Welcome to Calculus. 
I'm Professor Greist, we're about to begin lecture 13 on higher derivatives. 
In single variable calculus, the derivative of a function is another 
function of the same type. Therefore, one can iterate the 
construction. Computing rates of change, rates of 
change, rates of change. In this lesson, we'll consider such 
higher derivatives from the point of view of physical applications, considering 
examples, covering race car driving to robotics to ocean waves. 
You know and I know, what happens when you differentiate a polynomial, and when 
you keep differentiating a polynomial, higher and higher derivatives give you 
deeper information, but polynomials are finite. 
And that means, that eventually, derivatives will annihilate a polynomial 
and send it to zero. Higher derivatives come with a number of 
different notational conventions. You may be tempted to write f prime, f 
double prime, f triple prime, et cetera. This is not so good, in general. 
It might make your homework a little bit confusing to read. 
A bit better notation would be that which we used for Taylor series. 
Denoting the nth derivative of f using a super script n in parenthesis. 
Better still is the notation df, dx, d squared f, dx squared, d cubed f, dx 
cubed, et cetera. But this leads to the question, why is 
that notation used? Why is the second derivative not df 
squared, dx squared. Of course, the reason is that we're 
taking powers of the differentiation operator d by dx. 
Where are these higher derivatives to be found? 
How do we interpret them? One good example comes from dynamics. 
If x, as a function of time t, denotes the position of a moving body Than, one 
interpretation for the first derivative is as velocity. 
As I'm sure you've seen before, in your entry Physics. 
You may also know that the second derivative of position is indeed 
acceleration. it is the rate of change of velocity. 
what about the third derivative of position? 
What is that? What about the fourth derivative and 
higher still? Do these have physical Meanings. 
Well, they certainly do. The third derivative is sometimes called 
jerk. The fourth derivative, depending on, 
which side of the ocean you're on. Is either called snap or jounce. 
Now, maybe you've heard of these before. Maybe you haven't. 
When might you need to know something about this. 
Well, if you were working with any sort of interesting system involving dynamics, 
a good example would be the quadrotor robots that can fly extremely precisely. 
From what does that precision come? It comes from careful control Not only a 
velocity, an acceleration, but of jerk and snap as well. 
There are other interpretations for higher derivatives as well. 
One interesting interpretation comes from geometry and curvature. 
What is curvature? Well, you can certainly feel curvature 
when your driving down the road and you make a sharp turn. 
You can tell the difference between a looser and a tighter sort of turn. 
The higher the curvature, the more you feel. 
When you turn. But what is curvature, really? 
Well, it seems, from this physical explanation, to have something to do with 
radius. If we had a circle that was tangent to 
our road. The point at which we're turning. 
Then the radius, r, of that circle. Would seem to correlate somehow to 
curvature that you feel as you go around the smaller radius the higher the 
curvature and indeed at a point on curve one definition for curvature is as the 
reciprocal. Of the radius of the osculating circle. 
The circle that agrees with that curve to second order. 
We're going to do a somewhat lengthy derivation to get a formula for 
curvature. Consider a function Y equals F of X. 
And let's fit to it a circle that agrees up to second order. 
We'll say that the radius of that circle is some unknown R, and we'll set up a 
coordinate system X based at the centre. Of that circle. 
Now at some value of X, the graph of this circle, that is square root of R squared 
minus X squared, is going to agree with F and it's derivatives. 
At least the first two. Now let's do a little bit of work. 
If we differentiate. This equation, what do we get? 
we get f prime is equal to what? Well, we have to take the derivative of r 
squared minus x squared with one half. That's one half r squared minus x squared 
to the negative one half, times the derivative of what's on the inside 
negative 2x. There's a little bit of simplification 
that goes on. And we get negative x times quantity. 
R squared minus x squared to the negative 1/2. 
Now, this has to agree to second order. So let's compute another derivative. 
On the left we get f double prime, on the right what do we get? 
Well, we're going to have to use a product rule. 
First of all, we have minus quantity r squared minus x squared to the negative 
one half. Then we have plus negative x times the 
derivative of R squared minus X squared to the negative one half. 
