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. A related is concept is that of a 
relative rate of change. A relative rate is a normalization of a 
differential, something of the form d u over u. 
This is telling you something about what percentage change you have. 
Notice, it is unitless, and notice you can also get it in terms of the log of u. 
Let's put this idea to work. If we have a voltage across a variable 
resistor that is fixed. Then, how are the relative rates of 
resistance and current related? So in a simple circuit with voltage 
potential v, current flow i, and variable resistance r, these quantities are 
related by Ohm's law. That is V equals I times R. 
Now, if we differentiate both sides of this equation we get dV equals RdI plus 
IdR. And now, to compute the relative rate of 
change, we look at dV over V. That must be 0 since the voltage is 
fixed. So, dividing the right-hand side by v 
gives us what? Well, Rdi plus IdR over i times r, that 
splits up into d I over I. Plus d r over r, meaning that the 
relative rates of change of R and I are inversely related. 
Let's look at an example from geometry. How does the relative change in volume 
relate to that of surface area? In the specific case of a solid ball of 
radius r, how do the volume and the areas compare? 
The volume is 4 3rds pi r cubed. When we differentiate that we get dv 
equals 4 pie r squared d r. Now, for surface area, that function is 4 
pi r squared. Differentiating that gives da equals 8 pi 
Rdr. Now, look at the relative changes. 
We must take dV over V and divide that by dA over A to compare these relative 
rates. What is that? 
Well, that is going to be 3 dr over r divided by 2 dr over r. 
With a little bit of algebra, you can see that it's a constant 3 halves. 
I wonder if that holds for any other shapes. 
Let's take a cube with a side length s and the volume of that cube is s cubed. 
When we differentiate, we get dv equals 3s squared ds. 
The surface area is 6 s squared. Differentiating yields d a equals 12 s d 
s. If we again look at d v over v, and 
compare that to d a over a, we're going to get 3 d s over s over 2 ds over s. 
The relative rates of length cancel and we get, again, 3 halves. 
That means that the relative volume change is always 150% the relative area 
change. Well, I wonder if that has any meaning. 
Well, let's go back to economics and consider the important questions. 
That is, what is the difference between milk and wine? 
Well, it is in elasticity of demand. If p is the price and q is the quantity 
demanded at that price. Then, the elasticity e is defined to be 
the rate of change of the relative demand with respect to the relative price. 
That is, dq over q divided by dP over P. We're going to put a minus sign in front 
of this since and price and quantity are inversely related. 
When we do so, what do we see? Well, the higher the price is, the less 
you want something. The lower the price, the more you want. 
But it depends on elasticity. If e is between 0 and 1, we say the 
demand is inelastic. Changes in price don't change the 
quantity demanded so much, like milk. On the other hand, if e is larger than 1, 
then this is called elastic demand. If wine goes on sale I'm more likely to 
buy more and store it. That doesn't work for milk. 
Now when e is equals to 1, it is called perfectly elastic. 
Note that is not slope, but rather relative rates of change. 
Here's a question. How do you maximize the relative revenue 
with respect to relative change in price? Well, R, the revenue, is the price times 
the quantity, P times Q, or the area of this box, if you like. 
Then, differentiating this, we get dR equals PdQ plus QdP. 
Looking at relative revenue is dR over R. We've seen a very similar computation 
with our electric circuit. We get dQ over Q plus dP over P, and now, 
looking at the ratio Of relative revenue to relative price gives us what? 
Dq over q over dp over p plus dp over p over dp over p. 
That latter term simplifies to one. The former is negative e. 
So, we have maximal revenue when the elasticity equals one. 
So concluding, what do we mean by d x and d u? 
There is much more to the story than what we have said. 
For now, differentials are functions that implicate rates of change. 
When you get to multivariable calculus, you'll learn a different perspective 
involving differential forms. We've seen in this lesson that thinking 
of derivatives from a more abstract perspective, that of differentials, is 
very useful in several application demands. 
We will continue this process in our next lecture, by thinking of differentiation 
itself, as an abstract operator. In this more rarefied setting, we'll see 
that it deepens our understanding and leads to better computation, as well as 
comprehension. [BLANK_AUDIO]