Gravel is being dumped from a conveyor belt at a rate of 30 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 is the height of the pile increasing when the pile is 11 ft high? (Round your answer to two decimal places.)

Answer:Rate of increase in height =[tex]\frac{dh}{dt}=0.3156ft/min[/tex]Step-by-step explanation:we know that volume of a cone is given by[tex]V=\frac{1}{12}\pi d^{2}h[/tex]It is Given that diameter equals height thus we have[tex]V=\frac{1}{12}\pi h^{2}h\\\\V=\frac{1}{12}\pi h^{3}[/tex]Differentiating both sides with respect to time we get [tex]\frac{dV}{dt}=\frac{1}{12}\pi \frac{dh^{3}}{dt}\\\\\frac{dV}{dt}=\frac{1}{12}\pi(3h^{2}\frac{dh}{dt})[/tex]Applying values and solving for [tex]\frac{dh}{dt}[/tex] we get[tex]\frac{dh}{dt}=\frac{12\frac{dV}{dt}}{3\pi h^{2}}\\\\\frac{dh}{dt}=0.3156ft/min[/tex]