Water is leaking out of an inverted right conical tank at a rate of 0.0118 m^3/min. At the same time water is being pumped into the tank at a constant rate. The tank has height 6 meters and the diameter at the top is 4.5 meters. If the water level is rising at a rate of 0.21 m/min when the height of the water is 3.5 meters, find the rate at which water is being pumped into the tank.
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height 6 meters and the diameter at the top is 4.5 meters
h=6 and d=2r=4.5
so 6/4.5 = h/d =h/2r
6(2r)=4.5 h
12r=4.5h
r=(4.5/12)h
v=1/3 π r^2 h
substituting with r=(4.5/12)h
v=1/3 π ((4.5/12)h)^2 h
v=1/3 π ((4.5/12)h)^2 h
v=(20.25/432)π h^3
differentiating the v to get the change rate in the volume
dv/dh=(20.25/432) 3 π h^2
dv=(20.25/432) 3 π h^2 dh
dv=(60.75/432) π h^2 dh
dh=0.21 m/min
h=3.5 m
let R="rate at which the water is being pumped into the tank"
R-0.0118 = dv=(60.75/432) π h^2 dh
R-0.0118 = dv =(60.75/432) π (3.5)^2 (0.21)
R=(60.75/432) π (12.25) (0.21) + 0.0118 =1.15 m^3/min
h=6 and d=2r=4.5
so 6/4.5 = h/d =h/2r
6(2r)=4.5 h
12r=4.5h
r=(4.5/12)h
v=1/3 π r^2 h
substituting with r=(4.5/12)h
v=1/3 π ((4.5/12)h)^2 h
v=1/3 π ((4.5/12)h)^2 h
v=(20.25/432)π h^3
differentiating the v to get the change rate in the volume
dv/dh=(20.25/432) 3 π h^2
dv=(20.25/432) 3 π h^2 dh
dv=(60.75/432) π h^2 dh
dh=0.21 m/min
h=3.5 m
let R="rate at which the water is being pumped into the tank"
R-0.0118 = dv=(60.75/432) π h^2 dh
R-0.0118 = dv =(60.75/432) π (3.5)^2 (0.21)
R=(60.75/432) π (12.25) (0.21) + 0.0118 =1.15 m^3/min