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( I figured out “A” but can’t figure out “B” ) 12 points Previous Answers Larson

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Question

( 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/min

Explanation / 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