When the water is 2.5m deep, the section draining is just a rectangular prism with area 160m^2 and height 1/2 m.
So, draining at 1m^3/min, the water level is dropping at a rate of 1/120 m/min
When the water is y m deep (y<2), the side cross-section of the pool is a right triangle with legs y and 10y.
So, the volume of the water is
v = (1/2)(y)(10y)(8) = 40y^2 m^3
dv/dt = 80y dy/dt
so, at y=1,
-1 = 80*1 dy/dt
dy/dt = -1/80 ft/min
A rectangular swimming pool is 8 m wide and 20 m long. Its bottom is a sloping plane, the depth increasing from 1 m at the shallow end to 3 m at the deep end. Water is draining out of the pool at a rate of 1 m3/min. How fast is the surface of the water falling when the depth of the water at the deep end is (a) 2.5 m? (b) 1 m?
2 answers
typo alert
dy/dt at y=2.5 is -1/160, not -1/120
dy/dt at y=2.5 is -1/160, not -1/120
2 answers