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A swimming pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and 9 ft deep at is deepest point. A cross-section is shown in the figure. If the pool is being filled at a rate of 0.7 ft3/min, how fast is the water level rising when the depth at the deepest point is 5 ft? (Round your answer to five decimal places.)

The pool is an irregular shape and I have no formula to find the volume or height. I don't know where to start this since I do not have related rates formula. Can someone please help?

cross-section_of_pool

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  • $\begingroup$ Can you find the length of the line that marks the waterline in the figure? From there can you get to the area of the surface of the water? How does the change in volume, compare to the change in height and the surface area? $\endgroup$
    – Doug M
    Commented Oct 6, 2016 at 22:38

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Shifting the pool bottom to left as shown for easier calculation of area /volume. Basis is that triangle area removed/added with same base and height has same area and, so the volume.

enter image description here Representing time rate with respect to time by dots $$ 20 \;dV = x\; dz , \quad 20 \;\dot V= x \;\dot z $$ Equation of slant floor $$x=12+\dfrac{11 z}{3}$$ $$ 20 \dot V= (12+\dfrac{11 z}{3})\dot z $$ Given that $$ z=5, \dot V= 0.7 $$ $$ \dot z = \dfrac{20\times 0.7}{12+55/3} = 0.461538 $$ $ ft^3 $ per minute.

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Area of water surface when the depth at the deepest point is 5 ft:

$(\frac{5}{6}\cdot 6+12+\frac{5}{6}\cdot16)20=\frac{1820}{3}ft^2$

$ft/min=\frac{ft^3/min}{ft^2}$

so you just need to take

$\frac{0.7}{\frac{1820}{3}}=\frac{2.1}{1820}\approx0.00115384615ft/min$

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  • $\begingroup$ This is not correct method also there is a large error $\endgroup$ Commented Jul 10, 2020 at 13:06

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