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Welcome to Calculus. I'm Professor Ghrist.

Â We're about to begin Lecture 37 on work. Our applications of integrals have

Â focused primarily on geometric quantities.

Â Length, area, volume. But there are so many other applications.

Â Of a physical nature that we could explore.

Â In this lesson, we'll focus on one particular application.

Â Now, let's get to work. Recall the formula for the computation of

Â work. Work is force times distance.

Â If we take a weight let's say and pull it in the direction opposed to gravity then

Â we'd be doing positive work. If we let that weight drop than we would

Â be doing negative work. Well, what happens in a more interesting

Â case where, say, we're pulling that weight up over a pulley, using a rope.

Â A heavy rope. Then as we pull, there is less rope that

Â we have to support. The force, in other words, is not a

Â constant. What would we do in that case?

Â Well, we would. Think in terms of instantaneous

Â quantities and compute the work by integrating a work element dW.

Â Now, that is the challenge. Let's look at an example involving

Â springs the amount of force required to displace a spring varies in terms of how

Â far you've pulled it. The displacement acts.

Â The force varies with that. In what manner, well, it may be linear.

Â The simplest type of spring is called a linear spring.

Â It satisfies the law of the force as a function of displacement x.

Â Is a constant kappa times x. This is called the spring constant.

Â This is sometimes known as Hook's law. But not all springs are linear, you may

Â have a hard spring. Like in a shock absorber, where the

Â amount of work has a linear term, but then a positive, higher order term, as

Â well. Or you may have a soft spring which is

Â sub-linear, it has some linear term that is valid for very small values of x, but

Â then, decreases. This one gets easier to pull.

Â In either case, the work element, is, what?

Â Well, you need to, think about an instantaneous or infinitesimal change in

Â displacement, dx. And then, the work is that force.

Â F at x times the displacement, dx. And so we can compute the work done in

Â pulling spring by integrating this work element, f of x, dx.

Â Let's look at a slightly different example.

Â This one involving liquids and pumping. How much work does it take to pump liquid

Â from a tank? Let's say you pull it up and pump it out

Â at a height of h from the bottom of the tank.

Â How would you compute the work? Done in this case we're going to have to

Â make a few assumptions. First of all, let's assume a constant

Â weight density row for our fluid. Then our tank is going to have some

Â cross-sectional area. a is a function of x, where x is, oh

Â let's say distance from the bottom of the tank, Then what is the work element?

Â If we take one slice of fluid at this level x.

Â There's some force involved in moving that up.

Â How much? A distance, h minus x, that's the

Â distance from that slice to where the pump exits.

Â That's how high against gravity, we have to move in.

Â Well, what's the force in this case? It's the weight density, well, times the

Â volume element. Where the volume element.

Â Is equal to what? Well, the volume element, as we know, is

Â the cross-sectional area A of X times the thickness, dx.

Â Therefore, we can compute the work in this case by integrating the work element

Â by taking the integral of row times a of x times h minus x, dx.

Â But don't memorize this formula, when you see a problem of this form work it out

Â for yourself. You might be pumping the fluid up, you

Â might be pumping the fluid to a lower Not all.

Â You are going to have to work out the work element.

Â This type of reasoning is valid in many different contexts.

Â Let's say moving earth as opposed to pumping of fluid.

Â Let's say consider a hole that is being dug by two workers who take turns in

Â digging the dirt and moving it to the top.

Â How deep should the first worker dig in order to divide the amount of work done

Â evenly? Is it a fair deal to go half way down?

Â And then, let the second guy do the second half.

Â Or maybe not. Let's make some assumptions.

Â Again, this is in the case of a fluid. We'll assume constant weight density rho.

Â For the dirt. Let's say our hole has to go to a depth,

Â capital D and that the hole has some fixed cross sectional area, A.

Â We're not going to say what that shape is.

Â We're just going to keep track of the area of that cross-section.

Â In this case, what is the work element? Well, we have to take the force, that is,

Â the weight of the slice row times A times dx times the distance that that slice of

Â earth has to be moved. That distance is x if we denote by x.

Â The distance from the top of the hole to the layer where we're considering.

Â Then, what is the work? The work is the integral of the work

Â element. That is the integral of row Ax dx as x

Â goes from 0 to d. This is easily integrated to one half row

Â a d squared. That is the total amount of work to be

Â done in digging the hole. Now, if the two diggers divide the work

Â evenly then the first one has to do one half of that work, digging down to a

Â level D tilde. And so, let's integrate from 0 to D

Â tilde, this same function will get of course one half row A D tilde squared.

