Related rates fill rate problem

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The problem involves calculating the rate at which the water level in a trapezoidal swimming pool rises when filled at a rate of 1m³/min. The pool dimensions include a length of 50m, a width of 20m, and a depth that decreases linearly from 3m to 1m. The volume of water is modeled using different geometric formulas depending on the water height, initially as a triangular prism and later as a combination of a triangular and rectangular volume. The discussion highlights the need to establish the correct volume equations for varying water levels and how to derive the rate of change of the water height over time. Ultimately, the solution requires careful consideration of the changing geometry as the pool fills.
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Homework Statement



A swimming pool is 50m long and 20m wide. Its depth decreases linearly along the length from 3m to 1m. It is initially empty and is filled at a rate of 1m^3/min. How fast is the water level rising 90 minutes after the filling began?

The Attempt at a Solution



I'm really lost on this problem, so my attempt at a solution isn't much of one. I assume that I will be using two separate volume formulas: one for the area with the sloped bottom, and one for the rectangle above that. Beyond that I'm lost.
 
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Well, it is not so much two different formulas for different areas as it is for different times. Look at the pool from the side. It is a trapezoid with bases, at the ends, of length 1 and 3 and height, at the top of length 50. Its volume will be the area of that trapezoid times the width, 20.

But until the water has covered the bottom, its figure, on the side of the pool will be a right triangle with height x, the height of the water, and base determined by where the line giving the bottom crosses that height. Taking the origin of a coordinate system at the lowest point in the pool, positive x-axis up, The bottom of the pool is the line from (0, 0) to (2, 50). That has equation y= 25x. So the area of the side covered by water will be (1/2)(25x)(x)= (25/2)x^2. The volume will be (25/2)x^2(20)= 250x^2. That will be true for x, the height of the water, from 0 to 2. After that, it will be the total volume of the water, 250(4)= 1000, plus the volume of the rectangular solid above it, (20)(50)(x- 2).
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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