Calculating the Rate of Water Height Increase in a Filling Swimming Pool

In summary, the two quantities are the depth of water at the deep end (1.0 m) and the rate at which the water is being filled (.80m^3/min).
  • #1
notsosmartman
6
0
Ok, I've been trying to figure this problem out of an old textbook for a couple days but seem to get nowhere.

"A swimming pool with a rectangular surface 18 meters long and 12 meters wide is being filled at the rate of .80m^3/min. At one end it is 1.0 m deep and at another it is 2.5 m deep, with a constant slope between ends. How fast is the height of the water rising when the depth of water at the deep end is 1.0 m??"

Any suggestions on how to set the problem up? I would like to solve it myself, but would like a starting place!

Thanks, Caleb
 
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  • #2
Draw a sketch of a side view of the pool. From the side, the pool looks like a trapezoid, with vertical sides of 1 and 2.5 meters, and length 18 meters. If you put in coordinates for the two points at the end of the sloping bottom, you can find the equation of the line that forms the bottom edge of the pool. The volume of water in the pool between the times when the pool is empty and when there are 1.5 meters of water is the area of the triangle cross-section times the width of the pool.
 
  • #3
Thanks Mark,

But I'm still having difficulty understudying your explanation. Maybe a sketch will help me understand?

Thanks, Caleb.
 
  • #4
Yes, probably, but you should draw it. Just draw a sketch of the side view of the pool.
 
  • #5
HAHA, i have made so many damn sketches on my dry erase board... but yeah i ican post any drawing on here because its a mac and its new to me...
 
  • #6
Step 1, in any related rates problem, is to identify the two quantities that are to be related.

They pretty much tell us what one of those quantities is when they ask "how fast is the height of the water rising...?" And they strongly hint at what the other is when they say the pool is "being filled at the rate of .80m^3/min".
 

1. What is the concept of "Related Rates rising water"?

The concept of "Related Rates rising water" is a mathematical problem that involves finding the relationship between two changing quantities. In this case, it refers to the rate at which the water level is rising in a container, and how it relates to the rate at which water is being poured into the container.

2. How is the related rates formula used to solve problems involving rising water?

The related rates formula, which is derived from the chain rule in calculus, is used to find the rate of change of one quantity in relation to the rate of change of another quantity. In the case of rising water, this formula can be used to determine the rate at which the water level is increasing based on the rate at which water is being added to the container.

3. What are the steps for solving a related rates problem involving rising water?

The first step is to identify the changing quantities in the problem and assign them variables. Next, write out an equation that relates these quantities to each other. Then, use the related rates formula to find the derivative of the equation with respect to time. Finally, substitute in the given values and solve for the desired rate of change.

4. Can related rates problems involving rising water be solved without calculus?

No, related rates problems involving rising water require the use of calculus because they involve finding the derivative of a function with respect to time. This is necessary in order to determine the relationship between the changing quantities and their rates of change.

5. What are some real-life applications of related rates problems involving rising water?

Related rates problems involving rising water can be applied to real-life situations such as filling a swimming pool, draining a bathtub, or filling a water tank. They can also be used in engineering and physics to calculate the flow of fluids in pipes and channels.

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