Question

At a local sawmill sawdust is falling onto a pile at a rate of 0.5pi m^3/min.If the pile maintains the shape of a cone with the height equal to three times the radius of the base,how fast is the height of the pile increasing when the volume is 8pi.m^3?

Answer 1

Help!!

Answer 2

\[V=\frac{1}{3}\pi r^2h\]
\[h=3r, r=\frac{h}{3}\]
\[V=\frac{1}{3}\pi (\frac{h}{3})^2h=\frac{1}{27}\pi h^3\]

When Volume = 8pi
\[8\pi = \frac{1}{27} \pi h^3\]
\[8(27)=h^3\]
\[h=6\]
\[\frac{dV}{dt}=\frac{1}{9}\pi h^2\frac{dh}{dt}\]
\[.5\pi=\frac{1}{9}\pi (6^2)\frac{dh}{dt}\]
\[.5\pi =4\pi \frac{dh}{dt}\]
\[\frac{dh}{dt}=\frac{1}{8}\]