 Suppose we have a water tank that has the shape of an inverted circular cone. So let's stop there for a second. Basically what we're talking about is we have a tank that looks like an ice cream cone, right? It's conical in shape and inverted. We're just suggesting that the apex, the point of the cone is pointing down. So this looks like essentially an ice cream cone that we're gonna fill some water into this thing. Now, so some of these things we should say about this this conical water tank here is that the radius of the circle that's on the base of the cone, that's gonna be two meters in length. We also know that the height of this cone is going to be four meters in height right there. So this gives us the dimensions of this cone. That's gonna be important for us as we go forward. We don't exactly know why that is yet, but let's think about that for a moment. So suppose water is being pumped into the tank at a rate of two meters cube per minute. Then if that's the case, let's find the rate at which water, that, excuse me, at the rate at which the water level is rising when the water is exactly three meters deep. All right, so there's a couple of things to unravel with this situation. A graphic is concluded here to try to keep track of what's going on here and I'll explain what's gonna want in just a moment. This is gonna be an example of a related rates problem. Notice we're told that water is being pumped into the tank at a rate of two meters cube per minute. So a rate, notice here that the units form a ratio, meters cube per minute. If you look at that, for example, what is meters cube measuring? Meters cube would be measurement of volume because it's length by length, it's length cubed. That gives us a volume. So we have this volume measurement right here and then minutes, that's measuring time, okay? So this idea of two cubic meters per minute is a measurement of change of volume per time. This is giving us information about the change of volume. Let's say that V is the volume of the tank, then dV dt is telling us, I should say we know what the derivative volume with respect to time is. It's gonna be this two meters cube per minute, okay? So we see that the water level is rising, rising, rising, rising inside of our tank as time persists. But wait a second, the rate, so the volume is increasing, but the rate at which the water is rising also is not known. And it turns out it won't be constant in the situation. Notice, when you say words like rate, the word rising, rising would represent a change in height. These words are cues to us to indicate there's a question about derivatives here. Find the rate, let me rephrase that, find the derivative at which the water level is rising. So let's introduce some more variables here, right? So volume is gonna be characterized by a V, that kind of makes sense, and maybe we should write it. So V is gonna equal the volume, the volume, we're gonna measure this in cubic meters, let's say at time t, okay? Maybe I should have written on them blue for water, what have you, that's okay. But we also have this idea about the height, like how tall is the water inside the tank at any specific moment? So let H equal the height, the height of the water, we're gonna measure this in meters, that makes conversion from height to volume much easier. So the height of the water at a given moment of time, measured in t, we're gonna measure that meters, t, at the time, we're gonna measure in minutes as well, we got that from this derivative here, we'll measure that in minutes. And so what we then need to do, so let me pause for a second. So we know the derivative volume with respect to time, it wants to know how fast is the height increasing. So that's a question about the derivative of height with respect to time, that's what we need to figure out, that's the unknown rate. But not at all moments of time, specifically we need to know when the height is equal to three. What is the change of height? How rapidly is the height increasing? And you'll notice that with this cone, that when you're at the bottom, you're gonna see that this cone, the water level's gonna rise really quickly. But when you're near the top, it's gonna be much, much slower. I remember I was a child visiting an amusement park, in which case I saw that they had these glasses that you could buy at the various restaurants and such. They kind of looked like this really long beaker a little bit, right? In which case you had this very large opening at the top, you had a very large base at the bottom, and but then it had this really long neck. And I loved as this child, again, I was probably like seven, maybe eight years old at the time, I just loved to see them like fill it up with like a soda or whatever the soft drink was at the time. Cause at the beginning it would be like, it would fill up slow, right? And then when it hits this neck, it's like whoop, it fills up really quickly cause it's so skinny in comparison. And then it's slow again, that's when they stop filling up the drink. It was just so cool as a kid, at least it fascinated me at the time, because although the change of volume was the same, the soda fountain that was putting the root beer into the cup, it was spitting out the same amount of volume per minute as it was the whole time, no big deal there. But because of the shape of the container, the height, excuse me, the rate at which height was increasing changed, and that was sort of a fascinating thing to watch. It seemed cool as a child, I thought I needed it. Of course, kids can be very impressed by things like that. That same thing's happened with this conical tank right here, that the water level will increase slower, the higher we fill up the tank because of the wider base that will be in the tank at that moment, okay? So we see that we know something about derivative and we want to know something about different derivatives. So this is a related rates question. We need to find an equation that relates together the volume of our cone with the height of the cone, okay? And that makes me think of the volume formula for a cone. Notice that the volume of a cone is equal to one third pi R squared H where volume, V here represents the volume of the cone and H here represents the height of the cone, like we're talking about right here, but it also depends on the radius of the cone. Notice we are not given the derivative of radius. We don't know at what rate the radius is increasing, which clearly as the water goes up, the radius is gonna get bigger and bigger and bigger, okay? But we don't know the rate in which the radius is changing nor do we actually care about it. We don't need it. So how do we deal with that? How do we handle a related rates problem and our relationship between the quantities we know and wanna know about involves some other third party? It's like we don't need you R. Turns out we want to actually remove R from consideration because if we just take the derivative, if we just take the derivative with respect to time with both sides of the equation, the left hand side's easy, you get the derivative of volume. The right hand side, you can pull out the pi thirds, but you're gonna have to take the derivative of R squared H