Welcome to a lesson on defining a piecewise
smooth parameterization of a path in the xy-plane
to prepare for line integrals.
One skill needed to successfully evaluate line integrals
is to be able to define a smooth parameterization
of a given path in the xy-plane.
This means we want to be able to break up
a given curve and define it in terms of t
so each piece is smooth
while the value of t remain continuous.
So, let's take a look at our first example.
Our curve is this two by two square,
and since a square has four sides, we'll have a parameter
for each side.
So, we'll label this curve one, curve two, curve three,
and curve four.
So, for curve one, we go from the point zero zero
to the point two zero.
That's pretty straight forward.
If we let x equal t, and y equal zero,
and then define t on the closed interval zero to two
we would take the path of curve one.
And for curve two, we go from the point two zero
to the point two two.
Notice x is equal to two, and y increases from zero to two,
but we can't just put y equals t because the value of t
now has to start at two.
So if we start y at t, and then subtract two,
and then let t go from two to four, if t is four,
y would be equal to two, and then we take the path
of curve two.
Now, take a look at curve three.
Go from the point two two to the point zero two.
Now t has to start at four, and we'll see where it stops.
So we then created a formula with t in it,
where t starts at four and x goes from two to zero.
So if we let x equal two
minus, t, minus four,
let's see what that does.
If t is four, we subtract zero, that's good.
When t is five, we would subtract one, that's good.
And when t is six, we would subtract two,
that would give us zero.
So that's good.
And our equation for y would stay at y equals two.
So, t's our closed interval from four to six this time.
And for curve four, go from the point zero two
to the point zero zero.
x stays at zero.
We need an equation for y.
The t has to start at six.
So very similar we did up here.
y starts at two, and now we're going to subtract,
instead of just t, we'll have to have t minus six.
When t is six, we'd have two minus zero, when t is seven,
we have two minus one, so we stop at t equals eight.
So now go ahead and write this curve as a
vector value function, using these parameterizations.
So we'll have four pieces, the first one,
first piece will be t comma zero,
when t's on the interval from zero to two,
and then we'll have two comma, t minus two,
from two to four.
Here, we're going have two minus t plus four,
it's going to be six minus t, the x component.
Then y is two, this is from four to six.
For the fourth curve, zero comma, this is going to be
two minus t plus six, that'll be eight minus t.
And t is from six to eight.
Let's go take a look at another example.
Here they give a curve as an ellipse,
given by this equation here.
x squared divided by nine plus y squared
divided by four equals one.
So to parameterize this one,
we're actually going to use trig identity.
We know that cosine squared t
plus sine squared t is equal to one.
Then, using the given equation,
if we let x equal three cosine t,
this will give us nine cosine squared t
divided by nine, if we let y equal two sine t,
this will give us four sine squared divided by four,
that would satisfy the equation for this curve.
Now we have both x and y expressed as functions of t.
And if we trace out this curve starting here
and going around the ellipse,
we can let our vector value function
have an x component of three cosine t,
and a y component of two sine t for the given curve,
where t is on the closed interval from zero to two pi.
We should be all the way around the ellipse.
Let's go and take a look at one more.
So, for our last example,
we have a path that starts here at the origin,
follows the cubic function out to this point
of intersection, and then down, back to the origin
along the line y equals four x.
This point of intersection here is the point two eight.
So this will be curve one, and this will be curve two.
For curve one, we can let x equal t.
So t would have to be on the interval from zero to two,
and then, if x is equal to t, y would be equal to t cubed.
Now for the second curve, we know that t has to start
at the value of two.
And our path is along this line from the point two eight,
back to the origin, or zero zero.
So for x,
starts at two,
and then t also starts at two.
So we have two minus, and then t minus two.
When t is two, this would be zero, when t is three,
this would be two minus one.
When t is four, we'd have two minus two which would be zero
back at the origin, for the x-coordinate.
Now see if we can find an equation for y,
we know the y coordinate will start at eight
and decrease to zero along the line y equals four x.
So this slope here is going to be an important part
to this equation.
We'll let y start at eight, we want to subtract four
times, now if we come up with an expression in terms of t,
where t is on this interval, but this quantity takes on
the values from zero to two, it would satisfy
the values for y.
So if we put t minus two here, let's check it.
When t is two, we would be subtracting zero from eight.
When t is three, we'd have eight minus four.
When t is four, we'd have eight minus four times two,
which would give us zero.
So this works.
Let's go and write this as a vector value function.
The first curve, we have x equals t, y equals t cubed.
Closed interval from zero to two for t.
Let's go ahead and simplify our expression for curve two.
Here we'd have two minus t plus two, that's four minus t.
And then for the y component, we have eight, this would be
plus eight, so that would be 16.
And this would be minus four t.
And t's on the closed interval from two to four.
So as you can see, some of these formulas can be
a little tricky, when trying to determine how to
manipulate an expression for t to get the values
that we do need to make these formulas work.
I hope you found these examples helpful.