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ID: 2881208 • Letter: I
Question
IF YOUR GOING TO WRITE IT ON A PIECE OF PAPER, PLEASE BOX IN YOUR ANSWERS AND WRITE NEATLY.
A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet. 2 ft min 3 ft h ft 3 ft (a) If water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when h is 1.0 feet deep? ft/min (b) If he water is rising at a rate of 3/8 inch per minute when h 2.3 determine the rate at which water is being pumped into the trough. ft /min Need Help? Read it l Talk to a TutorExplanation / Answer
Volume of trough V = Area x length
Area = bh/2
Given b= h
So
V = h^2*l/2 = h^2*12/2 = 6h^2
a)
dV/dt = 12h*dh/dt
2 = 12*1.dh/dt
dh/dt = 1/6 feet per min = 2 inches per minute ==> 1/12
The water level is rising at 2 inches per minute when the level is at 1 foot.
b)
dV/dt = 12h*dh/dt
dV/dt = 12*2.3*3/8*1/12 = 0.8625 cubic feet pr minute
The water is being pumped in at 0.8625 cubic feet per minute