 Welcome back to our lecture series, Math 1210, Calculus I for students at Southern Utah University. As usual, I'll be your professor today, Dr. Angela Missildine. In lecture 29, we want to continue what we talked about in lecture 28. That is, we want to do more problems about related rates. And so before we start this lecture, I want to review some strategies of solving related rates problems that we saw previously, but actually kind of summarize them and put them in an official list. So these are things that I would want someone working on related rates problems to do. The first thing to do is actually to read the problem, but not just read the problem, you need to read it carefully. We're not reading it the same way we read a Harry Potter novel. We have to read it for understanding. Why are these words here? What's the context? What do they mean? What cues are telling me that this is such and such problem or disinformation, that information? So read the problem. You might have to read them more than once. That's perfectly fine. Step two is to then draw the picture as appropriate. Now, not every problem is going to need a picture, but most of the time there is a good geometric relationship that relates the information you're given. This can be based in geometry or physics or others, other situations. Draw a picture when appropriate. It helps you visualize these problems when you can actually see them. Another step that students often skip, which is extremely important, is to introduce notation. We need to use some variables, whether they be X, Y, and Z, A, B, and C, or maybe some other symbols with a mnemonic device. We should assign variables, which should be measuring some quantities. And these quantities should be functions of time. In related rates problems, these quantities are functions of time. So identify what are the quantities that change with respect to time and be specific. What are their units? Are we measuring these things in, say, feet or inches, centimeters, be specific? And also, it can be very useful to actually write out what it is, like X equals, it's the length, maybe the length I said, between such and such. You keep on going, right? It's better to be specific. It could be that as a student, maybe you're required to write this out because you're going to be graded upon it. But even if you're not being graded upon it, it helps you articulate exactly what does that variable mean. We're going to see in a second when we do an example in this video that if we're not careful with the variable, we can get some things mixed up that we don't want to happen. So it's very important to be very specific with our variables, what they're meaning. So with respect to the notation that you've introduced, figure out what information has been given, what information is expected, and when appropriate, make mention of what information are derivatives, like what derivatives do we know, what derivatives do we want? It's a critical step in a related rates problem. Once we've identified the variables we have in play and we know what derivative we need to find, then you're going to look for an equation that relates together these quantities. That's going to be important that oftentimes we can manipulate some type of geometric relationship or physical relationship or this will often, this, how we do that will depend on the problem in play here. Once we have this equation, we're then going to take the derivative of both sides of the equation with respect to time, which most likely time will not be a variable in the equation. So we have to take the derivative implicitly, a.k.a. we have to use the chain rule. Once you have down the derivatives, you can plug in the information that you know into this equation and then solve for the unknown derivative that's not already determined and that then solves the problem. That's how we work through a related rates problem. Let me demonstrate that to you with the following example. Now in terms of drawing the diagram, I took the derivative using the computer to help me produce a picture. So imagine I'm drawing this picture as I'm trying to solve this problem here. So don't fix it too much on the picture side of things here, but let's read through it. So a spotlight is on the ground, shines on a wall 12 meters away. If a man two meters tall walks from the spotlight towards the building at a rate of 1.6 meters per second, how fast is the length of his shadow on the building decreasing when he is four meters from the building? So let's try to process this. Let's go through it one by one by one, these sentences here. So we have a spotlight, it's on the ground, and it shines on a wall that's 12 meters away. So if I just look at that right there, what do I understand? There's again the spotlight, which at the moment not sure why that's significant. I guess it's shining, so the light is coming out of it. So the light must be significant to some degree. It's on the ground. And so that makes me think of it's really sitting on top of a horizontal line. And then there's a wall, which the wall has got to be perpendicular to the ground. So there's some type of right angle relationship going on here. And yeah, the light is shining. So if it's like shine, shine, you get something like that. Okay. So I'm not sure what the significance of the shining is yet, but we do know the distance between the light and the wall is going to be 12 meters. Okay. So if a man walks, or he's two meters tall, so we have our man right here, he's walking, he's walking from the spotlight towards the building, building, what building? When we heard a wall, it's okay, the wall must be a part of the building. Okay. He's walking at a rate of 1.6 meters per second. So that's a speed. I'm also looking at the units there, that's going to be some type of velocity. It's because we know he's walking towards the wall, it's a speed and the direction. So this is a derivative of the position of the man. So that's going to be relevant. I'm going to come back to that in a second. So how fast is the length of the shadow of the building decreasing? So look at the words there for a moment. Fast decreasing, that sounds like a rate of change. That's, that is a derivative, derivative of what? Derivative of length of his shadow. So this is where the light of, this is where the light of the spotlight comes into play here. It casts its light on the building. And so the man has a shadow on the building. As the man gets closer to the building, we see that then the angle of light will be different and then the shadow is different. So as he walks closer and closer to the building, we see the shadow is getting smaller. That's the basic idea. So let's come to this picture that's drawn here by the computer and label what we now know about our problem. Okay. So the man, the man is walking towards the wall. That's happening at 1.6 meters per