The last video was very abstract in general, and
I used you know, f of x and g of t, and h of t.
What I want to do in this video is do an actual example.
Let's say I have f of xy.
Let's say that f of xy is equal to xy.
And let's say that we have a path in the xy plane, or
a curve in the xy plane.
And I'm going to define my curve by x being equal to
cosine of t, and y being equal to sine of t.
And we're going to go from-- you know, we have to define
what are our boundaries in our t --and we're going to go from
t is equal to 0-- or t is going to be greater than or equal to
0 --and then less than or equal to.
We're going to deal in radians, pi over 2.
If this was degrees, that would be 90 degrees.
So that's our curve.
And immediately you might already know what this type
of a curve looks like.
And I'm going to draw that really fast right here in and
we'll try to visualize this.
I've actually drafted ahead of time so that
we can visualize this.
So this curve right here, if I were to just draw it in the
standard xy plane-- do that in a different color so we can
make the curve green; let's say that is y, and this is right
here x --so when t is equal to 0, x is going to be
equal to cosine of 0.
Cosine of 0 is 1, y is going to be equal to sine of 0, which
is 0, so t is equal to 0.
We're going to be at x equal to 1, that's cosine of 0, and y is
sine of 0, or y is going to be 0, so we're going
to be right there.
That's what t is equal to; t is equal to 0.
When t is equal to pi of 2, what's going to happen?
Cosine of pie over 2-- that's the angle; cosine
of pi over 2 --is 0.
Sine of pi over 2 is 1.
We're going to be at the point 0, 1.
So this is when we're at t is equal to pi over 2.
You might recognize what we're going to draw is actually the
first quadrant of the unit circle; when t is equal to pi
over 4, or 45 degrees, we're going to be at square root
of 2, square root of 2.
You can try it out for yourself, but we're just
going to have a curve that looks like this.
It's going to be the top right of a circle,
of the units circle.
It's going to have radius 1.
And we're going to go in that direction, from t is equal to
0, to t is equal to pi over 2.
That's what this curve looks like.
But our goal isn't here just to graph a parametric equation.
What we want to do is raise a fence out of this kind of base
and rise it to this surface.
So let's see if we can do that or at least visualize it first,
and then we'll use the tools we used in the last video.
So right here I've graphed this function, and I've rotated it a
little bit so you can see [UNINTELLIGIBLE]
case.
This right here-- let me get some dark colors out --that
right there is the x axis, that in the back is the y axis, and
the vertical axis is the z axis.
And this is actually 2, this is 1 right here, y equal 1
is right there, so this is graphed that way.
So if I were to graphs this contour in the xy plane, it
would be under this graph and it would go like something like
this--- let me see if I can draw it --it would look
something like this.
This would be on the xy plane.
This is the same exact graph, f of x is equal to xy.
This is f of x; f of xy is equal to xy.
That's both of these, I just rotated it.
In this situation that right there is now the x axis.
I rotated to the left, you can kind of imagine.
That right there is the x axis, that right there is the y
axis-- it was rotated closer to me --that's the z axis.
And then this curve, if I were to draw in this rotation, is
going to look like this: when t is equal to 0, we're at x is
equal to 1, y is equal to 0, and it's going to form a unit
circle, or half or quarter of a unit circle like that.
And when t is equal to pi over 2, we're going to get there.
And what we want to do is find the area of the
curtain that's defined.
So let's see, let's raise a curtain from this
curve up to f of xy.
So if we keep raising walls from this up to xy, we're
going to have a wall looks something like that.
Let me shade it in, color it in so it looks a little
bit more substantive.
So a wall that looks something like that.
If I were to try to do it here this would be under the
ceiling, but the wall look something like
that right there.
We want to find the area of that.
We want to find the area of this right here where the base
is defined by this curves, and then the ceiling is defined by
this surface here, xy, which I graphed and I rotated
in two situations.
Now in the last video we came up with a, well, you could
argue whether it's simple, but the is, well, let's just take
small arc lengths-- change in arc lengths, and multiply them
by the height at that point.
And those small change in arc lengths, we called them ds,
and then the height is just f of xy at that point.
And we'll take an infinite sum of these, from t is equal to 0
to t will equal pi over 2, and then that should give us
the area of this wall.
So we said is, well, to figure out the area of that we're just
going to take the integral from t is equal to o to t is equal
to pi over 2-- it doesn't make a lot of sense when I write it
like this --of f of xy times-- or let me even better, instead
of writing f of xy, let me just write the actual function.
Let's get a little bit more concrete.
So f of xy is xy times-- so the particular xy --times the
little change in our arc length at that point.
I'm going to be very hand-wavy here.
This is all a little bit review of the last video.
And we figured out in the last video this change in arc length
right here, ds, we figured out that we could rewrite that as
the square root of the dx versus-- or the derivative of x
with respect to t squared --plus the derivative of y with
respect to t squared, and then all of that times dt.
