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.

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 ?