 Hello and welcome to screencast about related rates. Related rates problem involves trig as well as fishing. All right, so let's say you are going to be reeling a perch in at a rate of one foot per second from a 0.5 feet above the water. At what rate is the angle between the line and the water changing when there is a total of 13 feet of line out? OK, so let's say here's the water, and it's blue, like my pen, and let's say we are five feet above the water. So pretend like this is a person sitting here fishing. So there's my person. Oh, that's a really bad drawing, but that's OK. All right, and then my fish. Let's say that somewhere out here. OK, there's my little fish. So then here's my line going out to my fish. And you'll notice I'm making somewhat of a right triangle. OK, so obviously you can pretend that this is actually a right angle and that sort of thing. So there's our fish. So here's our angle that we're trying to figure out. I'm going to call that theta, because it wants the angle between the line and the water. So that would be this angle right in here. And then let's see, we also know that we're eventually going to have 13 feet of line out. But hopefully, if you've ever gone fishing, or you can just picture what's going on here, you've got a certain amount of line, but then you're reeling it in. So this variable is actually changing. So I'm going to call this L for line. Only because it's not fixed, it's going to be changing. This five, though, is fixed. We're going to stay exactly five feet out of the water the whole time. This side here is also going to be changing. So let's go ahead and call that X, just in case we need it later. It doesn't seem totally relevant right now, but we may need it eventually. OK, so again, X and L are changing as we are reeling this in. So our triangle starts out with a really long hypotenuse, and then the hypotenuse is getting shorter and shorter and shorter. OK, so let's see. We need this theta, and I know the side that's opposite. And then I also know something about this L. So what is going on with this L? Well, it's telling us that it's being reeled in at a rate of one foot per second. So if we do dL dt, so that's the change in line per time, that's going to be one foot per second. OK, so I know something about this L, so I'd kind of like to be able to work that into my equation that I'm going to have to set up. So I need theta, I know 5, and I know something about L. So let's see, what would that be? So we've got a theta, we've got an angle, we've got the side opposite, and we've got the hypotenuse. So that, in this case, is a sign. So we know that the sign of theta equals opposite of hypotenuse, so that is 5 over L. OK, now we know we are given dL dt, and then we're trying to find at what rate is the angle changing. So remember, rate tells us that's a derivative. So we need to find d theta dt. And again, this is when L is 13. OK, but do not plug this 13 in yet. If you do, you go to do the derivative, you're going to end up with a 0. So you do not want to do that. So this, you don't only want to plug in at the very end. Same thing with our one foot per second. Plus, we don't have any derivatives in here yet to be able to plug that into. OK, I'm going to rewrite this quantity so I can see what's happening a little bit easier. So this is the same thing as 5L to the negative 1. OK, so now I want to do the derivative of both sides with respect to t, because again, we're talking about rates. So rates have to do with some sort of time. So the derivative of sine is cosine theta, and theta is changing with respect to time. So I'm going to have to tack on a d theta dt, thinking about the chain rule. Like theta is changing as my time is going by. Like as I'm reeling this thing in, this theta is going to be changing. OK, so that's where this piece comes from. Now let's do the derivative of this side. So that's going to be negative 5L to the negative 2. And then again, is the line changing with respect to time? Absolutely. So that's our dL dt. OK, fantastic. So we need d theta dt. We know L is 13. We know dL dt is, oh, and let's go back and look at this one again. If I'm reeling this in, is this rate a positive rate or a negative rate? And this rate actually is going to end up being a negative. I should put that in there to begin with. Whoopsie. So here, this is because the line is coming in. So that's kind of a negative idea when you think about it. If the line were going out, that would be a positive idea, because you'd be adding more line to it. All right, so I need to solve this big mass for dL dt, or sorry, for d theta dt. So I'm going to divide both sides by cosine. So d theta dt, and I'm also going to do some rewriting here. I see this negative exponent, so let's go ahead and kick that to the denominator. So I'm going to rewrite this as negative 5 dL dt over cosine of theta, because remember, I was dividing by that. And then I'm going to say this is L squared. OK, so now let's start plugging in some stuff that we know. So negative 5 times negative 1 over cosine theta. I don't think we know that one yet. And then L, I said, was 13, so that's going to be squared. OK, well, let's go back and look at our triangle. So cosine is adjacent over hypotenuse. And sure enough, yeah, we're definitely going to need that. So let me redraw my triangle over here. So I know L is 13. We can go ahead and pretend like that side is known. We know that that side is known now. So thinking back to your Pythagorean theorem then, if we call this side x, so 13 squared equals 5 squared plus x squared. And if you crunch out all that, x is going to end up giving us 12. Or maybe you remember that there's a 5, 12, 13 triangle. I don't know. I don't like to memorize stuff, so I'd prefer to figure it out myself. OK, so if we know that this side is 12, there's our theta. So we know cosine of theta is going to be 12 over 13. So that's going to give us, let's see, a 5 in the numerator. Then this is 12 over 13, and then times 13 squared. So one of those 13s will cancel as well. So this will end up giving us 5 over 12 times 13 is 156. And if you like, this is an exact value. That would be perfect. So the units of this is going to be radians per second. Now how do I know it's radians? Well, because that's how we always do trig in calculus is in radians. And then we know it's seconds because they gave us here the unit of our dL dt is in seconds. So if you like this answer, that's fine. If you want a decimal, this would be approximately 0.032 radians per second. And that would be the answer to this question. So again, just like with optimization, I think it's smart to draw a picture, see what's going on, and then set up your equations and do your derivatives from there. Thank you for watching.