Asked by Eric
Water is leaking out of a pool at a rate of 4 cubic inches per second, what is the rate of change of the height of the water at the instant the volume equals 20ft cubic feet
Answers
Answered by
Reiny
V= (l)(w)(h)
since the length and width do not change, only V and h.
dV/dt = (lw) dh/dt
We must have the same units,
20 cubic feet = 20(12^3) or 34560 cubic inches
when V = 34560 and dV/dt = -4
34560 = (lw) h
lw = 34560/h
-4 = 34560/h dh/dt
dh/dt = -h/8640 inches/second
Are you sure there were no other units given?
since the length and width do not change, only V and h.
dV/dt = (lw) dh/dt
We must have the same units,
20 cubic feet = 20(12^3) or 34560 cubic inches
when V = 34560 and dV/dt = -4
34560 = (lw) h
lw = 34560/h
-4 = 34560/h dh/dt
dh/dt = -h/8640 inches/second
Are you sure there were no other units given?
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