Asked by Sammy
Water, at the rate 10 ft^3/min, is pouring into a leaky cistern whose shape is a cone 16' deep and 8' in diameter at the top. At the time the water is 12' deep, the water level is observed to be rising 4''/min. How fast is the water leaking away?
Can anyone please give me some ideas to do it???THANKS A LOT!!!
Can anyone please give me some ideas to do it???THANKS A LOT!!!
Answers
Answered by
Reiny
at some time of t minutes, let the height of the water be h ft, and the radius of its surface be r feet.
Let the rate of leakage be x ft^3/min
Volume = (1/3)pi(r^2)h
by similar triangles r/h = 8/16
so r = h/2
Volume = (1/3)pi(h^2/4)(h)
= (1/12)pi(h^3)
then d(Volume)/dt = (1/4)pi(h^2)dh/dt
10+x = (1/4)pi(144)(1/3)
x = 12pi - 10 ft^3/min
Let the rate of leakage be x ft^3/min
Volume = (1/3)pi(r^2)h
by similar triangles r/h = 8/16
so r = h/2
Volume = (1/3)pi(h^2/4)(h)
= (1/12)pi(h^3)
then d(Volume)/dt = (1/4)pi(h^2)dh/dt
10+x = (1/4)pi(144)(1/3)
x = 12pi - 10 ft^3/min
Answered by
Teacher
The above is good, except for the fact that the diameter is given to be 8ft, so the radius is only 4ft.
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.