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.

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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.