 The difficulty with related rates is that they are as variable as the types of problems that you could write with them. So here's another example, a very classic problem. We have a lighthouse one kilometer from a shoreline that runs north-south. Now this is a lighthouse, so it has a light beam, and the light beam makes one revolution every six seconds. So the question is how rapidly is the beam moving along the shore when it touches a point two kilometers north of the lighthouse? So again, sketch a picture. You don't have to have a very artistic picture, so here's my attempt. And again, one important bit of analysis that you want to undertake is to identify what changes and what stays the same as the beam moves along the shoreline. So here I have my very artistically drawn lighthouse, and here's my beam of light, and it's hitting some point on the shore. You can tell it's shore because it's got all this gravel here, and here's the water. And what's going to happen? Well, as that beam swings around, it's going to illuminate different points along the shoreline. So my beam, as it moves, will do something like that. And the thing to notice here is that the distance between the lighthouse and the shore doesn't change. The lighthouse is one kilometer from the shoreline, it's always one kilometer from the shoreline. On the other hand, where the beam hits the shoreline does change over time. So that position should be a variable, and the other thing that is important here, the other thing that changes is the angle that the beam makes with some sort of perpendicular distance. So if I imagine this perpendicular distance to the shoreline, this angle changes over time. And what that means is that I can take the position and the angle, the position along the shoreline, and the angle that I'm making with this perpendicular distance. These should be variables, how about x and theta, but the distance to the shore is a constant one kilometer. And again, the identification of these variables is important because we are given some information here that looks like it might be a constant. I'm interested in a point two kilometers north of the lighthouse up here someplace. And it's important for setting this problem up to recognize that the position of the beam on the shoreline is a variable, even though we're interested when it happens to have a value equal to two kilometers. Now, what do we actually want to know? Well, we're interested in knowing how rapidly the beam is moving along the shore. So that's a rate of change of the beam's position with respect to probably time, because again, our time is something that is tied in with a changing amount. The light beam makes one revolution every six seconds. So that's going to indicate where the beam is pointing to, and that's going around once every six seconds. So presumably our rate of change is going to be with respect to time. And again, the point of interest is going to be when we have touched a point two kilometers north of the lighthouse, so up here someplace. So I can express that as I want to find the rate of change of the position with respect to time, rate of change of the beam's position, when x is equal to two, when I'm up here someplace. Now, I do have some extra information here. I have a rate of change of an angle with respect to time. So this is one revolution every six seconds. And here's the important thing to remember. The only angle measures in calculus are radians. Any other way of measuring an angle, degrees, revolutions, anything like that, any other measurement of an angle is going to make the problem more complicated. You want to always be talking in radians. Fortunately, it's easy to convert back and forth between the different types of measurements. So remember that once around a circle, one revolution is the same as two pi radians. So this rate of change, one revolution every six seconds, is the same as two pi radians every six seconds. So this rate of change that we're given, it's really two pi over six. And so I had the given information d theta dt is two pi over six. So now I have a couple of variables, x and theta, and I want to find any relationship at all between them. And a useful idea is that many of these problems come out of trigonometry, so look for the right triangle. Here's this beautiful right triangle right here. And in a right triangle, I know that there is a relationship between the angle and the side's opposite and adjacent. That's our tangent relationship. Tangent of theta is opposite over adjacent. That's x over one. I'll undertake the enormous amount of algebraic simplification we can do. And there's my relationship. I want the derivative with respect to t. So derivative tangent is secant squared. Don't forget the chain rule. Derivative of x is derivative of x. Again, don't forget the chain rule. And so if I want to find dx dt, I need to know d theta dt. Got it. And I need to know secant of theta. Maybe a little bit more complicated, but not too terribly difficult because I can use the other trigonometric identity. Tan squared theta plus one equals secant squared theta. I know the tangent of theta is two because I want x to be two. So at the theta I'm interested in, opposite is two. And that tells me secant is going to be plus or minus root of five. Since the point is to the north, I'm going to take the positive value. So going back to our relationship on the derivative, I know the secant of theta is square root of five. And I'll substitute in my value for secant, my value for d theta dt, and out pops the value for dx dt. One last step. We want to identify the units. dx dt is going to be units of x over units of t. And x is measured in kilometers. And t is measured in seconds. So dx dt is going to be measured in kilometers per second. Now one last reality check here. Note that dx dt is positive, which means that x is increasing as t increases. In other words, at a later point in time, x is bigger. Which in this context means that if I have the picture here at the instant that we're two kilometers north of the lighthouse, a little bit later that beam of light is going to be hitting up here someplace. Now our d theta dt is two pi over six. Again, what that's saying is theta is increasing with respect to time. Now if you look at how theta is being measured, I'm measuring theta from this horizontal line. So here I have some value of theta. A larger value of theta, a little bit later, theta is larger. A larger value of theta is going to open this up a little bit. And the beam is going to point out here someplace. And that's consistent with our determination that dx dt is positive. Because if I'm here now, a little bit later on, theta is a larger angle. And I'm pointing to a place further up the shoreline. So our signs, SIGN signs, are all consistent and we are in good shape.