Question

The volume V of a cone is increasing at the rate of 28pi cubic units per second. At the instant when the radius r of the cone is 3 units, its volume is 12pi cubic units and the radius is increasing at 1/2 unit per second.

Answer 1

At the instant when the radius of the cone is 3 units, what is the instantaneous rate of change of the area of its base with respect to its height???

Answer 2

\[\frac{ dV }{ dt }=28\pi\]
\[\frac{ dr }{ dt }=1/2\]

Answer 3

\[V=\frac{ 1 }{ 3 }\pi r^2h\]
\[V=12,r=3\]find h

Answer 4

\[V=\frac{ 1 }{ 3 }Ah,A=\frac{ 3V }{ h }\]
\[\frac{ dA }{ dt }=\frac{ -3V }{ h^2 }\]
\[\frac{ -3(12) }{ h^2 }\]
use the height above

Answer 5

h=4?

Answer 6

is there anther ? cos we didnot usethe rates above ,is there a next ?