second. Significant to this problem is going to be the length of the shadow. So what if we call the length of the shadow, this variable here, why? What is why is the length there? And so what we then need to figure out is how fast the length is decreasing. Essentially, what we have to figure out then is what is dy dt at a specific moment in time, which we'll come back to that in a second. So we need to figure out what is the change of y over time. Okay. That's somehow related to this distance or to this velocity right here, 1.6. So if we call y the vertical variable, then x is the horizontal one, but what does x need to be? It depends on a couple of things here. I guess what I'm trying to say is there's more than one option you can choose for x. So if we draw our picture right here, basically our man is just a stick, right? We have y over here. It turns out that we couldn't make x be the distance between the man and the wall. That actually makes a lot of sense for what x could be. Another option is we could make the distance from the spotlight to the man. That could also be x. And for reasons that'll become clear a little bit later in this problem, this is actually where I'm going to set x equal to. I want x to be the distance between the spotlight and the man. And as such, that means that the distance between the man and the spotlight or excuse me, the man to the wall is going to be 12 minus x. After all, the total distance between the spotlight and the wall is 12 meters. So we get that right here so we can fill that in. That's how we're going to decide to get this x right here. Okay. Then let's come back to this idea about the derivative. What's this derivative value going to be? Well, if the position, if the variable x is measuring the distance between the man and the spotlight right here, then as he walks towards the wall, x is going to be getting bigger and bigger and bigger as time elapses. This gives us in fact a positive 1.6. This always shows you a difference that if we had a different variable here, let's call this z, for example. If z is the distance between the man and the wall, then well, the relationship between z and x is this 12 minus x right here. But more importantly, when we take the derivative of z, that's going to be a negative 1.6 meters per second right there. And so the choice of variable makes a difference on other parts of the problem. So choosing one x versus another will affect whether we have a positive derivative or a negative derivative. This is why step number three was so important in this process. We need to be specific about the notation we're using. So here's our x value. And so how is then x related to y? Because we know something about the derivative of x. We know something about, or we want to know something about the derivative of y. How are these related together? So what are some possible approaches? Well, we could think of it as like, well, I know something about the horizontal side of this triangle. I know something about the vertical side of this triangle. It is a right triangle. If I knew something about the hypotenuse, then I could use the Pythagorean equation. That could be like a z or something. But I don't know anything about the hypotenuse. I don't think that's going to be the best approach. But we also would be like, oh, if I know the horizontal and the vertical side, I know the opposite and adjacent sides right here and here, in which case then if I knew the angle, then I could do something like tangent ratio. But I don't know anything about the angle. So that also doesn't seem like the best approach. It turns out that the approach we want to take here is actually to use similar triangles. Because why is the height of the man relevant? So we have this triangle that involves just the spotlight in the man. So the man is X meter or X meters away from the spotlight, and he himself is two meters. But then you have the entire triangle, the triangle associated to the shadow right here. In which case, these are similar triangles have the exact same angles. We know that the spotlight to the wall is exactly 12 meters. And then the other side is a y. And so these are related together in this fashion. So notice we're going to get two over X coincides with y over 12, for which case we could cross multiply and we get 24 equals XY. So just the usual cross multiplication there when you have this proportion problems, we get two times 12 times X times Y, for which then we could solve for Y and get Y equals 24 over X. Now we're going to take the derivative here with respect to time. So we're going to get that Y prime is equal to 24 over X, which as we're taking a derivative, I want to think of this as a power function. So I take it as X to the negative one. So we're going to get Y prime equals negative 24 X to negative two times X prime, or in other words, negative 24 X prime over X squared. So we then plug in the values we didn't know. So which derivative were we looking for? We're looking for the derivative when the distance between the man and the wall is four meters. Now that's not the X value. So if we try to plug in X equals four, we're going to get something that's incorrect. Instead, what we're looking for is over here, we see that if the man is four meters to the wall, it means he's eight meters from the spotlight. That's how we chose the X value. So that's again critical that we're very clear on what our variables mean here. So we need to figure out what the derivative is when X equals four. So plugging these things in here, we get negative 24 times X prime, which is going to be 1.6. And this sits above, of course, eight squared, for which then if we try to simplify this thing, I won't bother you with your arithmetic too much, but this will simplify to be a negative 0.6 meters per second. And we should expect this to be negative because after all, as the man gets closer and closer and closer to the wall, the shadow is going to get smaller, smaller, smaller. So therefore, since the shadow is decreasing, we should expect a negative derivative here. And this then illustrates the seven principles we saw on the previous slide. Notice what we did here. We read the problem, we drew a picture. Have the computer help me, of course. And over and over again, step number three, emphasize the notation. What does it mean? We figured out what information we knew, we figured out what information we needed to know. This step number five, we found an equation that was this proportion equation we had here with the similar triangles. We then took the derivative of set equation and then solved for the unknown derivative right here using the information we had that step number seven. If you follow these seven steps solving a related rates problem, you're going to be just fine. It's a good strategy. And just be patient with yourself. You can work through these story problems. I know you can.