 In this video, we're gonna provide the solution to question 21 for the practice final exam for math 1210, in which case we have a ladder that's 13 feet long, it's rested against a wall, and the bottom of the ladder is gonna be sliding away from the wall. So I'm gonna draw a picture to help us understand what's going on here. Here's gonna be the ground that the ladder's resting upon. Here's the wall that the ladder will be resting against and it's fair to assume that the wall is perpendicular to the ground. And in the color blue, I'm gonna put the ladder here against the wall. All right, so this is our ladder. What we know about the ladder is it's 13 feet long. It doesn't mention anything about a collapsing ladder or anything, so I think it's fair to assume that the length of the ladder will be constant 13 throughout the rest of it. We've seen problems like this before where the, since the ladder is sliding away from the wall, we're gonna let X be the distance between the wall and the bottom of the ladder. This should be increasing with time. Well, as the ladder slips, it also means the ladder will fall. Let Y be the height above the ground of the ladder. That is, let Y be the distance from the bottom that is the ground to the top of the ladder at a given moment of time. And so these quantities are gonna be related to each other by the typical Pythagorean relationship. X squared plus Y squared is equal to 13 squared, which 13 squared is 169. All right, so if the bottom of the ladder is sliding away from the wall at a rate of one quarter foot per second, how fast is the top of the ladder moving down? We'll get to the rest of it in a second. So what is this value here about the a quarter foot per second? This is describing the derivative here. So we get that DX, DT at this moment in time is gonna be one fourth of foot per second. So we know how quickly the X value is changing over time. We wanna know how fast it's falling down. So we have this question about what is DY over DT at the moment, at the moment. So let's go back to the sentence. How fast is the top of the ladder moving down? That would be DY, DT. When the bottom of the ladder is 12 feet away from the wall. So we're trying to figure out what is, what is DY, DT when X equals 12? So there's a couple of things we need to do here. So first we're gonna take the derivative implicitly to help us find a relationship between X prime and Y prime. So taking the derivative on the left-hand side, we're taking the derivative with respect to time. We're gonna get 2XX prime plus 2YY prime. On the right-hand side, you take the derivative of constants at zero. I'm gonna divide both sides of the equation by two. This gives me XX prime plus YY prime equals zero. And you just solve for Y prime. So we'll subtract XX prime from both sides. So YY prime equals negative XX prime. And then divide both sides by Y. We see that Y prime is gonna equal negative XX prime over Y. All right, so we have a formula for Y prime. We would plug in the value 12 for X. We can plug in the value one quarter for X prime, but we don't know what Y is. So we have to come back to our original relationship. That is this Pythagorean relationship. If X equals 12, we're gonna get 12 squared plus Y squared equals 169. We solve for Y. 12 squared is 144. And so we're going to subtract 144 from both sides of the equation. This gives us that Y squared equals 25. That is Y equals five. So with these values in hand, we can plug them into our formula. So we get that DY over DT at X equals 12 is gonna be negative 12 times one fourth over five here, which four goes into 12 three times. So we end up with negative three over five feet per second. And it ought to be negative because the ladder's falling. So the height above the ground is getting smaller over time.