Welcome to Calculus. I'm Professor Ghrist. We're about to begin lecture 37 on work. Our applications of integrals have focused primarily on geometric quantities length, area, volume, but there are so many other applications of a physical nature, that we could explore. In this lesson, we'll focus on one particular application. Now let's get to work. Recall the formula for the computation of work. Work is force times distance. If we take a weight, let's say, and pull it in the direction opposed to gravity, then we would be doing positive work. If we let that weight drop, then we would be doing negative work. Well, what happens in a more interesting case. Where say, we're pulling that weight up, over a pulley, using a rope, a heavy rope. Then, as we pull, there is less rope that we have to support. The force, in other words, is not a constant. What would we do in that case? Well, we would think in terms of instantaneous quantities and compute the work by integrating a work element dW, and that is the challenge. Let's look at an example involving springs. The amount of force required to displace a spring varies in terms of how far you've pulled it, the displacement, x. The force varies with that in what manner? Well, it may be linear, the simplest type of spring is called a linear spring. And it satisfies the law that the force is a function of displacement x, is a constant kappa times x. This is called the spring constant, this is sometimes known as Hooke's law. But not all springs are linear. You may have a hard spring, like in a shock absorber where the amount of work has a linear term but then a positive higher order term as well. Or you may have a soft spring which is sub linear. It has some linear term that is valid for very small values of x, but then decreases. The spring gets easier to pull. In either case, the work element is what? Well, we need to think about an instantaneous or infinitesimal change in displacement, dx. And then the work is that force, F at x, times the displacement, dx. And so we can compute the work done in pulling spring by integrating this work element, F(x)dx. Let's look at a slightly different example, this one involving liquids and pumping. How much work does it take to pump liquid from a tank? Let's say you pull it up and pump it out at a height of h from the bottom of the tank. How would you compute the work done in this case? We're going to have to make a few assumptions. First of all, let's assume a constant weight density, rho, for our fluid. Then, our tank is going to have some cross sectional area, A, as a function of x. Where x is, oh let's say distance from the bottom of the tank. Then what is the work element if we take one slice of fluid at this level, x. There's some force involved in moving that up. How much? The distance, h, minus x. That's the distance from that slice to where the pump exits. That's how high against gravity we have to move it. Now, what's the force in this case? It's the weight density, rho, times the volume element. Where the volume element is equal to what? Well, the volume element, as we know, is the cross-sectional area, A of x, times the thickness, d x. Therefore, we can compute the work, in this case, by integrating the work element. Letting the integral of rho times A of x times h minus x, dx. But don't memorize this formula when you see a problem of this form, work it out for yourself. You might be pumping the fluid up, you might be pumping the fluid to a lower level. You're going to have to work out the work element. This type of reasoning is valid in many different contexts. Let's say moving earth, as opposed to pumping a fluid. Consider a hole that is being dug by two workers who take turns in digging the dirt and moving it to the top. How deep should the first worker dig in order to divide the amount of work done evenly? Is it a fair deal to go halfway down? And then let the second guy do the second half or maybe not. Let's make some assumptions. Again, as in the case of a fluid, we'll assume a constant weight density row for the dirt. Let's say that our hole has to go to a depth capital D and that the hole has some fixed cross sectional area. A, we're not going to say what that shape is. We're just going to keep track of the area of that cross section. In this case, what is the work element? Well, we have to take the force. That is, the weight of the slice row times A times dx times the distance that that slice of earth has to be moved. That distance is x. If we denote by x the distance from the top of the hole to the layer where we're considering. Then what is the work? The work is the integral of the work element, that is the integral of row Axdx as x goes from 0 to D. This is easily integrated to one half rho A d squared. That is the total amount of work to be done digging the hole. Now if the two diggers divide the work evenly, then the first one has to do 1/2 of that work digging down to a level, D tilde. And so, let's integrate from zero to D tilde, this same function. We'll get, of course, one half rho A D tilde squared. If