Question

Sand falls from a conveyor belt at a rate of 30 m3m3/min onto the top of a conical pile. The height of the pile is always 3535 of the base diameter. Answer the following. a.) How fast is the height changing when the pile is 2 m high?

Answer 1

Explanation:

As the given data is as follows.

         h = $\frac{3}{5}d$

            = $\frac{3}{5} \times (2r)$

Also, we know that r = $\frac{4}{3}h$

and       Volume (V) = $\frac{1}{3} \pir^{2}h$

                         = $\frac{1}{3} \pi (\frac{4}{3}h)^{2} h$

                         = $\frac{16}{27} \pi h^{3}$

And,  $\frac{dV}{dt} = \frac{3 \times 16}{27} \pi h^{2} \frac{dh}{dt}$

         $\frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}$

Putting the given values into the above formula as follows.

       $\frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}$

       $30 m^{3}/min = \frac{16}{9} \pi (2)^{2} \frac{dh}{dt}$    

          [tex]\frac{dh}{dt} = 1.343 m/min

or,                        = 134.3 cm/min     (as 1 m = 100 cm)

thus, we can conclude that the height changing at 134.3 cm/min when the pile is 2 m high.