Swimming Pools and Related Rates along with Implicit Differentiation

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Homework Help Overview

The discussion revolves around a problem involving related rates and implicit differentiation in the context of a swimming pool's water level changes. The pool has specific dimensions and the rate at which water is being pumped in is provided, prompting questions about the rate of change of the water level at different depths.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply the formula for volume to find the rate of change of water level, but encounters difficulties with the varying dimensions of the pool as the water level changes. Other participants question how to define the volume for different water depths and suggest alternative approaches to express the volume without explicitly using length.

Discussion Status

Participants are exploring different interpretations of the volume calculation based on the water depth. Some have provided insights into how to approach the problem without relying solely on the pool's length, indicating a productive exchange of ideas. However, there is no explicit consensus on the method to be used for the varying water levels.

Contextual Notes

There is a lack of explicit equations provided for the volume at different water depths, leading to questions about how to handle the changes in geometry as the water level rises. The original poster also notes uncertainty regarding the calculation for the deeper end of the pool.

Salazar
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Homework Statement



A swimming pool is 40 feet long, 20 feet wide, 4 feet deep at the shallow end, and 9 feet deep at the deep end. Water is being pumped into the pool at 10 cubic feet per minute.
a. When the water is 3 feet deep at the shallow end, at what rate is the water level rising? b. When the water is 3 feet deep at the deep end, at what rate is the water level rising?

Homework Equations


None Given.

The Attempt at a Solution



I had [dV/dt] and the volume for part a is V = L*W*H and I had L and W, which were 40' and 20' respectively. So V = 800H and I used implicit differentiation to get [dV/dt] = 800[dH/dt], so [dH/dt] = 1/80 feet per minute.

For part be it seems straightforward, but I do not know how to get the length, since it has change. Picture of the pool is uploaded.
 

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Counting h from the top of the pool you have then
V = LWH for 0 < H < 4 ft and
V = something else for 4 ft < H < 9ft.
 
I understand that, but what is such something else for L, length?
 
The volume of the lower half is a triangles area times the width of the pool; you can give the triangle for example by one of the angles and the height, you do not need to use length explicitly at all.
 

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