Question

gravel is being dumped from a conveyor belt at a rate of 20 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. how fast (in ft/min) is the height of the pile increasing when the pile is 11 ft high? (round your answer to two decimal places.)

Answer 1

Answer:

0.21ft/min

Step-by-step explanation:

   $\sf Volume \ at \ given \ time = \dfrac{dV}{dt}= 20 \ cubic \ feet /min$

             h = d

             h = 2r

             $\sf r = \dfrac{h}{2}$

            Volume of heap = (1/3) πr²h

                     $\sf V =\dfrac{1}{3}\pi r^2h\\\\V =\dfrac{1}{3}\pi (\frac{h}{2})^2h\\\\V =\dfrac{1}{3}\pi \dfrac{h^2}{4}*h\\\\V=\dfrac{1}{12}\pi h^3$

Take derivative w.r.t time,

                     $\sf \dfrac{dV}{dt}=\dfrac{1}{12}\pi * 3h^2 \dfrac{dh}{dt}$

                     h = 11 ft

                     $\sf 20 = \dfrac{1}{12}\pi *3*11*11*\dfrac{dh}{dt}\\\\20=\dfrac{121}{4}\pi \dfrac{dh}{dt}$

               $\sf \dfrac{20*4}{121*\pi }=\dfrac{dh}{dt}\\\\ \dfrac{80}{120*3.14}=\dfrac{dh}{dt}$

                        $\sf \dfrac{dh}{dt}=0.21$

             The height is increasing at the rate of 0.21 ft/min

         

Answer 2

0.21ft/ml I think so