So I'm just rebuilding the formula that we
got in the last video.
So this expression can be rewritten as the integral
from t is equal to 0 to t is equal to pi over 2 times xy.
But you know what?
Right from the get go we want everything eventually
be in terms of t.
So instead of writing x times y, let's substitute
the parametric form.
So instead of x let's write cosine of t.
That is x.
x is equal to cosine of t on this curve.
That's how we define x, in terms of the parameter t.
And then times y, which we're saying is sine of t.
That's our y; all I did is rewrote xy in
terms of t times ds.
ds is this; it's the square root of the derivative of x
with respect to t squared plus the derivative of y with
respect to t squared.
All of that times dt.
And now we just have to find these two derivatives.
And it might seem really hard, but it's very easy for us to
find the derivative of x with respect to t and the derivative
of y with respect to t.
I can do it right or down here.
Let me lose our graphs for a little bit.
We know that the derivative of x with respect to t is just
going to be: what's the derivative of cosine of t?
Well, it's minus sine of t.
And the derivative of y was respect to t?
Derivative of a sine of anything is the cosine
of that anything.
So it's cosine of t.
And we can substitute these back into this equation.
So remember, we're just trying to find the area of this
curtain that has our curve here as kind of its base, and
has this function, this surface as it's ceiling.
So we go back down here, and let me rewrite
this whole thing.
So this becomes the integral from t is equal to o to t is
equal to pi over 2-- I don't like this color --of cosine of
t, sine of t, cosine times sine-- that's just the xy
--times ds, which is this expression right here.
And now we can write this as-- I'll go switch back to that
color I don't like --the derivative of x with respect to
t is minus sine of t, and we're going to square it, plus the
derivative of y with respect to t, that's cosine of t, and
we're going to square it-- let me make my radical a little bit
bigger --and then all of that times dt.
Now this still might seem like a really hard integral until
you realize that this right here, and when you take a
negative number and you squared it, this is the same thing.
Let me rewrite, do this in the side right here.
Minus sine of t squared plus the cosine of t squared, this
is equivalent to sine of t squared plus cosine
of t squared.
You lose the sign information when you square something;
it just becomes a positive.
So these two things are equivalent.
And this is the most basic trig identity.
This comes straight out of the unit circle definition: sine
squared plus cosine squared, this is just equal to 1.
So all this stuff under the radical sign is
just equal to 1.
And we're taking the square root of 1 which is just 1.
So all of this stuff right here will just become 1.
And so this whole crazy integral simplifies a good bit
and just equals the square root of t equals 0 to t is equal to
pi over 2 of-- and I'm going to switch these around just
because it will make it a little easier in the next step
--of sine of t times cosine of t, dt.
All I did, this whole thing equals 1, got rid of it,
and I just switched the order of that.
It'll make the next up a little bit easier to explain.
Now this integral-- You say sine times cosine, what's
the antiderivative of that?
And the first thing you should recognize is, hey, I have a
function or an expression here, and I have its derivative.
The derivative of sine is cosine of t.
So you might be able to a u substitution in your head;
it's a good skill to be able to do in your head.
But I'll do it very explicitly here.
So if you have something that's derivative, you
define that something as u.
So you say u is equal to sine of t and then du, dt, the
derivative of u with respect to t is equal to cosine of t.
Or if you multiply both sides by the differential dt, if
we're not going to be too rigorous, you get du is
equal to cosine of t, dt.
And notice right here I have a u.
And then cosine of t, dt, this thing right here, that
thing is equal to d of u.
And then we just have to redefine the boundaries.
When t is equal to 0-- I mean so this thing is going to turn
into the integral --instead of t is equal to 0, when t is
equal to 0 what is u equal to?
Sine of 0 is 0, so this goes from u is equal to 0.
When t is pie over 2 sine of pi over 2 is 1.
So when t is pi over 2, u is equal to 1.
So from is equal to 0 to u is equal to 1.
Just redid the boundaries in terms of u.
And then we have instead of sign of t, I'm
going to write u.
And instead of cosine of t, dt, I'm just going to write du.
And then this is a super-easy integral in terms of u.
This is just equal to: the antiderivative of u is u 1/2
times u squared-- we just raised the exponent and then
divided by that raised exponent --so 1/2 u squared, and we're
going to evaluate it from 0 to 1.
And so this is going to be equal to 1/2 times 1 squared
minus 1/2 times 0 squared, which is equal to 1/2 times 1
minus 0, which is equal to 1/2.
So we did all that work and we got a nice simple answer.
The area of this a curtain-- we just performed a line integral
--the area of this curtain along this curve right here
is-- let me do it in a darker color --on 1/2.
You know, if this was in centimeters, it would be
1/2 centimeters squared.
So I think that was you know, a pretty neat application
of the line integral.