please help find dh/dt

Question

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 16 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate (in ft/min) is the height of the pile changing when the pile is 2 feet high?

dh/dt=

Answer 1

Given:

dV/dt = 16 ft³ / min

D = 3h

Find:

@ h = 2 ft, dh/dt = ? ft / min

Solution:

Volume of a Cone:

V = πR² * h / 3

V = π ( D / 2 )² * h / 3

∴ V = πD² / 4 * h / 3

∴ V = πD² * h / 12

V = π ( 3h )² * h / 12

∴ V = 9πh² * h / 12

∴ V = 3πh³ / 4

d/dt( V ) = d/dt ( 3πh³ / 4 )

dV/dt = 9πh² / 4 * dh/dt

dh/dt = ( 4 * dV/dt ) / ( 9πh² )

dh/dt = ( 4 * 16 ft³ / min ) / ( 9π( 2 ft )²

∴ dh/dt = 16 / 9π ft/min ≈ 5.59 ft/min

I hope this helps! Message me in the comments if you have any questions, comments, or concerns!