 Hi everyone, let's take a look at a related rates problem that is going to involve proportions with similar triangles. A floodlight at ground level is illuminating a building 100 meters away. A girl 1.8 meters tall is walking away from the floodlight to the building along a path that is perpendicular to the building. If she walks at a rate of 1 meter per second, at what rate is her shadow on the building decreasing when she is 60 meters from the building? So let's draw it out. Here's the ground. Here's the building. We'll add windows to the building. And suppose here's the girl. Now she's 1.8 meters tall. I don't think she's going to grow during the course of this scenario. So that is a constant value that we can put into our diagram right away. Now we have this floodlight. So let's put that into our diagram and then we'll be able to get some variables. So the floodlight, suppose that's over here because remember the girl is walking away from the floodlight towards the building. And really what happens is she's going to have a shadow that's being cast here on the building. So we're trying to figure out the rate at which the length of that shadow is decreasing. So if you imagine as she gets closer and closer to the building, the angle, if you think of this angle here, let's draw this out. So there's an angle here of elevation. As she gets closer and closer to the building, that angle of the elevation is decreasing. Therefore her shadow and its height on the building is decreasing. So let's call the height of her shadow on the building Y. We know that she's walking this way towards the building. So really what's changing is her distance from the floodlight. So let's call that distance, the distance of her from the floodlight, let's call that X. And I think that's pretty much everything we need to get started. So let's go ahead and figure out what all of our variables are and what equation we're going to set up to solve. So we're told that the girl is walking towards the building away from the floodlight at a rate of one meter per second. So that becomes the rate that we're given. And since it's the distance of her from the floodlight, that's DXDT. And what we're trying to find is the rate at which the shadow that's being cast on the building is decreasing. So that's DYDT. Because of the fact that we know it's supposed to be decreasing, this better come out as a negative answer in the end. So we'll see if that works out. And we're trying to figure that out at the point in time when she is 60 meters from the building. Now the other thing that we've neglected to put in here is that we know that the spotlight is 100 meters from the building and that's a fixed distance. So if we're trying to figure out the rate at which her shadow is decreasing when she is 60 meters from the building, well that means the X has to be 40. So hopefully that makes sense to you. So what we're going to do is set up a proportion of the similar triangles. We have a little right triangle here and then we have the big one around it. So remember there's lots of different ways to set up similar triangles and proportions. So I think I'm just going to do it as 1.8 to X. So her height to the base of that little triangle and that's going to be equivalent to Y over 100. So we'll go ahead and set it up that way and we'll do our implicit differentiation from there. So to begin doing our implicit differentiation, let's go ahead and cross multiply. And because we have X times Y, we're going to have to do product rule over there. So remember we're going to have X then times the derivative of Y which would just be 1 dy dt. Remember we're differentiating with respect to t the time. Then we'll have the Y and we'll need to multiply by derivative of X which is going to be 1 dx dt. And that of course is equal to 0, the derivative of 180. So this is the point at which we'd want to start substituting in our values. We know X is 40 because we're talking about the point in time at which remember she was 60 meters from the building so it meant she was 40 meters, the X from the spotlight. We're trying to find dy dt. Now the Y we need to find, so this is where we're going to have to do a little side scratch work. We know dx dt, that's one, but we're going to have to do a little bit of work to find the Y. So let's do that. So to find the Y, we're going to go back to the proportion we set up and we're going to put in place of the X the 40. So we'd have 1.8 over 40 equals Y over 100 and that's going to allow us to solve for Y at this particular point in time in which X is 40. So if you go ahead and do that, let you try it on your own, you should get that Y is 4.5 if you want to give it a shot. So that's what we'll put over here in place of the Y. And then of course dx dt simply was 1. So if we go ahead and solve for dy dt now remember we're thinking that it should be a negative quantity because the height of that shadow is decreasing. So when we take the 4.5 over to the other side, there's your negative. We divide that by 40. Makes sense if we go ahead and finish that out. We get a decimal, a negative, 0.1125. So our conclusion then would be that when the girl is 60 meters from the building, or you could say 40 meters from the spotlight if you prefer, her shadow is decreasing at a rate of 0.1125 meters per second. Notice that when we write up our conclusion we do not include that negative in the answer. The negative is really just telling us that the the length of the shadow is decreasing, which we've already taken care of by saying the shadow is decreasing at a rate of 0.1125 meters per second. Please take note of little things like that because they do make a difference. Also again, as we've noticed in other problems, the units of measure, it's just meters per second because the height is a linear measurement which would have been measured in meters.