Question

A right conical container has slant height 10 cm and height 8 cm. The cone is base up, and water is leaking from the tip (bottom) at 2 cubic cm per minute. When the water is 6 cm high in the container, what is the rate that the radius of the water's surface is decreasing?

Answer 1

when H=6 cm
\[R=\frac{ 6 \times 6 }{ 8 }=\frac{ 9 }{ 2 } cm.\]
\[\frac{ dV }{ dt }=2 cm^3/ minute.\]
\[find~\frac{ dR }{ dt }\]