Related Rates, A Conical Tank Example: Consider a conical tank whose radius at the top is 4 feet and whose depth is 10 feet. It’s being ﬁlled with water at the rate of 2 cubic feet per minute. How fast is the water level rising when it is at depth 5 feet? As always, our ﬁrst step is to set up a diagram and variables. r h Figure 1: Illustration of example 2: inverted cone water tank. This diagram just helps us to start thinking about the problem. For instance, we see that because the cone is narrower at the bottom the rate of change of the depth will vary; we need to depict the water level. We also realize that it’s diﬃcult to draw useful and accurate diagrams of three dimensional ﬁgures — a simple schematic may be more helpful. The key here is to draw a two-dimensional cross-section. In the ﬁgure we’re looking at one half of a vertical slice of the tank. The height of the slice equals 10 feet, which is the height of the tank. The widest part of the slice is 4 feet, which is the distance from center to edge of the top of the tank. We’ll use the variable r will represent the distance from center to edge of the top of the water, and h will represent the height of the top of the water (which is also the depth of the water). We can ﬁnd the relationship between r and h from Fig. 2) using similar triangles: r 4 = . h 10 4 r 10 h Figure 2: Relating r and h. Our goal is to ﬁnd out how fast the water is rising when the tank is half full. What we know is that the volume of water in the tank is changing at a rate of 2 cubic feet per minute. We need equations relating the volume of water in the tank to its depth, h. The volume of a cone is 13 · base · height. From Fig. 1), the volume of this tank is given by: 1 V = · �� πr2 · �� h 3 base height This relates the volume to the height and radius, and we know the relation between the hight and the radius. We have one more piece of information that we can use: dV dt = 2. The question is: “What is dh dt when h = 5?” We’ve now translated all of the words in the original problem into formulas. Our word problem is now simply a calculus problem. We could do this by implicit diﬀerentiation, but it’s easy enough to solve for r in terms of h that there’s no need to. 2 r = h. 5 We plug this expression for r back into V to get: � �2 1 2 4 V = π h h= πh3 3 5 3(25) At this point we could solve for h, but that turns out to be a bad idea. Implicit diﬀerentiation is much easier. dV dt = = = dV dh dh dt � �2 π 2 dh 3h2 3 5 dt 4 πh2 h� 25 Now that we’ve calculated the rates of change we can plug in the numbers dV = 2 and h = 5: dt � � 4 2= π(5)2 h� 25 2 = 4πh� 1 h� = ft/min 2π We were given the rate at which the volume of water in the tank was changing and we used that to compute the rate at which the water in the tank was rising. At the heart of this calculation was the chain rule: dV dV dh = . dt dh dt Related rates problems are all about applying the chain rule to solve word problems.