Welcome to calculus. I'm professor Ghrist.

We're about to begin lecture 15 on differentials.

We begin this lecture with a question. What is dx?

You may have been waiting your whole life to find out, but you're not going to get

a complete answer right now. Sorry.

But along the way, we'll learn enough about differentials to be able to answer

some important questions about geometry, physics, and economics.

So what exactly do we mean when we write dx or du?

They're used all the time in calculus class.

One perspective is that there are infinitesimals, really, really small bits

of x or u, so that when you integrate du, what you get is u.

That's a reasonable intuition, but is that really what they mean?

Maybe they can note a rate of change so that according to the chain rule du and

dx are related by du, dx. Meaning that if du connotes the rate of

change of u, and dx, the rate of change of x.

Then they are related by du, dx. The rate of change of u with respect to

x. That's a reasonable interpretation.

It's not the full truth. The full truth is that these are

differential forms. We are not, however, going to cover that

in this single variable course. For now, let's think about what d is

doing. d is an operator, which means that if you

put in a function, f, you get out df, this differential that is telling you

something, about rates of change of outputs with respect to inputs.

Now, the fact that it's an operator, means, that we can differentiate an

equation. Something of the form, f equals g.

And what we will get out, is again, and equation, df equals dg.

Now, you've probably seen this before. It goes under the name of implicit

differentiation, and it's extremely useful.

For example, if we want to know the slope of the tangent line to a circle of radius

r centered at the origin in the x, y plane, what would we do?

We would write down the equation of that circle, x squared plus y squared equals r

squared, and then we would differentiate the entire equation, applying the d

operator gives us what? d of x squared is 2x, dx.

d of y squared is 2ydy. What is d of r squared?

Well, if r is a constant, then dr would be equal to 0, or if like,

differentiating a constant gives you 0. This equation, then, involving the

differentials dx and dy allows us to solve for the slope of the tangent line.

dy, dx using a little bit of algebra, we get that it is negative x over y.

A result that you can clearly see. Let's turn to an application in economics

this one dealing with substitution rates in economics, the marginal rate of

substation of a product x for a product y.

Is the rate at which a consumer is willing to exchange the good or product y

for the good x. This assumes a fixed utility.

I don't want to go too much into economics here, but let's take a moment

and explain what this means. This marginal rate of substitution, MRS,

can be expressed as minus dy over dx, along a curve of fixed utility, u.

So lets say we're looking at the xy plane.

I have some utility function u. When we set that equal to a constant it

gives us a curve in the xy plane. The marginal rate of substitution is

related to dy, dx, the slope of the tangent line.

Let's make this a little more specific. Let's say that x is the amount of coffee

I have, y, the amount of doughnuts that I have.

And my utility function u is y squared times quantity x minus 3.

In this case, differentiating that equation gives me du equals d of quantity

y squared times x minus 3. Now, since I'm along a fixed utility

curve u is a is a constant. So du is 0.

And on the right, applying the product rule I get 2y dy times x minus 3 plus y

squared times dx. And now, solving for the marginal rate of

substitution, that is minus dy over dx, gives me negative, negative y squared

over 2 y times quantity x minus 3. We can simplify removing the negative

signs and the y's, and we get a marginal rate of substitution of y over 2 times

quantity x minus 3. Now, what does that actually mean?

Well, what that actually means is that depending on my values of x and y, how

much coffee I have versus how many doughnuts I have.

My willingness to trade one for the other changes.

If I have an equal amount of coffee and donuts.

Then, that's one thing. I might be willing to trade one for

another. However, if the amount of coffee that I

have is sufficiently low. That is x is close to 3.

It doesn't matter how many doughnuts you're willing to give me.

I'm not going to give up any more coffee. Differentials often arise in related

rates problems. Here's a fun one.

Let's say that you have water flowing from a faucet in a smooth stream.

What I'd like to know is how does the cross-sectional shape change?

It seems as though the stream gets thinner as you go.

Let's assume that the cross sectional shape is circular.

We'll try to see what happens to the radius of that circle.

We'll start from the assumption that we have a steady stream, which means that

the flow rate through a cross-section must be constant.

What is the flow rate? Well, that's the area times the velocity

at that slice. If that's a constant, then we can write

that equation as pi r squared v equals c, where r is the radius and v is the

velocity. Differentiating this equation yields

what? Well, we're going to ignore the pi,

absorb that into the constant on the right.

And then, using the product rule, we get 2r dr times v plus r squared times dv

equals 0. Now, we can do a little bit of algebra to

get our hands on dr. That is negative r squared over 2rv times

dv. We can simplify that cancelling the r's

to negative r over 2v, dv. And now, let's say we wanted to know how

that radius is changing with respect to time.

The time rate of change would be dr, dt, which we can write as negative r over 2v

times dv, dt. What can we say now to simplify things?

Well, since d v d t is a constant, since the water is falling under the influence

of gravity, dv, dt is exactly g. And so, we get dr, dt equals negative r

times g over 2v. We could simplify that, knowing what we

know about v2, negative r times g over 2, quantity v not plus g times t.

Where v not is the initial velocity coming out from the faucet.

That still doesn't tell us exactly what the shape is, say as a function of

height. But we could either solve that

differential equation or use some conservation properties from physics to

get that shape if we wanted. We have manipulated the differentials to

give us information about time rates of change.