Asked by Arsh
A rectangular trough is 8 feet long, 2 feet aross the top and 4 feet deep, if water flows in At a rate of 2 cubic feet per minute. How fast is the surface rises when the water is 1 feet deep.?
Answers
Answered by
Reiny
At a given time of t minutes, let the height of the water be h ft, h < 4
V = (8)(2)(h) = 16h feet^3
dV/dt = 16 dh/dt
2 = 16dh/dt
dh/dt = 1/8 ft/min
notice that the fact that the container is 4 ft high, did not enter the picture
V = (8)(2)(h) = 16h feet^3
dV/dt = 16 dh/dt
2 = 16dh/dt
dh/dt = 1/8 ft/min
notice that the fact that the container is 4 ft high, did not enter the picture
Answered by
SALVATUNAC
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