# Solve Urgently Needed: Related Rates Swimming Pool Problem

• rum2563
In summary, the conversation discusses a problem involving a swimming pool with specific dimensions and an inclined bottom. The question asks for the rate at which the water level is rising when the water is 1 m deep at the end of the pool, while water is being pumped in at a rate of 2m^3/min. The conversation explores various approaches to solving the problem, including using ratios and formulas for volume and surface area. Ultimately, the correct answer is determined to be 1/48 m/min, and the conversation concludes with an expression of gratitude for help understanding the problem.
rum2563

## Homework Statement

A swimming pool is 24 m long by 8 m wide, 1 m deep at the shallow end and 3 m deep at the deep end, the bottom being an inclined plane. If water is pumped into the empty pool at a rate of 2m^3/min, then how fast is the water level rising at the moment when the water is 1 m deep at the end of the pool.

## Homework Equations

dh/dt = 1/As x dV/dt

As: area of exposed surface

## The Attempt at a Solution

Basically I used ratios to get the length of the pool when the height is 1 m deep.

1/3 = x/24
So x = 8 m.

Then As = 8*8 = 64 m^2

dh/dt = 2/64
= 1/32 m/min

I am not sure what I am doing wrong. I have my diagram attached. Please tell me what I am doing wrong. Also, if you have the answer please tell me because I need this urgently. I can use the answer and then work it back to get everything right.

Thanks.

#### Attachments

• diagram.JPG
4.2 KB · Views: 778
The surface area 'As' is a function of time as well, it's not just a constant you can factor out.

So would that I mean I have to use some other formula rather than the As one. Because my teacher told me that formula is good for any questions which have exposed area.

Volume(t)=(1/2)*As(t)*h(t) since the volume is 1/2 of a rectangular solid. So you have to use the product rule to get dV/dt. Your teacher may have exaggerated the usefulness of "dh/dt = 1/As x dV/dt". That's only good if As is a constant.

Last edited:
Um, I am sorry I cannot understand. Perhaps if you could explain without As because it's confusing to me.

Good idea. Write a formula for the volume of the pool without using As. V=(1/2)*h*(8m)*(h*24/3). h*24/3 is the length of the filled part. I got that from you 1/3=x/24 by replacing 1 with h. Now how is dV/dt related to dh/dt? Notice As=(8m)*(h*24/3).

I am extremely confused.

How did you get the formula for the volume of pool? Also, I understand that as the height increases, the volume increases too.

Exposed surface will be a rectangle, so area would be length by width.
I understand the width is 8 m and the length is (24h)/3.

But if we solve the volume formula you gave, we end up with the same conclusion that dh/dt is 1/32 m/min.

Please do explain. I really need to know how this works. Thanks.

I got the formula by multiplying the area of the triangular cross section of the pool by the width. Yes, you get the same answer of 1/32 m/min. Because that's the correct answer. There is also nothing wrong with dV/dt=(1/As)*dh/dt. You seemed to be so convinced you were wrong that you convinced me. Why do you think 1/32 m/min is wrong?

Dick said:
I got the formula by multiplying the area of the triangular cross section of the pool by the width. Yes, you get the same answer of 1/32 m/min. Because that's the correct answer. There is also nothing wrong with dV/dt=(1/As)*dh/dt. You seemed to be so convinced you were wrong that you convinced me. Why do you think 1/32 m/min is wrong?

LOL.The reason why I think it's wrong is because my teacher did not give me marks for the answer 1/32 m/min. That is why I think there must be some twist to it.

And when I showed my calculations for h/3 = x/24 the teacher also put a "x" mark there.

Maybe then I did it right, I am not sure.

rum2563 said:
LOL.The reason why I think it's wrong is because my teacher did not give me marks for the answer 1/32 m/min. That is why I think there must be some twist to it.

And when I showed my calculations for h/3 = x/24 the teacher also put a "x" mark there.

Maybe then I did it right, I am not sure.

Well, I think you did it right.

Here is a solution that I present only because you say the (1/32) solution didn’t go well with your instructor. My math, being what it is, I would treat it with suspicion.

Let b be the length of the surface of the water in the 24 meter direction.
Let h be the depth of the water at the deep end of the pool.
Let v be the volume of water in the pool at time t.

b = 12 h
v = (1/2) h (12 h) 8 = 48 h^2
dv/dt = 96 h dh/dt
dh/dt = 1/(96 h) dv/dt

At h = 1 and dv/dt = 2
dh/dt = 1/(96×1)×2 = 1/48

jimvoit said:
Here is a solution that I present only because you say the (1/32) solution didn’t go well with your instructor. My math, being what it is, I would treat it with suspicion.

Let b be the length of the surface of the water in the 24 meter direction.
Let h be the depth of the water at the deep end of the pool.
Let v be the volume of water in the pool at time t.

b = 12 h
v = (1/2) h (12 h) 8 = 48 h^2
dv/dt = 96 h dh/dt
dh/dt = 1/(96 h) dv/dt

At h = 1 and dv/dt = 2
dh/dt = 1/(96×1)×2 = 1/48

That would be fine, except I question the b=12h. At h=0, b=0. At h=3, b=24. So b=8h.

The bottom of the pool terminates at the shallow end at h=2, not h=3. At h=2, b=24.

Oooooooops. You've got it. I was looking at the attached JPG and reading the little tick mark near the 3m label as the depth of the slanted section. Silly me. Sorry. Thanks!

Thanks jimvoit. I finally get it. Your answer was very detailed and it helped me understand this question. The answer 1/48 seems better than the one I got, also the fact that it is 2m and not 3m for the ratios was a small mistake that I had made, but your post was really helpful.
You are the best.
Thanks again.

I’m glad we were able to help out. If there is a lesson here, it is that things can go wrong in at least 2 ways…First if the translation from the description of the physical situation to the math is faulted, and second if the math is faulted. The first way is a real headache because perfectly good math can lead to the wrong answer.

You can also go wrong by focusing on the big picture and forgetting to check the little details. As we did. Thanks again.

## What is the "Solve Urgently Needed: Related Rates Swimming Pool Problem"?

The "Solve Urgently Needed: Related Rates Swimming Pool Problem" is a mathematical problem that involves finding the rate of change of one variable with respect to another variable in a swimming pool scenario. This problem is often used in calculus and physics to help understand how different variables affect each other in real-life situations.

## What kind of information is needed to solve this problem?

In order to solve this problem, you will need to know the dimensions of the swimming pool, the rate at which the water level is changing, and the rate at which the water is being pumped in or drained out of the pool. You will also need to know the formula for finding the volume of a pool and how to take derivatives.

## How is this problem solved?

This problem is solved using the related rates method, which involves setting up an equation that relates the variables in the problem and then taking the derivative with respect to time. This will result in a related rates equation that can be solved for the desired rate of change.

## What are some common mistakes when solving this problem?

Some common mistakes when solving this problem include not properly setting up the related rates equation, using the wrong formula for finding the volume of the pool, and not considering the units of measurement when taking derivatives. It is important to double check all steps and units to ensure an accurate solution.

## How is this problem relevant to real-life scenarios?

This problem is relevant to real-life scenarios because it helps to understand how different variables in a particular situation are related and how changes in one variable affect the others. This can be applied to various situations such as filling a swimming pool, draining a water tank, or even tracking the growth of a population.

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