Yefim S. answered • 05/28/22
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Volume of water is v = (4 + 6)/2·4·h= 20h; dv/dt = 20dh/dt; dh/dt = (dv/dt)/20 = 2ft3/min/20ft2 = 0.1 ft/min
Les E.
asked • 05/28/22Calculus question below: Rates of change
The ends of a trough, 12 ft in length, are isosceles trapezoids with height 4 ft, upper base 6 ft and lower base 4 ft. Water is pumped into the trough at the rate of 2 ft^3/min. If the height of the water is 1 ft, how fast is the water level rising?
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Doug C.
Seems like question has not been interpreted correctly? 12 feet is the length of the trough which is a constant. h is the height of the water in the trough which is changing. At a given point in time one of the bases of the trapezoid cross section is 4, but the length of the other base depends on h. In fact the 2nd base will be 4 + 2(1/4)h (based on similar triangles). See: desmos.com/calculator/m62edog4b205/29/22