Welcome to a video
that will show how to determine where a curve
defined by a vectored value function is smooth.
First, a smooth curve does not have any sharp
corners or points,
so here are some examples of curves
that are completely smooth.
Again notice there are no sharp corners or points
on any of these curves whether it's a plain curve
or a space curve.
So a vector valued function is going to produce
a smooth curve on a given interval
if the derivative of the vector valued function
is not equal to the zero vector.
So if we have a vector valued function r of t
that has an x and y component
or an x, y, and z component
then if the derivative at that given value of t
does not equal zero for each component,
then that curve will be smooth.
The point where a curve has a sharp corner or point
is called a cusp or a node.
So looking at this curve here you can see
there are several places
where this curve has a sharp point or corner
as we see here, here, here and so on,
and these are called cusp or nodes.
And these will occur where each component
of the derivative of the vector valued function
will equal zero.
And anywhere in between those values it would be smooth.
So what we're going to do is determine
where the derivative of the vector valued function
is equal to zero.
That'll give us where the nodes occur.
And then the intervals of t in between these values
will be where the curve is smooth.
Let's go ahead and take a look at an example.
Here we want to determine the intervals
to which the given curve is smooth
on the closed interval from zero to two pi.
I provided the graph here so that we can tell
there are going to be places
where this curve has a sharp corner or point.
So what we do here is determine the derivative
of this vector valued function.
So the derivatives of each components equal to zero
and then determine for which values of t
both the x component and the y component
of the derivative would be equal to zero.
So it may be helpful to write this in a form
that's easier to find the derivative,
meaning I'm going to write this as three
and then cosine t to the third power.
This will remind me to apply the chain
when I find the derivative
and the y component will be three times sine t
raised to the third power.
Okay let's go ahead and find the derivative
of this vector valued function.
Remember that means find the derivative
of each component.
So cosine t is actually going to be our u.
When we apply the chain rules.
So we're going to multiply by this three
that'll give us nine u to the second times u prime.
Or nine cosine squared t
and then times u prime
the derivative of cosine t is going to be
negative sine t.
Now for the y component again we have
to apply the chain rule here
where sine t is equal to u
and u prime will be cosine t.
So we're going to have nine
sine squared t
times cosine t.
Cosine t is u prime.
Let's go ahead and clean this up.
This is going to be equal to negative nine,
cosine squared t
times sine t.
Then we'll have nine sine squared t,
cosine t.
And again we want to know when this is equal
to the zero vector,
so where the x component and the y component
are both equal to zero.
So what that means is we want to determine
when negative nine cosine squared t sine t
is equal to zero and when nine sine squared t
cosine t is equal to zero.
So we want the value of t
where both of these equations
would be equal to zero at the same time.
Now remember to solve these trig equations we want to have
it in terms of one trig function.
So using the identity cosine squared plus sine squared
equals one we can replace cosine squared t
with one minus sine squared t.
So we'd have negative nine
times one minus sine squared t
times sine t equals zero
and right here we can replace sine squared t
with one minus cosine squared t.
So using this equation here we want to know
for what values of t
would one minus sine squared t equals zero.
And when sine t equals zero.
And over here we want to know when one minus cosine
squared t is equal to zero
and when cosine t is equal to zero.
Let's go ahead and move the sine squared to the other side
we want to know when one equals sine squared t.
So if we square root both sides
we want to know when plus or minus one
is equal to sine t.
So an interval from zero to two pi
sine t would be equal to one
at t equals pi over two
and sine t would be equal to negative one
at three pi over two.
And then for sine equal to zero
that's going to occur at zero, pi, and two pi.
This isn't enough information to determine
where these nodes or cusps occur,
because these equations here
and these equations here must be zero
for the same values of t.
So let's go ahead and solve these equations.
Let's go ahead and add cosine squared t to both sides
so we have one equals cosine squared t.
Again we're going to square root both sides of the equation.
So we want to know when plus or minus one
is equal to cosine t.
Well cosine is equal to one
at zero and two pi
and cosine is equal to negative one at pi radiants.
And then cosine t is equal to zero
at pi over two
and also at three pi over two.
And in this case note that the values of t
that make the ex component of the derivative
equal to zero, and the values of t that make
the y component of the derivative equal to zero
are all the same.
It's not always the case but it is the case here.
So these are the values of t
where the derivative of the vector valued function
is equal to the zero vector.
So all the values in between these
the curve would be nice and smooth.
Let's go ahead and take this information
onto the next screen.
We had t equals zero,
t equals pi over two,
t equals pi,
three pi over two,
and two pi.
So that means the curve will be smooth
on the intervals between these values.
It'll be smooth on the open interval
from zero to pi over two
and then from pi over two to pi.
From pi to three pi over two
and then lastly from three pi over two.
Now keep in mind these intervals
are intervals of t
not x like we normally think of
when we talk about rectangular equations.
I hope you found this helpful,
thank you for watching.