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

Now, 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 acts.

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

The simplest type of spring is called a linear spring.

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

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

This is sometimes known as Hook'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. This one gets easier to pull.

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

Well, you 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 of 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 row for our fluid. Then our tank is going to have some

cross-sectional area. a is 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? A 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 in.

Well, what's the force in this case? It's the weight density, well, 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, dx.

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

by taking the integral of row 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 Not all.

You are 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 of fluid.

Let's say 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 half way down?

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

Or maybe not. Let's make some assumptions.

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

For the dirt. Let's say 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 Ax dx as x

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

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

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

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

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

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

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

squared. Canceling and solving for D tilde.

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

The first worker should dig a little more than 70% 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, and 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 row.

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 set up a coordinate system where y

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

weight of a slice through the 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.

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

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

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

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

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

That is, with a little arithmetic. 1 12th 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

gotta 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 weight density rho.

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

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

some height. Let's call that L.