Sand falls from a conveyer belt at a rate of 15m^3/min. How fast is the height changing when the pile is 2m high?

Question

This is a related rates problem using cones. I am in Calculus.

Answer 1

This problem can be solved if an assumption is made about the angle of repose for a sand pile.   According to Wikipedia, the angle of repose for dry sand is ~ 34°.   A little bit of trig shows that with this angle, the relationship between the radius of the circular base,r, and the height, h ,   is    r =  h tan(56°)

 

   The volume, V,   is   (1/3) π r² h.       =  (1/3) π [h tan(56°) ]²  h

 

    The time derivative of this is equal to the rate at which sand is being added.  So

 

  15 =  π h² [tan(56°)]² dh/dt

 

    Plugging in h =2  and solving for dh/dt   yields:

    

dh/dt =  2.17 m/min.