Which gives us say negative one half, then an R squared minus X squared to the 
negative three halves, and then a negative two X factor. 
There's a little bit of simplification we can do here by factoring out. 
A negative r squared minus x squared to the negative three halves, what we're 
left with is an r squared, a minus x squared, and a plus x squared. 
The negative x squared and the plus x squareds cancel and we're left with an r 
squared that we put out in front... So f double prime is negative r squared 
times quantity r squared minus x squared to the negative 3 halves. 
Now let's step back for a second. At the point of tangency, we know f prime 
and f double prime. What we don't know by the values of x and 
most importantly, r. That's two equations, two unknowns, we 
ought to be able to solve for r and get the curvature. 
Let's do some minipulations. Taking the first equation and squaring 
it. Gives us on the left f prime squared. 
On the right, negative x squared, that is positive x squared, times r squared minus 
x squares to the negative one. Now if we rearrange that a little bit... 
And then multiply through the f prime squared by the quantities on the left, we 
get f prime squared times r squared minus f prime squared times x squared equals x 
squared. If we rearrange again factor out the x 
squared term. Then solving for x squared gives us x 
squared equals f prime squared times r squared all over 1 plus f prime squared 
Not, we can substitute that into the second equation. 
The equation for F double prime and so eliminate the X variable. 
When we do that substitution, what are we going to get. 
We'll get that F double prime equals negative R squared times. 
R squared minus F prime squared, times R squared. 
Over one plus F prime squared. All of that to the negative three halves. 
Than we can factor out an R squared from that term on the right. 
Do a little bit of algebraic simplification and then some cancelling 
of the R squareds and we'll get that f double prime equals negative 1 over R 
times quantity 1 over 1 plus f prime squared all to the negative 3 halves. 
Once we get that Then, it's not going to be hard to solve for r. 
We'll get rid of that pesky negative sign by taking an absolute value and then 
solving, for 1 over r gives the curvature, kappa, 2b the absolute value 
of f double prime over quantity 1 plus f prime squared to the 3 halves. 
That's a complicated derivation and a complicated looking formula, but it's a 
very general formula for computing curvature. 
Now how might we use this? Well first of all, notice that curvature 
depends only on the first derivative and the second derivative. 
That makes sense since it's being captured by something to do with a circle 
and second order tangent see. If we were in a setting, let's say 
looking at ocean waves, where the derivatives were fairly small, then we 
could approximate the curvature by saying it's F double prime times what? 
Well, if that prime is small, we can use the binomial series. 
To expand it as 1 minus 3 halves f [UNKNOWN] squared plus some higher order 
terms in f prime, you could work out. And what those are, there are a few other 
things that we see from this formula, namely that the curvature is exactly 
equal to the second derivative at the maxima or the minima of a curve where f 
prime equals 0. That's probably something that you want 
to remember. Curvature is really just the second 
derivative when you're talking about a max or a min there are some other things 
that come out of this formula, mainly that the curvature is zero at an 
inflection point. And, as you might have already guessed, 
the sign of this second derivative determines the concavity of the graph. 
Whether it's concave up or concave down. Yet another physical interpretation of 
higher derivatives comes in the subject of elasticity. 
Consider a [UNKNOWN] beam with an uniform cross section. 
And soem static load, a weight that is applied that may vary with position x 
along the beam. What one often wants to know is, how much 
does the beam deflect this deflection u as a function of position x? 
Satisfies the following equation. The fourth derivative of the deflection 
is proportional to the load. Now, there are some constants of 
proportionality involved. One of them, E, is an elasticity based on 
the material the beam is made from. The other constant, I Is a moment of 
inertia related to the cross-section. You don't have to know what any of that 
is. We'll talk a little about moments of 
inertia later in the course. For now, the thing to know is that it's 
the fourth derivative, the deflection, that matters in this elasticity problem. 
But of course, our final interpretation for higher derivatives is in terms of 
Taylor series, that they provide the means to get better and better 
approximations to functions. Derivatives and iterated derivatives 
really are everywhere and all about. But what are they good for? 
In our next lesson, we'll consider one of the great applications of hired 
derivatives to optimization problems.