}^3/\text{min}
\]

So, the volume of water is increasing at a rate of **80 cubic feet per minute**.

---

### Part (c): Rate of Water Level Rising When Pool is Filled at 8 ft³/min

Here, we are given:
\[
\frac{dV}{dt} = 8 \, \text{ft}^3/\text{min}
\]

Using the equation from part (a) again:
\[
8 = 320 \frac{dh}{dt}
\]

Solving for \( \frac{dh}{dt} \):
\[
\frac{dh}{dt} = \frac{8}{320} = \frac{1}{40} \, \text{ft/min}
\]

So, the water level is rising at a rate of **\(\frac{1}{40}\) feet per minute**.

---

### Summary of Answers
1. **(a)** The equation relating \( V \), \( h \), \( \frac{dV}{dt} \), and \( \frac{dh}{dt} \) is \( \frac{dV}{dt} = 320 \frac{dh}{dt} \).
2. **(b)** The rate at which the volume of water is increasing when \( \frac{dh}{dt} = \frac{1}{4} \) ft/min is \( 80 \) ft³/min.
3. **(c)** The rate at which the water level is rising when the pool is filled at \( 8 \) ft³/min is \( \frac{1}{40} \) ft/min.

Would you like more details on any part of this solution? 

---

### Related Questions to Explore
1. How would the rate calculations change if the pool had a sloped bottom?
2. How would the rate \( \frac{dh}{dt} \) change if the base area of the pool was smaller?
3. What if the pool was filled at a variable rate instead of a constant rate?
4. How does the depth of the pool affect the rate calculations?
5. How would these calculations change if the pool had a circular shape?

### Tip
For similar problems involving rates of change, try identifying the relationship between variables first, then differentiate with respect to time. This often simplifies the calculation process.

A rectangular swimming pool 16 ft wide by 20 ft long and of uniform depth is being filled with water.

(a) If t is elapsed time, h is the height of the water, and V is the volume of the water, find equations relating V to h and dV/dt to dh/dt.

(b) At what rate is the volume of the water increasing if the water level is rising at 1/4 ft/min?

(c) At what rate is the water level rising if the pool is filled at a rate of 8 ft³/min?

Solve a related rates problem involving a rectangular swimming pool that is being filled with water. Learn how to relate volume, water height, and rates of change with step-by-step calculus-based solutions. Suitable for students in Grades 11-12.

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