Related Rates Swimming Pool Problem

[Starblazer]

The Question

A swimming pool is 50m long and 20m wide. Its depth decreases linearly along the length from 9m to 1m. It is initially empty and is filled at a rate of 1m^3/min. How fast is the water level rising 490 minutes after the filling begins.




We want
The water is rising at a rate of ____ m/min 490 min after the filling begins.

My Problem
I dont know how to incorperate 490 min into my solution.
I have worked out At time = 490 min water in pool will be 490 m^3

Variables
h(t) - height of water (depth)
dv/dt = 1 m^3/min (given)

Approach
When filling the pool, divide the pool into two sections
1. The rectangular part when height >= 8
2. The triangular prism at the bottom used when height <= 8
Area of Triangle = 1/2 * height * length
For triangle height = 8
Write equation in terms of height only
By Similar Triangles
length = 25/4 * height
Substitute to get
Area of Triangle = 1/2 * height * (25/4 * height)
= (25/8) * (height)^2

Volume = Width of Pool x Area of Triangle
Volume = 20 * (25/8) (height)^2
= 62.5 height^2
Diff. Volume Formula w.r.t. time
dv/dt = 125 height * dh/dt

rearrange to make dh/dt the subject

(dv/dt) / (125 height ) = dh/dt
dv/dt = 1 m^3/min (given)
So

1 / (125 height) = dh/dt

I dont know how to incorperate 490 min into my solution.
I have worked out At time = 490 min water in pool will be 490 m^3

Can you please assist?

 

[Deleted member 4993]

The Question

A swimming pool is 50m long and 20m wide. Its depth decreases linearly along the length from 9m to 1m. It is initially empty and is filled at a rate of 1m^3/min. How fast is the water level rising 490 minutes after the filling begins.


View attachment 2722

We want
The water is rising at a rate of ____ m/min 490 min after the filling begins.

My Problem
I dont know how to incorperate 490 min into my solution.
I have worked out At time = 490 min water in pool will be 490 m^3

Variables
h(t) - height of water (depth)
dv/dt = 1 m^3/min (given)

Approach
When filling the pool, divide the pool into two sections
1. The rectangular part when height >= 8
2. The triangular prism at the bottom used when height <= 8
Area of Triangle = 1/2 * height * length
For triangle height = 8
Write equation in terms of height only
By Similar Triangles
length = 25/4 * height
Substitute to get
Area of Triangle = 1/2 * height * (25/4 * height)
= (25/8) * (height)^2

Volume = Width of Pool x Area of Triangle
Volume = 20 * (25/8) (height)^2
= 62.5 height^2
Diff. Volume Formula w.r.t. time
dv/dt = 125 height * dh/dt

rearrange to make dh/dt the subject

(dv/dt) / (125 height ) = dh/dt
dv/dt = 1 m^3/min (given)
So

1 / (125 height) = dh/dt

I dont know how to incorperate 490 min into my solution.
I have worked out At time = 490 min water in pool will be 490 m^3

Can you please assist?

There are two parts to the depth of the pool.

The water will first fill-up the triangular part. Upto a depth of 8 meters (on the deep side), the depth will increase at a linearly changing rate
(i.e. \(\displaystyle \frac{dh}{dt} \ne constant \))

After the deep side has been filled upto 8 meters - the rest of the 1 meter will be filled uniformly (i.e. \(\displaystyle \frac{dh}{dt} = constant \)).

So what will be the \(\displaystyle \frac{dh}{dt} \) as the triangular part is filling up?

How long would it take to fill up the triangular part?

 

[Starblazer]

There are two parts to the depth of the pool.

The water will first fill-up the triangular part. Upto a depth of 8 meters (on the deep side), the depth will increase at a linearly changing rate
(i.e. \(\displaystyle \frac{dh}{dt} \ne constant \))

After the deep side has been filled upto 8 meters - the rest of the 1 meter will be filled uniformly (i.e. \(\displaystyle \frac{dh}{dt} = constant \)).

So what will be the \(\displaystyle \frac{dh}{dt} \) as the triangular part is filling up?

(dv/dt) / (125 height ) = dh/dt
dv/dt = 1 m^3/min (given)
So

1 / (125 height) = dh/dt

Is this on the right track ??

How long would it take to fill up the triangular part?

Area = 1/2 * 50m * 8m * 20m = 4000 m^3
Fill Rate = 1m^3 / min

Answer = 4000 min to fill triangular part

So time = 490 min will be in the triangular part

 

[Deleted member 4993]

There are two parts to the depth of the pool.

The water will first fill-up the triangular part. Upto a depth of 8 meters (on the deep side), the depth will increase at a linearly changing rate
(i.e. \(\displaystyle \frac{dh}{dt} \ne constant \))

After the deep side has been filled upto 8 meters - the rest of the 1 meter will be filled uniformly (i.e. \(\displaystyle \frac{dh}{dt} = constant \)).



(dv/dt) / (125 height ) = dh/dt
dv/dt = 1 m^3/min (given)
So

1 / (125 height) = dh/dt

Is this on the right track ??



Area = 1/2 * 50m * 8m * 20m = 4000 m^3
Fill Rate = 1m^3 / min

Answer = 4000 min to fill triangular part

So time = 490 min will be in the triangular part

At a height 'h' in the triangular part you can show that:

dV = (50/8) * h * 25 * dh

Now continue.....

 

[Starblazer]

I dont know how to get the two unknowns h and dh/dt.

 

[Deleted member 4993]

I dont know how to get the two unknowns h and dh/dt.

To find 'h', find the h of the triangular region when the volume is 491 m³.

After finding 'h', you will be able to calculate dh/dt (knowing dV/dt = 1).

 

[Starblazer]

Area of Triangle = 1/2 * height * length

Write equation in terms of height only
By Similar Triangles
50/8 = length/height
50 * height = 8 * length
50/8 * height = length
simplify to 25/4 = length

Therefore
length = 25/4 * height

Substitute to get
Area of Triangle = 1/2 * height * (25/4 * height)
= (25/8) * (height)^2

Volume = Width of Pool x Area of Triangle
Volume = 20 * (25/8) (height)^2
= 125/2 height^2

time = 490 min means 490 m^3 in Pool
So Find height for Volume of 490 m^3

490 m^3 = 125/2 height^2

Solve for height

490 / (125/2) = height^2

height^2 = (196/25)

height = square root (196/25) = (14/5)

dv/dt = (25/4) * h * 20 * dh/dt

dv/dt = 1 m^3 (given)

1 = (25/4) * (14/5) * 20 * dh/dt

1/((25/4) * (14/5) * 20) = dh/dt

dh/dt = 1/350 which is the correct answer


Thank you for your help

 

[Deleted member 4993]

Excellent work!!

One suggestion - dh/dt should have a unit of m/min