Calculus
posted by Z32 .
Water is being pumped into an inverted conical tank at a constant rate. The tank has height 9 meters and the diameter at the top is 4 meters. If the water level is rising at a rate of 28 centimeters per minute when the height of the water is 3.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
I got it down to pi/4 * (4/9)^2 *(350)^2 * 28 = 532131.4956 but it's not right. Any suggestions?

let the height of the water be h cm
let the radius of the water level be r cm
by similar triangles, r/h = 2/9 , r = 2h/9
V = (1/3)pi(r^2)h
= (1/3)pi(4/81)(h^2)h = (4pi/243)h^3
dV/dt = (4pi/81)(h^2)dh/dt
= (4pi/81)(350)(28) = 1520.375 cm^3/min
check my calculations 
I did it again!
forgot to square the 350 in the last line
(4pi/81)(350)^2(28) = 532131.4964 cm^3/min
which makes it exactly the same as your answer!
MMMHHHH ? 
What the heck? haha.
I was going crazy could I could've swore I did my calculations right. Oh well, thanks for the help man. I appreciate it.