 In this video, we provide the solution to question number 15 for practice exam number three for math 1210, in which case we have a related rates problem. A ship is traveling due north six miles east of the Maryland coast at a constant speed of 15 miles per hour on the shore is a lighthouse at what rate is the distance between the ship and the lighthouse increasing when the ship is eight miles north of the lighthouse. When I confess, these are not the best graphics ever, but this is our boat. It's a triangle and this is our lighthouse, which is a circle. Looking from above, it might not be so unreasonable. I mean, this is probably not what most boats look like nowadays, but hey, it worked. It's sufficient. It doesn't have to be a great picture. It just has to be an understandable picture that we can use to help us solve the problem. So what are some things we know about this coastline? So we're able to assume from the picture that the coastline is just is abstractly a straight line. There's going to be wiggles in any real coastline, but we don't have to worry about that. We can just pretend like it's a straight line for the most part. We're also relatively speaking, assuming the lighthouse is on the coastline. Clearly, that's not the best place for a lighthouse, put a little bit off. But when you're measuring thing in miles, that difference of a few hundred feet is not significant enough to worry too much about it. So our model is sufficient for our purposes. What do we now know? Well, for a typical for these, what we know is that the boat is going north here. It's traveling north along the coastline. So the distance, the horizontal distance here between the coastline, the boat doesn't actually change. This is constantly going to be six miles the whole time. So that is not a variable. That's going to be a constant. What is changing, though, is as time elapses, the boat's getting farther and farther up the coastline. And therefore the distance between the two is enlarging. And that's actually what we're asked about, right? What at what rate is the distance between the two increasing? That's the derivative we care about. Let's give some names here. Let's call the horizontal distance, which would be the east-west distance here. Let's call that X. Let's call the north-south distance. We'll call that Y. And then the distance between the lighthouse and the boat, we'll call that Z. So we get X, Y, Z, this and this is a right triangle here. This is useful because this gives us a relationship. These things will be related to each other by the Pythagorean relationship that X squared plus Y squared equals Z. Now, some things we're going to be aware of. Of course, if we take the derivative of this thing, we can do that. I'm not going to do that yet. I want to first plug in the fact that X is always going to equal 6. Because there's no variability to X, that means its derivative is going to be constant. So if you plug that in, you're going to get 36 plus Y squared equals Z squared. If you take the derivative of both sides, 36 plus Y squared prime is equal to Z squared prime. When it comes to related rates problems, we're taking the derivative with respect to time, which is not any of these variables X, Y and Z. The derivative of constant will be zero. The derivative of Y squared will be 2Y, Y prime. The derivative of Z squared will be 2Z, Z prime, for which we can cancel out the two, just divide that from both sides. And what is it we're looking for? We're looking for the rate in which the distance between the ship and the lighthouse is increasing. That's going to be Z prime. That's what we care about right here. So solving for Z prime, we see that Z prime is going to equal Y prime over Z. So what can we plug in here now? Well, Y, we know, is going to at this moment, we're looking for the moment where the distance is eight miles north. So we get Y equals eight. We can plug that in for Y. What about Y prime? Well, Y prime would be the speed, the boats going up the coastline, which that was also given to us as 15 miles per hour. Notice that Y would be increasing over time. So that's a positive 15 miles per hour. So we get positive 15 right there. And then the last thing is Z, what's the Z coordinate here? We don't actually know it, but we can figure it out from the original equation right here, right? So we see that when Y is eight, we need to figure out Z. So we get 36 plus eight squared is 64. That's equal to 100. So Z squared equals 100. Take the square of both sides, you get Z equals 10. So we make that substitution in over here. And let's see, I could actually simplify that one right here. 10, of course, is the same thing as two times five. Five goes into 15 three times, two goes into eight four times. We get four times three, which is 12. And so therefore we can conclude that the derivative of the change of Z with respect to time at the moment that Y equals eight is equal to 12 miles per hour.