 Imagine that a 10 foot long ladder is resting against a vertical wall, which I mean as long as you're not in Italy, I don't think you have to worry too much about the wall not being vertical, but we have a vertical wall that is it's going to form a 90 degree angle with the ground. That's all we really care about. And so we have some ladder that's resting against the wall, maybe someone's taking down their Christmas decorations, what have you, doesn't matter. Imagine that the bottom of the ladder is actually slipping away, right? So when you're taking down those Christmas lights, maybe there's some snow in January, so it's starting to slip away from the wall. Now assuming that the ladder stays next to the wall, so it's just a slow slip away from the wall. And so this part is then sliding down. Imagine it's slipping the bottom of the ladder slipping away from the wall at a rate of one foot per second. If that's the case, how fast is the top of the ladder sliding down the wall, particularly at the moment that the ladder is exactly six feet away from the wall? So let's try to unravel what this story problem means a little bit using the idea of related rates. Remember, in calculus, rates make us think about derivatives, because derivatives are rates of change. Related rates just means there's an equation that connects to seemingly unconnected derivatives. That's not the case. They are related to each other. So what are we trying to figure out? What do we know? What are we trying to figure out? So if the bottom of the ladder is sliding away from the wall, one could think of the variable a variable being the distance between the wall and the bottom of the ladder. So let's take that, for example, let's take x to equal that distance. It's a distance, so it needs to be measured in some length. So measured in feet. Those are the units that are in play here. So the distance in feet between the wall and ladder. That's what we see from our diagram. That's what this variable x is going to represent. And so if the ladder is sliding away from the wall, this means as time elapses, x is going to get bigger, bigger, bigger, bigger, x is increasing. That tells me that it's going to have a positive derivative. Or what we're told specifically here, the rate on which the ladder is sliding away from the wall to how it's sliding on the ice, it's at a rate of one foot per second. So this tells us that the derivative of x with respect to time is equal to one foot per second. So we're measuring distance here in feet, time is being measured in seconds. So that's actually slipping very quickly away, so it must be a very slick piece of ice. Well, if the ladder stays next to the wall here, as this portion slides away, this portion is going to have to kind of fall down. And so this introduces our next variable in consideration here. Let's suppose that y equals the distance, well, we'll just use the word height here. It'll be the height of the ladder, let's say above the ground. So essentially x is measuring a horizontal distance and y is measuring a vertical distance, which is why I use the numbers x and y. That coincides with our usual notions of geometry. So how fast, when you see words like fast, fast makes you think of speed, which makes you think of velocity. Velocity is the derivative of motion of the position function. So when you ask something like how fast you're asking about a rate of change, you're asking about derivative. So we're asking a question about derivatives, which is why this is a related rates problem. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is six feet from the wall? OK, we'll come to that in a second. So what we're being asked here is we're asked, so the stuff in blue is what was given to us, we need to figure out what is the derivative of y with respect to t at the moment that the bottom of the ladder is six feet away from the wall. So we specifically know x is going to equal six. So we need to figure out the derivative of y with respect to t when x is equal to six. So we have the given rate and we have the unknown rate. We need to relate them together now. How are we going to relate these things together? Well, the diagram is actually really suggestive of what we should do, because when you look at this, this forms a right triangle. So maybe some type of trigonometric argument could be useful. I don't know anything about angle measures, so I don't want to really do sine, cosine, tangent or anything like that. But we do know that the ladder is 10 feet, that this ladder is non-collapsible. It's not going to change its length as it's falling here. So we know the ladder, the hypotenuse of the ladder, excuse me, the hypotenuse of the triangle is the ladder. It's going to be 10 feet long. And we also know at this moment in time, like if we just freeze time right now, x is going to be six feet from the wall. So just by the Pythagorean equation, we can relate these quantities together. And so that's going to be the equation we want, that x squared plus y squared is equal to 10 squared, or in other words, 100. Now, I don't want to plug in x equals six yet because I need that x and y to be related variables. So we don't want to plug in x equals six because that takes away the variability of x. We're allowing y to vary, and they're connected to each other by this related equation. These quantities x and y are related to each other. Now, I'm able to put a constant in for the ladder length because that won't ever change. No matter what moment of time we are in this situation, the ladder will always be 10 feet long. Therefore, 10 squared will always be 100. So we want to hesitate plugging in the number x equals six until we've taken the derivatives because derivatives are rates of change. If I were to plug prematurely the value x equals six in here, what I would be saying is that the distance between the wall and the ladder is always six feet. Well, if the distance is always six feet, then the height above the ground will always be, well, it's going to be eight feet, but we'll get to that in just a second. And so there won't be any variability, there won't be any change, but that would defy the reality that in fact the distance between the wall and the ladder is increasing by a foot per second, okay? So we need to have an equation that relates the two variables together. Once you've done that, we then can take the derivative of both sides of the equation with respect to time. We're going to take the derivative. We'll just use the prime notation to simplify that. So on the left-hand side, we get x squared plus y squared prime. On the right-hand side, you take the derivative of 100, it's a constant, it's going to go to zero, okay? On the right-hand side, excuse me, on the left-hand side, we take the derivative, we can separate the sum, so we're going to take x squared prime plus y squared prime, this is equal to zero. If you take the derivative of x squared, realize here that we're taking the derivative with respect to time, we're not taking the derivative with respect to y, we're not taking the derivative with respect to x. So the derivative of x squared is going to be two x times x prime. We need to have that inner derivative because of the chain rule. Similarly, when we take the derivative of y squared, we're going to get a two y, y prime, you cannot forget those derivatives, otherwise your calculation will be inaccurate. We're trying to figure out what y prime is, so if you want to solve for y prime, we can subtract the two x, x prime from both sides of the equation. This is going to end up giving us a two y, y prime equals negative two x prime, and then divide both sides of the equation by two y, and then we've now solved for the y prime that we need. We're going to see that y prime is equal to negative two x, x prime over two y. Simplifying that, notice the twos can cancel on top and bottom, and so now we're in a position where we see that the derivative is going to equal a negative x, x prime over y. All right, so now let's fill in the information that we now know because what we need, what we're after is we're after the derivative of y with respect to t at the moment that x equals six. So we can plug in the information we have. We know that x at this moment in time is going to be six feet, okay? We also know that the derivative of x with respect to time is always going to be one foot per second, okay? And then in the denominator, we have to do y. y is going to be measured in some type of feet, but what is that measurement? Come up to the original equation. We kind of doodled over it, but the original equation was x squared plus y squared equals 100. If we plug in x equals six now, we're going to get six squared plus y squared equals 100. Solving for this, we get that y is equal to the square root of 100 minus 36. That is, it's the square root of 64. That is, it's eight. It needs to be a positive distance. So eight feet is what we're looking for right there. So when we plug that in for y, we're going to end up with eight right here. And so simplifying that fraction, we end up with negative three fourths, or if you prefer negative 0.75, and this would then be feet per second. Notice, notice that the number we ended up with is a negative three quarters. What does that mean? What, why is it negative? Well, coming back to our picture, right? As the distance from the wall to the ladder is increasing because it's getting farther away. The bottom of the ladder is getting farther away from the wall. The portion that's touching the wall, the y is measuring the height above the ground. So this is getting smaller and smaller and smaller as the ladder is falling. And so we see that in this case, the velocity of the falling ladder turned out to be negative because the distance between the ground and the top of the ladder is getting smaller. That's what our variable y was measuring, the height. The height is getting smaller over time as the top is getting closer and closer to the ground. And so this gives us an example of how we can solve a related rates problem that might involve a right triangle relationship that is the Pythagorean relationship.