Welcome to calculus. I'm Professor Greist and we're about
to begin lecture 35 on Arc Length. >> After ascending to the heights
of infinite dimensional spaces, we're going to come back down to Earth and
consider simple, one-dimensional curves. As we now know, one dimensional
volume is really just length. And we'll begin this lesson by
considering the length element. >> We begin our discussion of arc length
with that formula that you know is coming. Namely, that the arc length is
the integral of the arc length element. The question is how do you
determine arc length element? Well, we think of it, of course, as a very small infinite
piece of the curve. Let's say that we knew the graph and
the plane and we set up a right triangle
with base dx and height dy. Then, what would the length
of this hypotenuse b, well, from Pythagoras it would be the square
root of d x squared plus d y squared. That's simple enough, but
how do we make that work in practice? Well, in the one case, where we
have an implicit curve that is for example some curve where y is
defined as a function of x and we're looking at the graph. Then when we write out this formula,
what do we have? Well, from the chain roll
we can express dy as dydx times dx, when we square that and
follow through with the algebra, we can factor out a dx squared
from under that square root sign. Of course,
the square root of dx squared is dx. And we're left with dL is the square
root of 1+(dy/dx) squared, Times the x at simple manipulation
of differentials actually works. For another example, maybe we don't have a curve that's
expressed in terms of a graft. Maybe we have a Parametric Curve,
where the x and y coordinates are given
by sum function of t. Such curves can swoop around and
cross themselves in interesting ways. In this case, applying the same
differential analysis, what do we have? Well, if x and y are given as functions
of t and dx is dx,dt times dt, dy is dy,dt times dt. Once again,
we see that there's a dt squared under that square root that can
be factored and pulled out and we obtain an arc like element
that is the square root of dx dt squared plus dy dt squared times dt. Let's see this put to work in an example. Compute the length of a simple
circle of radius r in the plane. We'll do this parametrically, writing x as r times cosine of t and
y is r times cosine of t. We have to be careful to say the on t r,
t is going from 0 to 2 pi, let's say. Then, in this case, the arc length
element is square root of dxdt squared, plus dydt squared, times dt. When we differentiate, cosine and
sine, we get, very simply, r squared sine squared t,
plus r squared cosine squared t dt. We see, that it's easy to
factor out an r squared, and applying the simplification sine squared,
plus cosine squared, equals 1. We have an arc length element,
that is r dt. T. And to compute the length, we integrate
r dt as t goes from 0 to 2 pi. And that gives us, simply,
2 pi r, the answer. That we all know. For an example,
involving an implicit curve or a graph, consider equation y=ln(sin x) as x goes from pi/4 to pi/2. I don't know what a graph of that
looks like; but I don't need to. Since I know the formula for
the arch length element. It is square root of 1 +
(dy / dx) squared times dx. Now when I compute the derivative of
log of sine of x, what do I need to do? Well, from the chain rule,
that's 1 / sine of x, times the derivative of sine,
namely cosine. We can simplify that a little bit. 1 + COT squared x is of
course CSC squared of x. Taking the square root gives an arc
length element of CSCx times dx. Now to compute the length,
we need simply to integrate CSC x dx,
as x goes from pi over 4 to pi over 2. I may or
may not be able to remember the integral of cosecant but
I can assure you that it is minus log of absolute value
cosecant x plus cotangent x. When we substitute in those limits,
pi over pi/4 to pi/2. Then we add minus log of let's see,
CSC to pi/2, that's one and then plus is zero there, some other terms. I don't know,
that's too much to do in my head. But I think that the final answer is log
of quantity one plus square root of 2. One of the sad truths about arc
length problems is that, so often, integrals get messy,
because of that square root. And they can be pretty
difficult to evaluate. But it's not all bad news. There are some very interesting problems
that we can solve with integrals. One is the Catenary. That is the shape of a hanging cable. Here's a fact that comes from physics. The rate of change of
the slope of such a hanging cable is proportional
to it's length element. Now what does that mean? If we set up some x and
y coordinates to describe the curve. Then, what this statement claims
is that the slope of the cable, which we can interpret as dy dx. Changes, that is has derivative that
is proportional to the length element. That means we could write it as kappa
times dl, where kappa is some constant. Now how do we make sense of this
differential equation and then solve it. It's got two derivatives. But if we let U be equal to dy dx. Then we could write this as,
du dx equals kappa times the length element, in this case,
the square root of one plus u squared. That is a differential equation, that I
think we're going to be able to solve. Indeed, this is clearly
a separable differential equation. And so, moving root 1 plus u squared
on the left, we have on the right, kappa dx,
integrating both sides yields what? Well, on the left we're going
to have to apply a substitution. And I think this is an ideal candidate for
