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Darkiekurdo
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You're right.Dick said:
Thanks a lot though! I know sometimes I look a little bit stupid but I hope that's the age.
Calculating Water Flow Rate in Sloped Reservoir | Simple Calculus Question
- Thread starter Darkiekurdo
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SUMMARY
The discussion focuses on calculating the water flow rate in a sloped reservoir with a square cross-section, specifically a truncated pyramid shape. The volume of the reservoir is expressed as V = (1/3)h(p² + p(p + 2h) + (p + 2h)²). The rate at which the water surface rises, dh/dt, is derived as dh/dt = c / (p + 2h)², where p is the side length of the base, h is the depth of the water, and c is the inflow rate. For the specific values of p = 17, h = 4, and c = 35, the calculation yields the rate of rise of the water surface.
PREREQUISITES- Understanding of calculus, specifically differentiation
- Familiarity with the concept of volume for geometric shapes
- Knowledge of the relationship between flow rate and volume change over time
- Basic understanding of truncated pyramids and their properties
- Study the differentiation of volume functions in calculus
- Learn about the geometric properties of truncated pyramids
- Explore fluid dynamics principles related to flow rates
- Practice solving related rate problems in calculus
Students studying calculus, particularly those focusing on related rates, as well as educators and tutors looking for examples of practical applications of calculus in fluid dynamics.
Thanks a lot though! I know sometimes I look a little bit stupid but I hope that's the age.