



 
Hi Stella. The second part of the question looks easier than the first but it isn't. In the second part of the question, you only have liquid flowing one way (into the funnel), but you can see in the first part of the question there is a flow into the funnel and another one out. However, they are both constant flow rates and don't depend on the height, so you can lump them together. Thus the first part of the question is equivalent to an constant inbound flow rate of 180 cm^3/s. The flow rate is volume per unit time. The rate you are interested in is distance (height) per unit time. So you need something that connects the height of the liquid to the volume. A right circular cone, inverted in this manner with height h equal to the radius has the volume v given by:
So when we differentiate with respect to time, we get an equation involving dv/dt (that's the flow rate) and dh/dt (that's what you want: the vertical speed of the water level).
Constant factors come out: dv/dt = (π / 3) d/dt (h^{3}) And the exponent rule says to bring down the 3 from h^{3} as a factor and reduce the exponent by 1, so we get 3h^{2}, leaving just the dh/dt:
So we've done the calculus part. Now we just solve for what we want (dh/dt):
So given a value for dv/dt (the net flow rate) and a particular height h that we are interested in, we can find the change in height over time (that is the vertical speed of the water level). Cheers,  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 