 Let's take a look at yet another related rates problem. This one involves a cone, which definitely has a certain way in which you want to approach the solution. In this problem, we have an inverted conical tank with an altitude of 16 meters and a radius of 4 meters. So that's the dimensions of the entire tank. Water is flowing into the tank at a rate of 2 cubic meters per minute. How fast is the water level rising when the water is 5 meters deep? So let's go ahead and draw ourselves a visual of this. So here we have our inverted conical tank with the dimensions shown. And we have water flowing into the tank. So obviously, the water is going to be filling up inside the tank. Now, this is a cone. So what is going to be changing are both the height of how much water is inside the cone, but also the radius is going to be changing. So in this case, we have two variables to deal with. So let's think about what it is we're given. So in this case, we are told that the water is flowing into the tank at a rate of 2 cubic meters per minute. Notice the units of measure again. That's a volume measurement, the cubic meters. So the rate we are given is a volume measurement, dv dt. And we are going to keep it as a positive 2 because the water is flowing into the tank and accumulating. And what we're trying to find is dy dt because y represents the height. And we want to know how fast is the water level rising when y is 5 meters. So recall how you get volume of a cone. Remember, volume is going to be 1 third pi r squared and then the height, which we're calling y. So the difficulty is if we were to leave this as it is and do the implicit differentiation, we would end up doing product rule because we have two variables. So if you think through how that would look, when you got to the point in the product rule that you'd be doing the derivative of r, you'd have to throw in dr dt. And we don't know dr dt. Now the dy dt we're trying to find, so that's OK. But the issue becomes the dr dt. And that's why when you have a cone problem such as this, what we need to do is really get this equation now only in terms of y because it's the dy dt that we want to leave in that equation because that's what we're trying to find. We essentially need to eliminate the r variable out of the equation. The way in which you do that is with proportions. So if you take a look at your cone, we're going to set up a little proportion of the different dimensions and we're going to express r in terms of y. So let's take a look at what we have. We could do 4, which is the radius of the entire cone, over 16, the height of the entire cone. And we could set that equivalent to r over y. Now remember with proportions, there's lots of different equivalent ways to set them up. So I'm just going to do it this way. 4 over 16 equals r over y. So when we solve this out, and remember we want to solve for r in terms of y. So if we divide both sides by 16, we have 1 fourth y is equal to r. So what we're going to do is take that and we're going to substitute it into our volume equation. So remember what the volume equation was, 1 third pi r squared y, we're going to put 1 quarter of y in place of r. And notice how we're doing all this before we even do any differentiating. So when you simplify this and get it down to something nice and tidy, we have pi over 48 y cubed. And notice how simple that's going to be to do the derivative of. If we go ahead and do our implicit differentiation, we'd have dV dt equals 3 pi over 48 and we can simplify that later if you'd like. Y squared dy dt. Now remember we do not substitute in any other values, any other variable values until we have our implicit differentiation finished, which we do now. So we know dV dt was 2. Let's go ahead and simplify that fraction and make it pi over 16. R, or y rather, remember that was five. So when we square that we have 25. And there's the dy dt that we're looking for. So if we go ahead and rearrange this to solve for dy dt, you can try it on your own. You should have that it's 32 over 25 pi. Of course you're welcome to use that as a decimal if you'd like. The decimal value is approximately 0.407. So we would have as our final conclusion that when the water is five meters deep, the water level is rising at a rate of approximately 0.407 meters per minute. Now remember the units of measure, what we were trying to find was dy dt and y represented a height, which is a linear measurement, which we would be measuring in meters. And that's why this final rate we found is meters per minute.