Asked by George
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
Answered by
Steve
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.