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
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!!!
2 answers
The above is good, except for the fact that the diameter is given to be 8ft, so the radius is only 4ft.
2 answers