Â If that has to be half of the work Then we set it equal to 1 quarter row A D

Â squared. Canceling and solving for D tilde.

Â We obtain D over the square root of 2. That means to divide the word evenly.

Â The first worker should dig a little more than 70% the way down.

Â Because of the extra work. That is required to get to the bottom.

Â Let's consider a similar example, this one involving moving bricks and building

Â a pyramid. Let's say that you start off with a

Â collection of bricks that are at the ground level.

Â How much work would it take to stack them into a pyramid?

Â We'll make the standard assumptions, that all of your bricks start off at the

Â bottom, and we're building a pyramid with a square base.

Â And, in order to use calculus, we're going to assume that the bricks are very.

Â Very small and of a constant weight density row.

Â Then in this case modeling our pyramid by some smooth object, let's say of side

Â length s at the bottom and of height h. We'll set up a coordinate system where y

Â is the distance form the bottom. Than in this case the work element is the

Â weight of a slice through the pyramid at height y times y, the distance that you

Â have to raise the level of the bricks. Now that force element is really rho

Â times dV. The volume element and that again can be

Â computed as we've done in the past. As s squared the area of the base scaled

Â by the factor of 1 minus y over h quantity squared times the thickness dy.

Â And now, we have something that we can integrate in order to compute the work.

Â The work is the integral as y goes from 0 to h row s squared y times quantity 1

Â minus 2y over h plus y squared over h squared dy.

Â That's a simple polynomial integral, that gets us in the end rho times s squared

Â times h squared times 1 half minus 2 3rds plus 1 4th.

Â That is, with a little arithmetic. 1 12th rho, s squared, h squared.

Â If you compare that to the volume of the pyramid, you see that it's one quarter

Â the volume, times h. That h factor is telling you that you've

Â gotta raise those bricks up to that level.

Â And, that takes work. Let's look at some examples that we began

Â the lesson with. Thinking in terms of ropes.

Â How much work is it to pull a rope up over a pulley.

Â Well, we have to make some assumptions about how much the rope weighs.

Â Let's assume we have a length L, and weight density rho.

Â Now, let's. Think if we were to break this rope up

Â into infinitesimal pieces of length dL. Then we would have to raise that up by

Â some height. Let's call that L.

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Then the work element is row times the length element dL.

Â That's giving you the weight. Times the distance, capital l, that we

Â have to move. To get the work, we integrate this work

Â element. That's a simple integral, going from 0 to

Â little l, it gives us 1 half rho times little l squared.

Â Now, we've assumed that that rope just hangs off the edge.

Â What if the rope were a bit longer? Let's say that there's a table, a height

Â h, from the pulley, and there's a long segment of rope that maybe rests in a

Â bucket at the bottom. How much work would be done, in this

Â case? Well, we're going to have to use our

Â collective head, and think, about what we've done, and how to generalize it.

Â If little L is less than or equal to h, our previous analysis holds.

Â In a case where little L is bigger than h, then, let's think.

Â It is equivalent to computing the work done in pulling up a rope of length h.

Â We've already solved that problem. That's one half row h squared.

Â All of the leftover rope is at the bottom of the bucket, and we have to move it

Â wholesale to the top. We can consider that as just a fixed

Â weight. That is equal to rho times the leftover

Â length, l minus h, times the distance traveled, h.

Â That gives us a final answer of one half, rho, h, times quantity 2 l minus h.

Â There's really no end, to the type of tricky problems one can come up with.

Â One of the classics is the leaky bucket problem.

Â How much work does it take to pull up a bucket whose weight, is changing?

Â As a function of the height, x, off the ground.

Â In this or any similar tricky setting what you need to do is determine how much

Â weight you are pulling as a function of lets say x the height off of the ground.

Â Once you have that weight, then you can compute the work element by considering:

Â What am I integrating with respect to? On this case x, how much force is there

Â at that particular value of X? That's going to be f of x when you

Â multiply f of x by dx, you get the work element which you need to integrate to

Â get the work. When you do your homework assignment for

Â this lesson, you'll get a lot of practice setting up these kinds of problems.

Â But in all settings, there is no single formula for you to memorize.

Â Don't copy formulas for a book and try to memorize them.

Â Work out, dw, for yourself based on the particulars of the problem and then

Â integrate. The techniques that we've developed for

Â computing a work element are broadly applicable to many problems in the

Â Physical Sciences. The engineering sciences and the social

Â sciences. In our next lesson, we'll fly through a

Â collection of such applications and show how to compute elements.

Â