# Related rates fill rate problem

• kentarch
In summary, the swimming pool has a linearly decreasing depth from 3m to 1m along its length of 50m. It is initially empty and is being filled at a rate of 1m^3/min. To find the rate at which the water level is rising after 90 minutes, we need to use two different volume formulas based on the time. The first formula is for the area with the sloped bottom, which is a trapezoid with bases of 1m and 3m and a height of 50m. The second formula is for the rectangular area above the sloped bottom, with dimensions of 20m and 50m. By plugging in the given values, we can
kentarch

## Homework Statement

A swimming pool is 50m long and 20m wide. Its depth decreases linearly along the length from 3m to 1m. It is initially empty and is filled at a rate of 1m^3/min. How fast is the water level rising 90 minutes after the filling began?

## The Attempt at a Solution

I'm really lost on this problem, so my attempt at a solution isn't much of one. I assume that I will be using two separate volume formulas: one for the area with the sloped bottom, and one for the rectangle above that. Beyond that I'm lost.

Well, it is not so much two different formulas for different areas as it is for different times. Look at the pool from the side. It is a trapezoid with bases, at the ends, of length 1 and 3 and height, at the top of length 50. Its volume will be the area of that trapezoid times the width, 20.

But until the water has covered the bottom, its figure, on the side of the pool will be a right triangle with height x, the height of the water, and base determined by where the line giving the bottom crosses that height. Taking the origin of a coordinate system at the lowest point in the pool, positive x-axis up, The bottom of the pool is the line from (0, 0) to (2, 50). That has equation y= 25x. So the area of the side covered by water will be $(1/2)(25x)(x)= (25/2)x^2$. The volume will be $(25/2)x^2(20)= 250x^2$. That will be true for x, the height of the water, from 0 to 2. After that, it will be the total volume of the water, $250(4)= 1000$, plus the volume of the rectangular solid above it, (20)(50)(x- 2).

## 1. What is a related rates fill rate problem?

A related rates fill rate problem is a type of math problem that involves finding the rate at which the volume of a container is changing as a function of time. This problem usually involves multiple variables that are related to each other in a specific way, and the goal is to use calculus to find the rate of change of one variable with respect to another.

## 2. How do you solve a related rates fill rate problem?

To solve a related rates fill rate problem, you first need to identify all of the variables involved and the relationships between them. Then, you can use the chain rule and other calculus techniques to find the derivative of the volume function with respect to time. Finally, you can plug in the given values and solve for the unknown rate of change.

## 3. What are some common real-world examples of related rates fill rate problems?

Some common real-world examples of related rates fill rate problems include filling a swimming pool with water, filling a cylindrical tank with oil, or draining a conical tank of water. These problems involve finding the rate at which the volume of the container is changing as the height of the liquid inside changes.

## 4. What is the difference between a related rates fill rate problem and a standard related rates problem?

The main difference between these two types of problems is that a related rates fill rate problem involves finding the rate of change of the volume of a container, while a standard related rates problem involves finding the rate of change of any given quantity. In a fill rate problem, the volume is usually a function of time, while in a standard problem, the two quantities involved are usually related in some other way.

## 5. How can understanding related rates fill rate problems be useful in real-life situations?

Understanding how to solve related rates fill rate problems can be useful in many real-life situations, especially in fields such as engineering, physics, and economics. These types of problems require critical thinking and the application of calculus principles, which are valuable skills in various industries. Additionally, being able to analyze and solve these types of problems can help with decision-making and problem-solving in everyday life.

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