a trigonometric substitution. If we let u be SINH t,
then du is COSH t dt. And we have, on the left,
the integral of COSH t dt over square root of
1 plus SINH squared t. The right integrates simply,
to kappa x plus a constant. On the left we have some great
cancellation 1 plus SINH squared is COSH squared. The square root removes the square,
the COSH's cancel and we're left with the integral
of d t namely, t. If we substitute back in for
u, we see that this is really ARCSINH of u equal to
kappa x plus a constant. If we take the SINH of both sides then we obtain u as SINH (Kx + C). If we set up our coordinates so that X=0, right at the minimal point of the curve at the bottom, then since SINH of 0 equals 0, we can say that that constant
of integration vanishes and the slope at x equal zero
that is u vanishes also. And therefore,
we have u equal SINH kappa x. Now this is not exactly what
we were trying to find. Since recall, u is the derivative, dy dx. How do we get y for
the shape of the curve? Well, we integrate
the integral of SINH is COSH, we have to pull out a factor of kappa,
and so we get y is one over kappa, COSH kappa x. Plus some constant of integration. Why not? That would be how high off
the ground the bottom hangs. All this tells us the wonderful fact that
the catenary hangs as a hyperbolic cosine. That is the shape of a hanging cable or
chain. With this shape decided,
let's compute the length. It is of course the integral
of the length element. And in this case
integrating the square root of 1 + dy/dx squared dx
is going to work for us. Let's set the limits to be x
goes from negative l to l, where l is half of the projected
length to the x-axis. Then, in this case,
we need to substitute in for d y d x. But that is SINH of kappa x, leading to an integral of the square root
of one plus SINH squared of kappa x. We can substitute in COSH squared for
one plus SINH squared. The square root goes away, and we're left
with the integral of COSH kappa x, dx. That is, of course,
SINH of kappa x divided by kappa, evaluating from negative
l to l gives us 2 over kappa times the hyperbolic
sin of kappa times l. Now what does that mean? Well, one of the things that that implies,
is that we have exponential growth in this length of the chain,
given linear growth in l, the width. That's a bit of a surprising result. But it falls out easily
from what we've done. Let's take a look at
an example involving a spiral. We're going to use a parametric
representation, like we did for the circle. But with a 1 / t coefficient,
are in front, that causes you to spiral
in as t increases. We'll let t begin at 2 pi. So that we start to the right. The length element,
in this parametric case, Is the square root of d x d t squared,
plus d y d t squared, d t. Now, this is going to involve
a fair bit of differentiation using the quotient rule and
what we know about sines and cosines. I'm going to leave most of
the simplification work to you here. But do note that when you
multiply these terms out, there's quite a bit of cancellation
between the cross terms and using the fact that sine squared
plus cosine squared equals 1. In the end, you get the square root of t
squared plus 1, all over t squared dt. Now integrating that is not gonna be so
bad, maybe, let's try. What we need to do is integrate
this t goes from 2 pi to infinity. Now I don't see a really good
substitution at this point. But what I can do is say that the t squared is the dominant
term under that square root. And pulling that out and simplifying,
we get 1/t plus something that's in big O of 1/t squared. Now, I know that, that leading-ordered
term leads to a divergent integral. When you integrate 1 over t from
a constant to infinity, that diverges. We can follow the steps as
we've done back in lecture 27. But what this implies,
is that we have an infinite spiral. That's a little bit of a surprise. There's an infinite amount of length
in that simple looking spiral. There are more surprises still. Let's say, that you looked
a spiral that was not smooth, but rather based on squares descending. Lets say, that at each step you
are multiplying the length of the last segment, by some factor, r. Well, what would you do in this case? In this case,
integration becomes simply summation and you would add up the lengths
of all these segments. We recognize in this our old
friend the geometric series. And so in this multiplicative case, it's
possible to have a finite length spiral. Now other stranger things could happen. Let's say, we took a line segment and
removed the middle third and replaced it with two line segments
each with length one-third. Then, we would have a new curve,
whose length is four thirds. Now, if we took each of those resulting
segments, removed the middle third and replaced with two smaller segments,
as shown. Then we would be multiplying
the length of the entire curve again by a factor of four-thirds. Now, I think you can guess
what we're going to do. We're going to continue
this process inductively. At each stage,
multiplying by a factor of four-thirds. In the end, we obtain an interesting
object called a fractal curve, whose length is infinite. Even though the curve is abounded extent. This is the kind of interesting thing
that can happen when you start using your imagination. >> And so we see, that even some simple one dimensional
curves can pack some surprises. Be sure to internalize the idea and
computation of a length element. We're gonna need it in our next lesson, where we compute surface area.