rate of change, volume of a funnel, height, radius, time

[red and white kop!]

A funnel has a circular top of diameter 20 cm and a height of 30 cm. When the depth of liquid in the funnel is 12 cm, the liquid is dripping from the funnel at a rate of 0.2 (cm^3)(s^(-1)). At what rate is the depth of the liquid in the funnel decreasing at this instant?

Alright so at this instant the volume of water in the funnel is equal to that of a cone with a circular base, having height 12 cm and diameter 8 cm.
Therefore volume= (1/3)(pi)((r)^2)(h) = (1/3)(pi)(16)(h)
Therefore dV/dh = (16/3)(pi)
and we know that dV/dt = -0.2
so dh/dt = (dV/dt)/(dV/dh) = -.0119

But the right answer is (the depth of the liquid in the funnel is decreasing at a rate of) 0.0040 cm (s^(-1))

Where did I go wrong?

 

[tkhunny]

Right up front. How did r^2 turn into 16?

You have Diameter/Height = 20/30 = 2/3, so Radius/Height = 1/3

Rather that r^2 = 16, give it a try with r^2 = ((1/3)h)^2.

 

[red and white kop!]

thanks!