Parametric Arclength | MIT 18.01SC Single Variable Calculus, Fall 2010

Hi.
Welcome back to recitation.
We've been talking in class a little bit about parametric
equations and arc length.
So let's do an example of a problem where you compute an
arc length of a curve given by some parametric equations.
So in particular, I have here the parametric equations y
equals t minus 1 over t, and x equals t plus 1 over t for 1
less than or equal to t, less than or equal to 2.
So those parametric equations trace out some piece of a
curve in the plane.
And what I'd like you to do is write down an integral whose
value is equal to the arc length of that curve.
So the integral you're going to get is going to be pretty
hard to evaluate.
So I wouldn't recommend you spend a whole lot of time
trying to evaluate it.
But just so we see that we, you know, we can do this, and
so we have the integral whose value is the arc length.
So take a few minutes, work on that, pause the video, come
back, and we can work on it together.

All right.
So I hope you've had some luck working this problem.
Let's get started on it together.
So in order to compute the arc length by an integral, we need
to figure out what the element of arc length is, that
little piece ds.
So this was true when we had our curve given in rectangular
coordinates, and it's also true when our curve is given
in parametric coordinates.
So in general, the little element of arc length is given
as ds equals the square root of dx squared plus dy squared.
Now, when our curve is given in rectangular coordinates,
usually what we do here is we factor out a dx squared, and
then we have dx by dy, the derivative of y
with respect to x.
But here, we don't have x or y given in terms
of the other one.
We have them both given in terms of this parameter, t.
So the parameter t is the thing, is our sort of
independent variable.
It's the thing we're going to want to end up integrating
with respect to.
So what we actually want to do in this case, is we want to
factor out a dt from everywhere.
So if you factor out a dt, what this becomes--
or dt squared, I guess I should say--
is that you get dx dt squared plus dy dt squared, the whole
thing square root dt.
OK.
So this is just a little algebraic manipulation
involving differentials.
And what this gives me is an expression that will, you
know, x is a function of t. x is given as a function of t,
right here.
So dx dt is a function of t, and similarly, y is given as a
function of t.
So dy dt is a function of t.
So this whole expression is a function of t, so we have a
function of t dt.
So it's all set up to be integrated.
So in order to actually, you know, compute with this
formula, well, we just take our y, and we take our x, and
we take their derivatives.
So from--
well, I guess we could start with a y, since it's on top
over there.
So dy dt.
Well, you know.
You just take a derivative.
Right?
So you have y is equal to t minus 1 over t.
And you take a derivative of that.
Derivative of t is just 1, derivative of 1 over t is
minus 1 over t squared, so this becomes-- but there's an
extra minus sign there, so this becomes 1
plus 1 over t squared.
And similarly, dx by dt is equal to 1
minus 1 over t squared.
So you have these two derivatives.
So to compute arc length, you plug them into this formula
for the element of arc length, and then you'll integrate.
So OK.
So we have to put them into that formula.
So what do we get?
We have ds is equal to the square root of, well, so dx dt
squared plus dy dt squared.
So that's 1 minus 1 over t squared squared, plus 1 plus 1
over t squared squared.

dt outside.
OK.
And so now we, you know, maybe we can simplify this a little.
I would probably expand out, at this point.
So if you expand this out, this is--
well, let's see.
The first one is just is 1 minus 2 over t squared plus 1
over t to the fourth, plus--
that's the first one we expanded out, and the second
one, we get 1 plus 2 over t squared plus 1
over t to the fourth.

All right.
And if we put all this together, say, over a common
denominator, and simplify it a little bit, we can see that
this is equal to--
over t squared terms cancel out, and we end up with
something like square root of t to the fourth plus 1 over t
squared dt.
OK.
So this is all this ds that we've been playing with.
So the actual arc length is what we get when
we integrate ds.
So we need appropriate bounds.
So where do we get those bounds?
Well, we go back to the question, and here, the
question had to have told us.
Right?
If the question didn't tell us, then it didn't actually
describe a piece of curve, or a finite piece of curve,
really, is the point.
So the question tells us that we want this from
t equals 1 to 2.
So the arc length, we just integrate our little element
of arc length over the appropriate interval.
So that's t from 1 to 2 of the square root of t to the fourth
plus 1 over t squared dt.
All right.
So as I said at the beginning, this is not an integral that's
readily susceptible to the techniques that we've learned.
You know, if you were interested, you could try some
numerical methods on it.
I'm sure, you know, computer algebra software can spit out,
at worse, a good numerical
approximation, if you're curious.
One more thing I want to say about this question, is that
we did all this without ever trying to draw the curve, or
think about what the curve looks like.
So it's not that hard to get some sort of basic sense of
the behavior of this curve.
For example, when t gets very, very large, we see that what
happens is that x and y, well, the 1 over t gets small, so x
and y get close to each other.
But x is just a tiny, tiny bit bigger.
So for, as t gets very, very big, you see that the point on
the curve is getting very close to the line y equals x,
but not actually touching it.
And you can, you know, you can try other things, as t gets
close to 0 from the positive side, as t gets close to 0
from the negative side.
So this has some sort of asymptote, it seems like.
There's some other things you could check.
You could try to show that the x value is always either--
well, when t is positive, it's always at least 2, or when t
is negative, it's always at least negative 2, and there's
a sort of gap in between.
You could analyze its properties there.
The other thing you could do, is you could try and solve
these equations to eliminate t, and get just rectangular
equations in terms of x and y.
So I'm not going to do that for you.
I'll give you a hint about how I would go about it.
Which is, you could start off-- so you want
to eliminate t.
So I think the nicest way to do that is if you add these
two equations.
You'll get that the 1 over t's cancel out, and you get 2t
equals x plus y, or t equals x plus y over 2.
And then you can take, and you could substitute it back into
one of these equations.
And when you substitute it back, then you'll have an
equation that just involves x and y, which, with a little
more simplification, takes a form that you should already
be familiar with.
OK.
And yes.
Arc length.
This is how we do arc length in parametric form.
Works out pretty nice, except in this case, we get an
integral we can't evaluate at the very end.
Too bad.
All right.
I'll leave it with that.