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Thank you for taking my question. Brushing up for end-of-semester Calculus exams, can't get the answer to this question:

Sand falls from a conveyor belt at the rate of 20 meters cubed per minute onto the top of a conical pile. The height of the pile is always 2 times the base diameter. How fast is the height changing when the pile is 6 meters high?

Thanks, John

The volume of a cone is (1/3)bh*pi where b is the base radius and h is the height.
Since h is given to be twice the base diameter, and the base diameter is twice the radius,
we can see that h = 4b.

That makes the equation be V = (4/3)b*pi.
The surface area is the rate at which the volume is changing.
The derivative will give this, and that is A = 4b*pi.

Taking the derivative gives us dA = 8b*pi*db.
Since the surface area is changing at 20 meter cubed per minute, put that in and get
20 = 8b*pi*db.

When the height is 6 meters, and that is twice the base diameter,
the base diameter is then 3 meters.  This means the radius is 3/2 = 1.5.

Putting that in gives 20 = 8*1.5*pi*db, which is the same as 20 = 12*pi*db.
This can be solved for db giving db = 5/(3*pi), and that is meters per minute.  

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Scott A Wilson


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