Tank Cone 2
 1/2" Bulkhead NBR Gasket Polypropylene Fitting for plastic cone tank NPT Rubber US $4.25 
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Can anyone help me with this related rates problem?
I cant figure it out at all:
A tank of water in the shape of a cone is leaking water at a constant rate of 2 ft^3 per hour. the base radius of the tank is 5ft and the height is 14ft. V=1/3pi*r^2*h
a. at what rate is the depth of the water in the tank changing when the depth of the water is 6 ft?
b. at what rate is the radius of the top of the water in the tank changing when the depth of the water is 6ft?
A tank of water in the shape of a cone is leaking water at a constant rate of 2 ft^3 per hour. the base radius of the tank is 5ft and the height is 14ft. V=1/3pi*r^2*h
V =(1/3)π r² h
Taking the tank as an inverted cone with vertex downward
Let α= semi vertical angle of the cone
Relation between the height and the radius at any moment is given by tanα= h/r =H/R
Where' h' and 'r' is the depth of water and radius at water level at any time and H = total height and R= Base radius of the conical tank.
Hence h/r =H/R =14/5
hence h= 2.8 r and r= h/2.8
Hence V =(1/3)π r² h =(1/3)π(h/2.8 )²h
dV/dt = π h² /2.8² dh/dt
Given dV/dt = 2 ft³ per hour
2 =π h² /2.8² dh/dt
or 2 =π h² /2.8² dh/dt=2 ×2.8² /(π h²)
a. at what rate is the depth of the water in the tank changing when the depth of the water is 6 ft?
dh/dt=2 ×2.8² /(π h²)
When h = 6 ft
dh/dt = 2 ×2.8² /(π×6²)=0.1386 ft/hr
b. at what rate is the radius of the top of the water in the tank changing when the depth of the water is 6ft?
r= h/2.8 Hence dr/dt =(dh/dt)/2.8
Hence dr /dt (when h = 6 ft ) = 0.1386 /2.8=0.0495 ft/hr
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