Posted by ibranian on .
San falls from a conveyor belt at the rate of 10 m^3/min onto the top of a conical pile. The height of the pile is always three-eighths of the base diameter. How fast are (a) height and (b) radius changing when the pile is 4 m height? Answer in cm/min.
let height of cone be h m
let radius of cone be r m
but h = (3/4)(2r) = (3/2)r OR r = (2/3)h
or 2h = 3r -----> dh/dt = (3/2) dr/dt
V = (1/3)π r^2 h
= (1/3)π (4h^2/9) h
= (4/27) π h^3
dV/dt = (4/9) π h^2 dh/dt
when dV/dt = 10 and h = 4
10 = (4/9)π(16) dh/dt
dh/dt = 90/(64π) m/sec
= appr .4476 m/sec or appr 45 cm/sec
sub 90/(64π) into dh/dt = (3/2) dr/dt to find the other answer.