Calculus
posted by Trevor .
Water is being poured into an inverted cone(vertex down) of radius 4 in and height 10 in at a rate of 3 in^3/sec. Find the rate at which the height of the water is increasing when the water is 5 in high.

at water depth y, the radius of the surface is 2y/5, so y = 5r/2
v = 1/3 pi r^2 h = 1/3 pi (25y^2/4)(y) = 25pi/12 y^3
dv/dt = 25pi/12 * 3y^2 dy/dt = 25pi/4 y^2 dy/dt
3 = 25pi/4 * 25 dy/dt
dy/dt = 12 / 625pi
Seems kinda slow  better double check my algebra 
Ahh. I see  I did the wrong substitution
r = 2y/5
Make that change and follow it through