I need help answering this question, Example 2 is shown below the original problem for reference. Thank you!

Water is pouring into a conical tank at the rate of 8 cubic feet per minute. If the height of the tank is 9 feet and the radius of its circular opening is 4 feet, how fast is the water level rising when the water is 4 feet deep? feet per minute See Example 2 page 136 for a sketch and a similar example.Figure 3 __ __ EXAMPLE 2 Water is pouring into a conical tank at the rate of 8 cubic feet per minute. It' the height ofthe tank is 12 feet and the radius ofits circular opening is 6 feet, how fast is the water level rising when the water is 4 feet deep? SOLUTION Denote the depth of the water by h and let r be the corresponding radius of the surface of the water (see Figure 3). We are given that the volume. V.of water in the tank is increasing at the rate of 8 cubic feet per minute; that is. dV/dt = 8. We want to know how fast the water is rising (that is, dhfdt) at the instant when h = 4. We need to find an equation relating V and h; we will then differentiate it to get a relationship between dV/dt and dh/dt'. The formula for the volume of water in the tanks l/ = inrgh, contains the unwanted variable 1"; it is unwanted because we do not know its rate dr/dt. However, by similar triangles (see the marginal box), we have r/h = 6/12, so r = J's/2. Substituting this in V = %7rth gives 12 Now we differentiate implicitly, keeping in mind that both V and h depend on t. We obtain dV 3wh dh _ who dh 12 dt 4 di Now that we have a relationship between dV/de and dh/dr, and not earlier, we consider the situation when / = 4. Substituting h = 4 and dV/di = 8, we obtain 8 = "(4) dh 4 dt from which dh dt _ = 0.637 T When the depth of the water is 4 feet, the water level is rising at 0.637 foot per minute. If you think about Example 2 for a moment, you realize that the water level will rise more and more slowly as time goes on. For example, when h = 10 8 = - w(10) dh dr so dh/di = 32/100# = 0.102 foot per minute. What we are really saying is that the acceleration d'h/dr is negative. We can calculate an expression for it. At any time , 8 = The dh 4 di 50 dt If we differentiate implicitly again, we get 0 =mid-h dh dr2 + - (2h dh) from which -2 dh This is clearly negative