 Welcome back to our last video in lecture seven. Let's look at one more example of using integration to help us with these variable force problems in order to calculate the work. We'll do some more examples of this in lecture eight, but this will be the last one for this official lecture part right here. And so this is the proverbial leaky bucket problem that calculus students absolutely love. Imagine we have a bucket that weighs four pounds and a rope of negligible weight. What I mean here is that the weight of the rope here is so small that we're not gonna consider it. I mean, clearly that's sort of a simplifying assumption that isn't real. You know, it's kind of like, oh, assume there's no air drag or there's no friction. These are things that people do in physics types problems all the time. The thing is we just don't wanna worry about the weight of the rope. We've seen in a previous example how one actually would calculate the work of lifting a rope. So we'll defer to that example if you wanted to. Let's just focus on this bucket that contains four pounds of water. And it's being drawn, the bucket's being drawn up a well and the well is 80 feet deep, okay? Now the bucket's full, the bucket itself is four pounds, but the bucket is contained. It's large enough that it contains 40 pounds of water. And how fast are we pulling it up? It's being pulled up at a rate of two feet per second. But the water has a leak in it. It's losing water. And it's leaking at a rate of 2.2 pounds per second. So assuming we pull the bucket up at this constant rate, how much work is done to pull up this bucket? Now, to be aware that this problem is very similar to lifting up the rope. That is, as we pull it, as we lift it, it's getting lighter so its weight's going down and the force is variable. And so how does one try to accommodate for such a thing? Well, the strategy of accumulation here is to first pick an axis. We have to pick some axis of line that we're gonna integrate along. So for this one, we're gonna choose the bottom of the well to be X equals zero and then we're gonna go upward, go up to be our positive X direction. We could have done it so that the origin was the top of the well. We point down. It doesn't really matter. You'll get two different integrals that actually give you the same quantity. The geometric analysis is what's really important here. And as I did in the previous example, I did X equals zero at the top. I wanna do an example of X equals zero at the bottom. So you see there's not much difference here but you could try it out yourself, see how things are different. And so what we're gonna do is we're gonna take our axis and subdivide it into teeny tiny little pieces, right? And so what we're gonna do is we subdivide and then we have to assume that the problem is easy, which in mathematics the easy, what we mean by easy is we mean it's linear, right? We wanna approximate this with linear approximations. So what do we mean by easy in this case here? Well, we're gonna assume that for this time interval as it goes from Xi minus one to Xi, let's assume that the weight is constant. Let's kind of assume that there's no leakage between these two markers, that when it gets to Xi minus one, it squirts. Then it goes to Xi, it squirts again. Then we go to Xi plus one, it squirts again. Let's assume the mass is only lost and we have those little markers there. Is that realistic? No, but it's a simplifying approximation for which integration will correct any error that we introduce into it. So what we have to do as we work with this work problem, of course, is we're going to work, oh, work is the integral of force times distance, right? So we have to figure out what's our distance function, what is our force function go from there? So our force function in this situation, F of X, it's gonna be weight. How heavy is this thing at location X? Well, there's two aspects to it. There's the weight that comes from the bucket, right? And there's the weight that comes from the water. Now, the weight that comes from the bucket is gonna be constant. The bucket always weighs four pounds. That doesn't change as we go through this exercise. But what's the weight of the water? Well, the weight of the water, it starts off at 40 pounds. And then we lose something, right? Sorry, 40 pounds. Then we lose something. I know that's not super specific, but when you're working with a story problem, it's not like your equations, your functions have to be picture perfect from the very beginning. Just kind of write down as your mind is thinking about something. It's like, okay, it's gonna be 40 pounds, but you take away something. What are we taking away? Well, it loses water over time. And so we can say something like, okay, a little bit better than something. We're gonna lose 0.2 pounds per second. Well, you should times that by time, t. So if t is measuring in seconds, seconds maybe since the start of this exercise, right? I want you to start lifting the bucket. So we get this 0.2t. Now that's not quite the right, I mean, that works for us, but we wanna find a function with respect to x because we're gonna integrate with respect to distance, not with respect to time. So how do you write time as a function of x? Well, that's where this rate comes into play, right? So the distance with respect to time is our distance x is gonna equal two feet per second. So if you take two feet per second, and we multiply that by time, which is in seconds, that'll give us how much feet we've done. Now solving for t, this tells us that t is gonna equal x over two or one half x if you prefer. So the distance that we travel divided by two gives us the time. And so if we plug that in here, right? We end up with four plus 40 minus 30. 0.2 times one half times x, like so, which of course will give us a 0.1, I guess it's a 0.1. And so our force function, summarizing what we did right here, the 40 plus 40 plus four is 44. So we get 44 minus 0.1x. This is our force function for which we want to integrate. And this will give us the force, the weight of the bucket at any location. So when you're at location zero, this thing will weigh 44 pounds. When we're at location, say one, right? It would be 44 minus 0.1, so 43.9. Then if you do with the very top of this thing, right? It's 80 feet tall. We could take 44 minus 0.1 times eight. 0.1 times eight, you know, we could figure out exactly how much water is in the bucket at the very end. I don't really care about that too much. Just be aware, that's what we've done right here. So once we have the force function, because we're assuming weight is constant from this right here, these two intervals, we want to go from this interval to this interval. And what's that distance? That distance is gonna be delta x. So now in a position where we can set up our integral that we need to compute, the work is equal to the integral force. Force is gonna be 44 minus 0.1 x. That's the weight. Then the distance it has to travel is dx. Now we're left with the bounds, right? What are the bounds of integration right here and right here? Well, we have to go from the, where can x live here? x could be at the bottom of the well, which represents x equals zero. It can go to the top of the well, which would be x equals 80 feet. And so now we just have to compute the anti-derivative from here. The anti-derivative 44 minus 0.1 x would be 44 x minus 0.05 x squared from zero to 80. Plugging in zero makes it disappear, so we'll just plug in the 80. And we end up with 44 times 80. And then minus 0.05 times 80 squared. Let's see, 44 times 80, that's 3,520. And then 80 squared, that's 6,400 times that by 0.05. That should give you 320, I believe. And so then this adds up to be 3,200 foot-pounds of work. And so notice that the actual calculation of the integral is nothing challenging. There's not hard integrals right here. It's setting this thing up, that is the significant thing, right? Why was the integral this baby right here, right? If you can understand the argument going on there, that's really what we want you to get from these types of exercises. I don't really care about work that much, you know? I'm an mathematician, I'm not a physicist, but many of you watching here are, you know, people who want to be physicists or chemists or engineers or other types of scientists or even math students themselves, right? These applications do have reach into our real lives, whether it's professional or even personal lives as well. And it turns out the calculus is everywhere. We can use it to solve so many problems. It just takes this strategy of take your problem, subdivide it into small pieces. It's like if you have like a six-foot sub that you want to eat, right? You can't just eat in one bite. You got to take your sub and slice it up into small little pieces, pieces that are digestible. You eat the first bite, you eat the second bite, you eat the third bite. And you do this for all the bites there are. You subdivide the problem, solve the easy subdivision because you're adding some little assumption that makes it simple. And then when you take the limit as it goes to infinity, the number of subdivisions goes to infinity, the error will be removed from the problem. It's a beautiful strategy. We'll see some more of this in the next coming lecture videos as well. And I guess it's the end of lecture seven. Thanks for watching. If you like what you've seen, please hit the like button. Feel free to subscribe for more updates of these type of math lectures in the future. Post any comments. Post any comments if you have questions or want to know more about this topic and I'll be glad to talk to you about them. Otherwise, I'll see you all next time. Have a great day everyone. Bye.