1. **State the problem:** We have a swimming pool with varying depth and a trapezoidal cross-section. The pool is being filled at a rate of 0.8 ft³/min. We want to find how fast the water level is rising (in ft/min) when the water depth at the deepest point is 5 ft. 2. **Understand the geometry:** The pool is 20 ft wide and 40 ft long. The depth varies from 3 ft at the shallow end to 9 ft at the deepest point. The cross-section is trapezoidal with a sloped bottom. 3. **Determine the shape of the water cross-section at 5 ft depth:** Since the deepest point is 9 ft, and water is at 5 ft, the water fills part of the pool. The water surface is horizontal, and the water volume depends on the cross-sectional area of water times the length (40 ft). 4. **Find the width of the water surface at 5 ft depth:** The bottom slopes from 3 ft to 9 ft over 20 ft width, so slope = $$\frac{9-3}{20} = \frac{6}{20} = 0.3$$ ft depth per ft width. At 5 ft depth, the horizontal distance from the shallow end to the water edge is: $$x = \frac{5 - 3}{0.3} = \frac{2}{0.3} = 6.6667 \text{ ft}$$ 5. **Calculate the cross-sectional area of water:** The water cross-section is a trapezoid with: - One base = 20 ft (width of pool) - Other base = 6.6667 ft (water width at 5 ft depth) - Height = 5 ft (water depth) Area $$A = \frac{(b_1 + b_2)}{2} \times h = \frac{(20 + 6.6667)}{2} \times 5 = 13.33335 \times 5 = 66.66675 \text{ ft}^2$$ 6. **Volume of water:** $$V = A \times \text{length} = 66.66675 \times 40 = 2666.67 \text{ ft}^3$$ 7. **Relate volume and water depth:** Let $h$ be the water depth at the deepest point. The width of water surface at depth $h$ is: $$w = \frac{h - 3}{0.3}$$ Cross-sectional area as a function of $h$: $$A(h) = \frac{(20 + w)}{2} \times h = \frac{(20 + \frac{h - 3}{0.3})}{2} \times h$$ Simplify: $$A(h) = \frac{(20 + \frac{h - 3}{0.3})}{2} h = \frac{(20 + 3.3333(h - 3))}{2} h = \frac{(20 + 3.3333h - 10)}{2} h = \frac{(10 + 3.3333h)}{2} h$$ $$A(h) = \frac{10 + 3.3333h}{2} h = 5h + 1.66665 h^2$$ 8. **Volume as a function of $h$:** $$V(h) = A(h) \times 40 = (5h + 1.66665 h^2) \times 40 = 200h + 66.666 h^2$$ 9. **Differentiate volume with respect to time $t$:** $$\frac{dV}{dt} = 200 \frac{dh}{dt} + 2 \times 66.666 h \frac{dh}{dt} = (200 + 133.332 h) \frac{dh}{dt}$$ 10. **Given:** $$\frac{dV}{dt} = 0.8 \text{ ft}^3/\text{min}$$ and $$h = 5 \text{ ft}$$ Substitute values: $$0.8 = (200 + 133.332 \times 5) \frac{dh}{dt} = (200 + 666.66) \frac{dh}{dt} = 866.66 \frac{dh}{dt}$$ 11. **Solve for $$\frac{dh}{dt}$$:** $$\frac{dh}{dt} = \frac{0.8}{866.66} = 0.000922 \text{ ft/min}$$ **Final answer:** The water level is rising at approximately **0.00092 ft/min** when the water depth is 5 ft.