 So, one of the things you learn when you get a PhD in mathematics is that every calculus textbook must include some version of the following problem. This dictate is enforced by a secret cabal of mathematicians who can control every aspect of our lives, and is led by Professor J. No secret cabal of mathematicians exist. Go about your business. So here's the situation. We have a 25 foot ladder resting against the wall, and the base of the ladder is moving away from the wall at 3 feet per second, while the ladder itself maintains contact with the wall. So how rapidly is the top of the ladder falling when the base is 8 feet away from the top of the wall? Now, while these drawings in perspective are great for showing what actually happens, they're not so good for representing what actually happens. So let's change our viewpoint so that we can get a better picture of what's really going on. So, it helps to identify the variables and constants. So let's let our ladder slide a little bit. So we see that the distance to the top of the ladder and the distance to the base of the ladder from the wall are variable quantities. So we need to represent them as variables. So let y be the height of the top of the ladder, and x be the distance of the ladder from the base of the wall. And we'd like to find some relationship between these variables and maybe some other things. Well, notice that the length of the ladder doesn't change, and that x, y, and the ladder itself form three sides of a right triangle. And that means the Pythagorean theorem tells us that x squared plus y squared equals 25 squared. So we're told that the base of the ladder is pulled away from the wall at 3 feet per second. And so this is a rate of change of x with respect to time, and so we're differentiating with respect to time, t. We want to know how rapidly the top of the ladder is falling when the base is 8 feet away from the wall, and so that means we want to find dy dt when x is equal to 8. So we have our relationship between x and y, so let's differentiate with respect to time. We want x equal to 8, and since x squared plus y squared equals 25 squared, then we can find y. So we have a derivative relationship as well as x equals 8, y equals square root 561, and the x dt equals 3. So I'll substitute those into my equation and solve for dy dt. And don't forget those units. y is measured in feet, t is measured in seconds, so dy dt is going to be a value in feet per second.