Accuracy of derivative with change of water height in conical tank??

Question

With regard to a related rates problem where water is pumped into or flowing out of a conical tank and you're asked to determine the rate of change in the height of the water at a particular point, given a constant volume flow in/out rate, the formula and derivative is pretty straight forward and trivial. However, my 86 year old father, who dabbles with calculus, says the calculations are wrong (using the derivative). For example, say you have a cone with height of 16m and radius of 4m, with water flowing into the cone at 2m^3 /min. What is the rate of change in height when the water is at 2m of depth? Using the derivative this equals 2.546479089m/min. However, if you measure 2m^3 volume from the 2m mark up, you would get that 2m^3 from the 2m mark to 3.378344166m. If you subtract 2 from 3.378344166 you get 1.378344166, which means the height change was actually 1.378344166 m/min. So, how do I explain that the derivative is a measure of the instantaneous rate of change at a point and how can I show that this is truly the change at the 2m mark and not the 1.378344166 m/min he arrived at? Thank you.

Answer 1

Well, you have a challenge. Any physical measurement that he'd like to make would have to involve measurable quantities, so that automatically would be cruder than the INSTANTANEOUS values--not to mention that the measuring would have to be instantaneous, too.

 

As a still functioning retired math teacher only 8 days from my 81st birthday anniversary, I urge you to be gracious with your 86 y.o. grandpa. If the Calculus you performed is correct, stand by it.  And let's not go overboard on carrying out calculations to quite so many decimal places!