that has to be half of the work then we set it equal to one quarter rho A D squared, cancelling and solving for D tilde, we obtain D over the square root of 2. That means to divide the work evenly, the first worker should dig a little more than 70% of the way down because of the extra work that is required to get to the bottom. Let's consider a similar example. This one involving moving bricks and building a pyramid. Let's say that you start off with a collection of bricks that are at the ground level. How much work would it take to stack them into a pyramid? We'll make the standard assumptions that all of your bricks start off at the bottom, that we're building a pyramid with a square base and in order to use calculus, we're going to assume that the bricks are very, very small and of a constant weight density rho. Then in this case modeling our pyramid by some smooth object. Let's say of side length s at the bottom and of height h. We'll setup a coordinate system where y is the distance from the bottom, then in this case the work element is the weight of a slice through pyramid at height y, times y, the distance that you have to raise the level of the bricks. Now that force element is really rho times dV. The volume element, and that, again, can be computed as we've done in the past as S squared, the area of the base, scaled by the factor of 1 minus y over h, quantity squared times the thickness dy. Now we have something that we can integrate in order to compute the work. Work is the integral. As y goes from zero to h, rho s squared y times quantity 1 minus 2 y over h plus y squared over h squared, dy. That's a simple polynomial integral that gives us in the end row times squared times times each squared times one half minus two thirds plus one fourth. That is with a little arithmetic, one-twelfth rho s squared h squared. If you compare that to the volume of the pyramid, you see that it's one quarter the volume times h. That h factor is telling you that you've got to raise those bricks up to that level and that takes work. Let's look at some examples that we began the lesson with thinking in terms of ropes. How much work is it to pull a rope up over a pulley? Well we have to make some assumptions about how much the rope weighs let's assume we have a length l and a weight density rho. Now, let's think. If we were to break this rope up into infinite decimal pieces of length dl, then, we would have to raise that up by some height, let's call it L. Then, the work element is rho times the length element dL. That's giving you the weight. Times the distance capital L that we have to move, to get the work we integrate this work element, that's a simple integral. Going from 0 to little l, it gives us one half row times little l squared, now, we've assumed that that rope just hangs off the edge. What if the rope were a big longer? Let's say that there's a table a height h from the pulley and there's a long segment of rope that maybe rests in a bucket at the bottom. How much work would be done in this case? We're going to have to use our collective head and think about what we've done and how to generalize it. If little l is less than or equal to h, our previous analysis holds. In the case where little l is bigger than h, then let's think. It is equivalent to computing the work done in pulling up a rope of length h. We've already solved that problem. That's one half rho h squared, all of the leftover rope is at the bottom of the bucket. And we have to move it wholesale to the top. We can consider that as just a fixed weight that is equal to rho times the leftover length l minus h times the distance travelled h. That gives us a final answer of one half rho h times quantity 2l minus h. There's really no end to the type of tricky problems one can come up with. One of the classics is the leaky bucket problem. How much work does it take to pull up a bucket whose weight is changing as a function of the height, x, off the ground? In this or any similar tricky setting, what you need to do is determine how much weight you are pulling as a function of lets say x, the height off of the ground. Once you have that weight, then you can compute, the work element, by considering, what am I integrating with respect to? Well, in this case, x. How much force is there at that particular value of x? That's gonna be F(x). When you multiply F(x) by dx, you get the work element, which you need to integrate to get the work. When you do your homework assignment for this lesson, you'll get a lot of practice setting up these kinds of problems. But, in all settings, there is no single formula for you to memorize. Don't copy formulas from a book and try to memorize them, work out dW for yourself, based on the particulars of the problem, and then integrate. The techniques that we've developed for computing a work element are broadly applicable to many problems in the physical sciences, the engineering sciences, and the social sciences. In our next lesson, we'll fly through a collection of such applications and show how to compute elements.