Question
A water tank has the shape of an inverted right circular cone whose base radius is equal to half its height. Water is being pumped in at 2 cubic meters per second. How fast is the water rising when the water is 3 meters deep?
Answers
radius = (1/2)h
r = (1/2)h
r = h/2
V = (1/3)πr^2 h
= (1/3)π(h^2/4)h
= (1/12)πh^3
dV/dt = (1/4)πh^2 dh/dt
when h = 3, dV/dt = 2
2 = (1/4)π(9) dh/dt
dh/dt = 8/(9π) m/s
Check my calculations, I did not write them out first.
r = (1/2)h
r = h/2
V = (1/3)πr^2 h
= (1/3)π(h^2/4)h
= (1/12)πh^3
dV/dt = (1/4)πh^2 dh/dt
when h = 3, dV/dt = 2
2 = (1/4)π(9) dh/dt
dh/dt = 8/(9π) m/s
Check my calculations, I did not write them out first.
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