2:12

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.

12:28

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.