jake_f Posted November 18, 2009 Posted November 18, 2009 Please point out where I got wrong. The question: At a sand gravel plant, sand is falling off a conveyor belt and onto a conical pile at the rate of 10m^3 / min. The diameter of the base of the cone is approximately 3 times the altitude. At what rate is the height of the pile changing when it is 15 meters high? My solution: Let V be the volume of the conical pile, and the height be h, and the radius, r. Given, dv/dt = 10. Since the diameter of the base is approximately 3 times the altitude, h= 6r, and r = 2.5 m h=6r dh/dr = 6 v=1/3 πr^2h v=1/3 πr^2(6r) v=2 πr^3 dv/dr = 6πr^2 dv/dt= dr/dt X dv/dr dr/dt= 10 / 6π(2.5)^2 = approximately 0.0848826 m/min dh/dt = dr/dt X dh/dr =0.0848826 X 6 =0.5092956 m/min (my wrong Ans) But the answer should be 0.0063m/min. Please explain to me where I got wrong.
Mr Skeptic Posted November 18, 2009 Posted November 18, 2009 Please point out where I got wrong.... Since the diameter of the base is approximately 3 times the altitude, h= 6r, and r = 2.5 m h=6r dh/dr = 6 Wouldn't that be 2r = 3h ?
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