 Another problem that appears in every calculus book might be called the lighthouse problem. And this is because a secret cabal of math- There is no secret cabal of mathematicians who control the world. But I'll have to leave the exposure of this group to another day. So the standard lighthouse problem looks something like this. We have a lighthouse some distance from the shore, and the beam turns at some rate, and we want to find how rapidly the beam is moving along the shoreline when the beam meets the shoreline at a point. So again, while this perspective view is how we might actually see things, it's not the most useful view, and so we might shift our viewpoint. And this will help us identify what is variable and what is constant. So the distance between the lighthouse and the shoreline is a constant one kilometer. On the other hand, the distance from the lighthouse to the point where the beam meets the shore is variable. And so is the distance along the shoreline. And so that suggests we might let X be the distance along the shoreline and S the distance to the point where the beam meets the shoreline. So let's go ahead and write down our relationship between X and S, where we use the fact that S, X, and 1 are the lengths of the hypotenuse and two sides of a right triangle. Since we want to know how fast the beam is moving along the shoreline, it seems that we want to find dx dt, so we differentiate with respect to t. And that's great, except there's a problem. We don't know ds dt, and so that means we need to find another relationship. And this is not unusual when solving real-world problems. It may be that our first idea gets us to a point and we realize we need a little bit more information. And so rather than scrapping all of the work that we've done so far, we might see if we can find that extra information. So let's go back to our schematic diagram. Notice that S, 1, and the shoreline form a right triangle. We've already used the Pythagorean theorem, but we could also use a trigonometric function. And here we might see that 1 over S is equal to the cosine of theta, where theta is the angle between the perpendicular to the shore and the beam. So from 1 over S equals cosine theta, we can differentiate with respect to t. And now I have a formula for ds dt. Except now I need to know what d theta dt is. So let's think about that. d theta dt is the rate of change of the angle with respect to time. If we knew how rapidly this angle was changing, we'd know d theta dt. But we do know how rapidly the angle is changing. The beam rotates once every three seconds. Now it's very important to remember if you're working calculus, every angular measurement must be in radians. So we need to convert this amount. Since the beam rotates once every 30 seconds, then the rate of change of the angle with respect to time, we can express that as one rotation for every three seconds. But one rotation is the same as two pi radians. So since they are equal, I can replace the one with the other. And I can leave the three seconds alone. And that gives me my d theta dt of 2 pi over 3 radians per second. So that tells us d theta dt. We also know S is two kilometers because that's the distance to where the beam meets the shoreline. But we do need to know x. We can use the Pythagorean theorem for that. And we also need to know sine of theta, and we can use our right triangle for that. So substituting in these values gives us ds dt. And then once we know ds dt, we can substitute it in to find dx dt. Now after you've solved a problem, a good habit to get into is to look at your solution and ask yourself, could I have done this a different, possibly more efficient way? And in fact, there is a more direct solution if we start with the trigonometric relationship. There are several trigonometric relationships we could use, but because we're interested in the rate of change of x, we should use one that involves x directly. And in this case, we can use tangent of theta, differentiating with respect to t. Now we need secant theta, which we can get once we know cosine of theta. When the beam is two kilometers away, then cosine of theta will be, which gives us secant theta equal to two. We've already figured out d theta dt is two pi over three radians per second. And so that allows us to find dx dt.