 Hello and welcome to this video, which is an example of a basic related rates question. So here we're given that the edges of a square are growing at a constant rate of two inches per second. And we want to know at what rate is the area of the square changing when a side is 10 inches long. So the picture here is that we have a square and try to draw it perfectly square. But it's not static. It could be growing. For example, when you take your mouse on your computer screen and select an area, then drag on it. It's increasing. The sides are increasing. So this side's expanding, this side's expanding. And so is this kind of expanding out this way. And we want to know at what rate is the area of the stuff on the inside expanding as you pull those sides out. So the sides are increasing at a constant rate of two inches per second. This is a related rates question because we are being asked to find the rate of change in one quantity given the rate of change in another quantity. So in the quick review screencast, we stated a little workflow for how to think about and think through these related rates questions. The very first step of which is to identify the variable whose rate of change you want. So what variable is that? Well, it says at what rate is the area changing? Okay, so that is the area is the variable whose rate of change we want. Now secondly, we need to relate this variable back to the other variables in the problem. There's really only one other variable that shows up explicitly in this problem. That's the side length. Now thinking back to what we know about geometry, it's very easy to pull this out. How does area of a square relate to the length of a side? Well, simply that a area we'll call it a is equal to the square of side lengths. So let's call the side length x and so a is equal to x squared. Again, a is area, x is the length of one of the sides and all the sides are the same size because it's a square. So we've identified the variable whose rate of change we're looking for and we've related that variable back to all the other variables in the problem. Now we need to start thinking about a common third input variable to these two guys here. So what do a and x have in common? Well they're both growing. They're both changing with respect to time and so we actually have a common third variable here and that's time. Both a and x are functions of time in some way or another. So let's call that time t and we'll measure it in seconds because that's the time units that are given to us in the problem. So a and x are not just static or unchanging. They are changing with time so they're both related back to time variables. So now that helps us get ready to do a calculation that will give us the rate that we want. Okay so I'm just going to copy down what we already said which is something that a is equal to x squared. Now in order to solve the problem I need to get a derivative into the picture here. I'm being asked to find the rate at which the area is changing right when the edge length is 10 inches. So I need to find a rate of change and that's a derivative. Now derivative with respect to what? Well a is a function of x but it's really a function of time as well and x is also a function of time. So I'm going to take the derivative on both sides of this equation with respect to time. So the derivative with respect to time of a will be equal to the derivative with respect to time of x squared. And remember again I can't state this enough that x and a are functions of time. And so now when I move towards getting the derivative I'm going to need to use implicit differentiation because here neither function that I'm taking the derivative of a or x squared is explicitly a function of t. It's implicit not totally sure exactly how. So on the left hand side the derivative with respect to t of a is just d, a, d, t. Now we want to keep our eye on that quantity because that is the rate of change that we're being asked to find. It's the how fast is the area changing with respect to time. So I'll keep an eye on that. Now on the other side of the equation I'm taking the derivative of x squared with respect to time. And so using implicit differentiation which remember is really just the chain rule I would take the derivative of what I have and then multiply by the derivative of the quote on quote inside function which in this case is x. And so that's going to give me dx, dt. All right. So it's just like a regular derivative not caring about what the variables say but then I have to multiply by the derivative of x with respect to time because x is really a function. It's not just a variable law to itself. So now I'm ready to begin the final stage of answering this question. I now have identified what I want to find. That's the derivative of a with respect to time. And I have that related to another derivative namely the derivative of the side length with respect to time. Now the last step is to put in everything that I know every other bit of information that's given to me in the problem statement and try to solve for the thing in the green. Now technically speaking I don't want just any derivative of a. I want the derivative of a with respect to time when x is equal to 10. The exact moment that the square is 10 inches on a side. And that's going to help me set up the rest of the equation. So this would be two times 10. I'm going to lock the changing picture here in at one moment when x equals 10. And then dx dt remember was constant. That's always two. Always two inches per second. And so this comes out to be 40. Now let's make sure we got the units correct on this as well. Just to go in here. 10 is in inches. That's a side length. Two is given in inches per second. And inches times inches per second give me square inches per second. And that is a correct unit for the derivative of an area with respect to time. So in this fairly simple problem we see the workflow played out pretty well. We start by identifying the quantity whose derivative we're asked to find whose rate of change we're asked to find. We relate it to the other quantities in the problem through some form of an equation. We look for a common third variable to all those other variables. Usually that is time and the variables that you see in the problem like area and side length and so forth are often functions of time. And then we simply set up the equation, take the derivative with respect to time using implicit differentiation, put in what we know, and then solve for the derivative we want. Thanks for watching.