Related Rate Problem Including a Cone

[visoccer8]

A reservoir in the shape of an inverted cone has a radius of 2 meters at the top and a depth of 6 meters. Wine is poured into the reservoir at a rate of 1 m^3/sec. At what rate is the depth of the wine increasing when the depth is 4 meters?

Help?

 

[LCKurtz]

Have you written a formula for the volume of the wine in terms of its depth?

 

[StalkerM]

dV/dt=1 we are asked to find dh/dt when h is 4m.
The volume and the height are related by:
V=(1/3)*Pi*R^2*h
We have to eliminate R.
Sketch the inverted cone: if h is the wine level (and r is the radius of the cone at that h) and the cone height is 9 you can use similar triangles:
r/h=2/6 and so:
r=h/3
The expression is now:
V=(1/3)*Pi*(h/3)^2 *h
you can now differentiate each side.