Sand is poured onto a surface at 13 cm3/sec, forming a conical pile whose base diameter is always equal to its altitude. How fast is the altitude of the pile increasing when the pile is 1 cm high? Note that the volume of a cone is 13πr2h where r is the radius of the base and h is the height of the cone.

Answer:Altitude of the pile will increase by 16.56 cm per second.Step-by-step explanation:Sand is poured onto a surface at the rate = 13 cm³ per secondOr [tex]\frac{dV}{dt}=13[/tex]It forms a conical pile with a diameter d cm and height of the pile = h cmHere d = hVolume of the pile [tex]V=\frac{1}{3}\times \pi  r^{2}h[/tex]cm³per sec.Since h = d = 2r [r is the radius of the circular base]r = [tex]\frac{h}{2}[/tex][tex]V=\frac{1}{3}\pi  (\frac{h}{2})^{2}h[/tex][tex]V=\frac{1}{3}\pi \frac{(h^{2})}{4}(h)[/tex][tex]V=\frac{1}{12}\pi  h^{3}[/tex][tex]\frac{dV}{dt}=\frac{1}{12}\pi \times 3(h)^{2}\frac{dh}{dt}[/tex][tex]\frac{dV}{dt}=\frac{1}{4}\pi \times h^{2}\times \frac{dh}{dt}[/tex]Since [tex]\frac{dV}{dt}=13[/tex] cm³per sec.13 = [tex]\frac{1}{4}\pi  (1)^{2}\frac{dh}{dt}[/tex] [For h = 1 cm][tex]\frac{dh}{dt}=\frac{13\times4}{\pi }[/tex][tex]\frac{dh}{dt}=\frac{52}{3.14}[/tex][tex]\frac{dh}{dt}=16.56[/tex]cm per second.Therefore, altitude of the pile will increase by 16.56 cm per second.