Question
A container in the shape of an inverted cone has radius 6 ft and height 12 ft. It is being drained at 2〖ft〗^3/min. Find the rate of change of the height of the liquid in the cone when the height is 3 feet. The ratio of the radius to the height remains constant.
Answers
when the contents have depth y, the radius of the surface of the liquid is y/2
So,
v = pi/3 r^2 y = pi/3 (y/2)^2 y = pi/12 y^3
dv/dt = pi/4 r^2 dy/dt
Now just plug in your numbers.
So,
v = pi/3 r^2 y = pi/3 (y/2)^2 y = pi/12 y^3
dv/dt = pi/4 r^2 dy/dt
Now just plug in your numbers.
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