Asked by daani
A 24ft high conical water tank has its vertex on the ground and radius of the base is 10 ft. If water flows into the tank at a rate of 20 ft3/min, how fast is the depth of water increasing when the depth of the water is 20 ft?
Answers
Answered by
MathMate
The vertex is on the ground, so the tank is in a funnel position.
Let the water height be h, then the radius of the surface of water is r(h)=10h/24=5h/12
The volume at a height of h is
V(h)=(π/3)r(h)² h
=(π/3)(5h/12)² h
=(25π/432)h³
Differentiate with respect to time, t
dV(h)/dt
=(25π/432)*3h²dh/dt
=(25π/144)h² dh/dt
Since dV(h)/dt is known (=20 ft³/min), you can solve for dh/dt.
Note that the unit of dh/dt is in ft/min.
Let the water height be h, then the radius of the surface of water is r(h)=10h/24=5h/12
The volume at a height of h is
V(h)=(π/3)r(h)² h
=(π/3)(5h/12)² h
=(25π/432)h³
Differentiate with respect to time, t
dV(h)/dt
=(25π/432)*3h²dh/dt
=(25π/144)h² dh/dt
Since dV(h)/dt is known (=20 ft³/min), you can solve for dh/dt.
Note that the unit of dh/dt is in ft/min.
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