Asked by leone
A container in the shape of a right circular cone with vertex angle a right angle is partially filled with water.
a) Suppose water is added at the rate of 3 cu.cm./sec. How fast is the water level rising when the height h = 2cm.?
b) Suppose instead no water is added, but water is being lost by evaporation. Show the level falls at a constant rate.
a) Suppose water is added at the rate of 3 cu.cm./sec. How fast is the water level rising when the height h = 2cm.?
b) Suppose instead no water is added, but water is being lost by evaporation. Show the level falls at a constant rate.
Answers
Answered by
Reiny
Since the vertex angle is 90°, the radius of the cone must be equal to the height of the cone, so the radius of the water level = height of the water level , so
r = h
Vol = (1/3)πr^2h
= (1/3)πh^3
d(vol)/dt = πh^2 dh/dt
3 = π(2^2) dh/dt
dh/dt = 3/(2π) cm/sec
b) was there no rate of evaporation given?
r = h
Vol = (1/3)πr^2h
= (1/3)πh^3
d(vol)/dt = πh^2 dh/dt
3 = π(2^2) dh/dt
dh/dt = 3/(2π) cm/sec
b) was there no rate of evaporation given?
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