in which case the product rule's gonna come into play. You're gonna end up with this two R R prime H plus R squared H prime for which we know V prime, we want to know H prime, but we don't know R prime. So there's like this extra derivative we don't know nor do we need to know. And so it would be nice if we could remove it from consideration. How can we do that? It turns out there is a way to do it and that's what I wanna demonstrate in this video. It turns out that using an argument of similar triangles can be very useful in this situation because if you look at the entire cone, right, the whole tank itself, what we saw earlier is that the height of the cone, which is a measurement of H, right, was four, we also know that the radius of that circle on the top, that was equal to two meters, that was mentioned before. And so this is a measurement of R, we're at the very top of the cone. But we also, if we take a specific level of water, right, that water level has some specific height H and some specific radius R. These triangles that you see here, there's this red triangle right here and then there's this larger triangle right here. These two triangles we call similar triangles, meaning that the similar triangles have all the same angle measurements. I mean, because after all, you'll notice that if we just look at the right triangle, for example, right, you see this right triangle and you see this right triangle right here. I should use a different color. So we have this right triangle as well. Well, they're both right triangles, so they're both gonna have right angles. So those are both 90-degree angle measurements right there. Also, this angle right here is shared by both triangles, so they're gonna have that measurement the same. And given that these triangles will have an angle sum of exactly 180 degrees, we can subtract the known angle measures to find out the missing angle measure to see that the third side will, those will be congruent to each other as well. So these triangles have the same angle measurements, which mean that they will have proportional sides. We can set up a proportion between these two triangles here. So we have the little triangle, which has a height h and a radius r. And then we have the larger triangle, which will have a height of four and a radius of two. And so because these things are proportional to each other, we get that the ratio h over r is gonna equal two over four. What did I do there? So r corresponds to two, so those become the numerators of our fraction. We also see that h coincides with four, in which case those become the denominators of a fraction. This is how we can set up a proportion. But since we now have a proportion, we can solve for the radius r. My strategy would be just to cross multiply, like so. We end up with four r is equal to two h. We wanna solve for r, so divide both sides by four. We're gonna get that r equals two h over four, or in other words, r equals one half h. What, no matter how high up the cone you are, the radius will be exactly half of the current height. This right here then allows us to substitute out the radius in this formula. So using that substitution, we can remove this intermediate variable and get that volume equals pi thirds. Instead of the radius, we're gonna get one half height squared, and we times that by height. Let's try to simplify this thing a little bit. In which case one half height squared will give us h squared over four, times that by another h. Then we're gonna end up with pi h cubed over 12, four times three. Now we're in a position where we can take the derivative of both sides. Take the derivative on the left hand side, take the derivative on the right hand side. The derivative on the left, you take the derivative of volume with respect to time, you're gonna end up with v prime, just for short. On the right hand side, pi twelts is a constant, so we can factor it out, so we get pi twelts, and then you're gonna times that by h cubed prime, which we're taking the derivative of h cubed with respect not to h, we're taking the derivative with respect to time. So the chain rule comes into play here. It comes into play. We need to have the outer derivative and the inner derivative. So taking the derivative here, we're gonna get pi twelts, we're gonna get three h squared times h prime, for which then the three cancels in the 12, leaving the four that we had from before. So maybe we should have left it factored. Didn't I say it's best to leave denominators factored? I should follow my own advice, right? So we get that v prime is going to equal pi h squared h prime all over four. And so what is it we're after again? As a reminder, we're looking for the derivative of height when the height is equal to three. So we could solve for h prime right here. It would be pretty easy to do. Let's do exactly that. H prime is going to equal four v prime over h squared pi. And this is the general formula for h prime, which h prime will depend on how quickly is the volume changing with the current height. We specifically wanna know what is the change of height with respect to time at the moment that the height is equal to three meters. So we can plug in the change of volume, which we knew, right? That was given by an assumption. We were told that the change of volume is two meters cube per minute. So we get two meters cube per minute, like so. In the denominator, we have to plug in for h. We're gonna plug in a three m, it's three meters. We're gonna square that times that by pi. I do like to put in the units. It's not always a necessary thing, but it is helpful. It helps you to make sure your formula doesn't even make sense. Because you'll notice the numerator has a meters cubed, right? And we're divided by a meter squared. So when you cancel that, you'll be left with a meter and nothing cancels out with the per minute. So in the end, this thing is gonna look like meters per minute, which is great because we're looking at the change of height with respect to time. Height's measured in meters, time's measured in minutes. So this does have the right units we would expect here. And so then continuing on here, excuse me, we see that we're gonna get four times two, which is eight. We're gonna get three squared, which is nine. And so we see that the change of height at this moment in time is gonna be eight over nine pi, which again, an approximation for these story problems is probably appropriate. In which case, put that in your calculator, eight divided by nine pi. That would be approximately 0.28 meters per minute. And so we can see that when it comes to our related rates problems, sometimes the equation that relates the two variables together might involve some intermediate variables. In which case you'll need a second or third, depends how many variables you have, you might need another relationship so you can substitute out that intermediate variable. In this case, it was radius. We didn't actually need to do anything with radius because we didn't know the derivative of the radius whatsoever. We needed to use a similar triangle argument so we can remove the radius from consideration and just focus on the relationship between volume and height of this particular comb.