( I figured out “A” but can’t figure out “B” ) 12 points Previous Answers Larson
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( I figured out “A” but can’t figure out “B” ) 12 points Previous Answers LarsonET5 3.7.024 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 min 3 ft 3 ft h ft 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. 5/51 ft/min (b) If the 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. 27/20 × ft3/minExplanation / Answer
a) Let V(t) is the volume of water in the trough at time t.
We know dV/dt=2 ft^3/min since the amount of water in the trough is increasing at the rate of 2ft^3/min.
Let h is the depth of the water in the trough at time t.
The volume is the length of the trough times the area of the triangle that is filled.
V=12*((1/2)*base *height)
using similar triangle base/h=3/3 implies base=h
V=12*((1/2)*h *h)=6h^2
dV/dt=12h dh/dt
plug dV/dt=2 ft^/min,when h=1.7 ft
2=12*1.7 dh/dt
dh/dt=20/(12*17)
=5/(3*17
=5/51
dh/dt=5/51 ft/min
the water level is rising when h=1.7 feet deep is 5/51 ft/min
b) We have given dh/dt=3/8 inch per minute when =2.3
We know that dV/dt=12h dh/dt from part (a)
converting 3/8 inches into feet
1 inch =1/12 feet
3/8 inches =3/8*1/12=1/32 feet
dh/dt=3/8 inch per minute=1/32 feet/min
dV/dt=12h dh/dt
=12*(2.3)*(1/32)
=27.6/32
=276/320
=138/160
=69/80
=0.8625 ft^3/min
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