Related Rates Problem: Swimming Pool Depth and Filling Rate Calculation

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SUMMARY

The discussion focuses on calculating the rate of increase in water depth at the deep end of a swimming pool, which is being filled at a rate of 1000 gallons per minute. The pool measures 50 feet in length, 20 feet in width, and has a depth that varies from 2 feet to 12 feet. To solve the related rates problem, participants are advised to find the volume of the pool using the area of the triangular and rectangular cross sections, applying the volume formula and the relationship between volume and depth changes.

PREREQUISITES
  • Understanding of related rates in calculus
  • Familiarity with volume calculations for irregular shapes
  • Knowledge of the Pythagorean theorem
  • Basic unit conversions (gallons to cubic feet)
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  • Learn how to derive volume formulas for triangular prisms
  • Study the application of the chain rule in related rates problems
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Related Rates.. again...?

A swimming pool is 50 feet long and 20 feet wide. Its depth varies uniformly from 2 feet at the shallow end to 12 feet at the deep end. (The figure shows a cross-section of the pool.) Suppose that the pool is being filled at the rate of 1000 gal/min. At what rate is the depth of water at the deep end increasing when the depth there is 6 feet? (One gallon of water occupies a volume of approximately 0.1337 cubic feet.)


So first of all, I can't even picture what this is supposed to look like when I draw it. Supposedly this is a related rates problem, but this doesn't go with any of the equations I've used so far (pretty much just volume equation and pythagorean theorem equation).. I don't know what I'm supposed to do, could anyone at least help me get started or walk me through? I'd appreciate it..
 
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Just to get you started, the shape of the swimming pool is similar to a wedge or a triangular block with a cuboidal top . How can you find the volume of this block given the dimensions ?
Hint:Find the area of the triangular and rectangular cross sections of the pool.

Arun
 

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