 Let's look now at an example of a related rate problem that's going to involve some trigonometry. We have an airplane flying east at 500 feet per second at a constant altitude of 4,000 feet. A search light on the ground lies directly under the path of the plane. If the light is to be kept on the plane, how fast is the search light revolving when the horizontal distance of the plane from the search light is 2,000 feet? So let's try to draw a picture, but I'm not going to attempt to draw an airplane for you. So here's the ground, and here's the airplane. I'm just representing as a dot because I cannot draw. Now we know that plane is flying at a constant altitude of 4,000 feet. So that's a quantity that's not going to change in our problem. So we can go ahead and put it into our diagram right away. Now the plane is flying east, which let's assume is to the right. So the plane is flying this way. Now remember, there's a search light on the ground. Let's put it over here. Put it in red so we can see it. And we know that as the plane is flying east, we want to know how fast is the search light revolving when the horizontal distance of the plane from the search light is 2,000 feet. So it sounds like we're going to need a variable to represent this horizontal distance of the plane from the search light. So let's go ahead and call that x. So think about this for a second, and I gave you a hint that it's going to involve some trigonometry. If we're talking about how fast is a search light revolving, if you imagine the search light revolving, it's creating an angle. So let's think of this as like a right triangle, and we have here an angle theta being described as that search light revolves. So let's think about the items that were given. We knew this 4,000 was a constant. We know that we're trying to find how fast is the search light revolving when the plane happens to be 2,000 feet from the search light, and we know the plane is flying east at 500 feet per second. Well, that horizontal distance x is really the distance from the plane as it flies east from the search light. So we can assume that dx dt is going to be 500 feet per second, and what we're trying to find is d theta dt. How fast is the search light revolving that angle? At the point in time when x is 2,000 feet. So if you take a look at your right triangle and the parts that you're given, try to think of what trig ratio we could set up. We could easily do a tangent ratio and set it up as tangent of theta. Remember tangent is opposite over adjacent. So that's going to be 400 or 4,000 rather over x. And that's what we're going to go ahead and do our implicit differentiation on. Now we will get to the point that we're going to, we know what x should be at the particular point in time in which we're finding d theta dt. We are going to have to figure out what theta is at that point in time. So we'll get to that in a second. So if we go ahead and do our implicit differentiation, remember derivative of tangent is secant squared, but remember we need to multiply that by d theta dt. Over on the right side, our derivative of 4,000 over x is going to be negative 4,000 over x squared dx dt. So now that we've done our implicit differentiation, we can go ahead and substitute in the values we know. We're going to have to find theta. So we'll do that in just a second. d theta dt is what we're trying to find. So that will stay as d theta dt. The x remember is 2,000. So we'll be substituting that in there. And then dx dt is 500. So really we need to go ahead and do a little side mass to figure out what theta is at this point in time. And we're trying to find d theta dt in the end. So let's go ahead and try to find theta first. So you have a little side work to do. So remember we knew tangent of theta was 4,000. And we're talking right now at the point in time when x is 2,000. So that's something you can simply solve on your calculator by using inverse tangent. Remember to be in radiant mode. So if you go ahead and try that, make sure you get the correct answer. It should come out as approximately 1.107 and that would be radians. And we're going to want to store that. And I'm going to store it as a in my calculator for angle. But that is a value you will need the full decimal value as you work through the remainder of the problem. So you are storing it in your calculator to use, but it's perfectly fine to round it off to thousands as you're documenting your work. So as we substitute everything in, we'd be doing secant squared of 1.107. D theta dt remember is what we're trying to solve for. We'd have negative 4,000 over 2,000 squared. And dx dt is 500. So let's go ahead and do the secant squared of 1.107. Definitely you'll be using that on your calculator. Secant is reciprocal of cosine. So get out your calculator and make sure you can do it correctly. You'll need to do cosine of 1.107. And actually you'll want to pull up whatever you stored it as. So I called it a. So do cosine of that. Get your answer for that. Do 1 divided by it. Get that answer. And then square it. What do you get? You should notice this quantity right come comes out as a nice perfect five. Now, if you had not stored that value, chances are you will not get it as nice as just five. That's actually how on the AP exam we can figure out if students stored their values to use or not. Because these problems are often rigged to come out really, really nicely if you store your decimals. All right, that's why you want to get in the habit of doing that. So simply five d theta dt. How nice that worked out. On the other side, if you simplify all this, well that just comes out as negative a half easy. So what we end up with is d theta dt is a negative one-tenth. So for our final answer and conclusion, we would have that when the horizontal distance of the plane from the search light is 2000 feet, the light is revolving at a rate of one-tenth radians per second.