3.1. Related Rates http://www.ck12.org
Solution:
We first note that this problem presents some challenges that the other examples did not. Bothrandhare functions
oft, and so implicit differentiation of theV(t)function is going to produce several variables in the formr(t),drdt,
h(t), anddhdt, and that will give us too many variables to solve fordhdt. So we need to find a way to eliminateranddrdt.
When we differentiate the original equation,V= ( 1 / 3 )πr^2 h,we get
dV
dt =1
3 π(h)(^2 r)dr
dt+1
3 πr2 dh
dt.The difficulty here is that we have no information about the radius when the water level is at 6 feet. So we need
to relate the radius a quantity that we do know something about. Starting with the original equation, let’s find a
relationship betweenhandr.Letr 1 be the radius of the surface of the water as it flows out of the tank.
Note that the two triangles are similar and thus corresponding parts are proportional. In particular,
r 1
h =8
20
r 1 =^820 h=^25 h.Now we can solve the problem by substitutingr 1 = ( 2 h/ 5 )into the original equation:
V=^13 π( 2 h
5) 2
h=^475 πh^3.