Welcome to our video on
arc length using polar equations.
If the polar equation is differentiable on the
close interval from alpha to beta,
then the length of the graph from alpha to beta
is given as def integral from alpha to beta
of the square root of f of theta squared
plus f prime of theta squared d theta,
or we can replace f of theta with r
and f prime of theta with d r d theta.
So these two equations are the same,
just written with different notation.
This formula can be derived using the
arc length formula for parametric equations,
but we're not going to take the time to do that
in this video.
We're going to go ahead and look at an example.
Here, we will determine the arc length of
r equals one minus sine data, already graphed here,
on the interval from zero to two pi.
So we want to find the arc length of this entire curve.
However, if we want, we can take it--
however this curve does have symmetry,
if we could find the arc length of half of this curve
and then double it to find the total arc length.
And let's go ahead and do it that way.
So we'll set it up to find the arc length from here to here
and then we'll multiply this by two.
So we'll start integrating here
and then we'll stop integrating here.
So our lower limit of integration is going to be
negative pi over two,
and our upper limit of integration is going to be
positive pi over two.
Notice when theta is negative pi over two,
we'd have one minus negative one,
so r is equal to two, so this corresponds
to this point right here,
and then when theta is pi over two, we have one minus one,
which would be zero, so we're at the pole right here.
So this arc length is going to equal two,
because we're doubling it,
times a def integral from negative pi over two,
pi over two,
of the square root of r squared.
So we have one minus sine theta squared
plus d r d theta squared.
Well, the derivative of r, with respects to theta,
would be negative cosine theta.
Square that
and all this integrated with respects to theta.
So let's go ahead and start to simply our radicand.
So if we square this, we'll have one minus two sine theta
plus sine squared theta.
This will be plus cosine squared theta.
And notice we have sine squared theta
plus cosine squared theta, that's equal to one.
So we can combine this one with this one
and we'll have the square root of two minus two sine theta.
Let's copy this over to a new slide.
Now we're not going to be able to integrate this
in its current form.
We're going to have to multiply and divide this
by its conjugate.
So were going to have the square root of
two plus two sine theta,
divided by the square root of two plus two sine theta.
And if it helps, we can put this over one.
Let's go ahead and find this product.
The numerator is going to be the square root of four
minus four sine squared theta.
Our denominator's going to stay the same.
Let's go ahead and factor
four minus four sine squared theta.
So notice we have four times the quantity
one minus sine squared theta.
The one minus sine squared theta is the same
as cosine squared theta.
So our numerator becomes the square root of four
co sine squared theta, which simplifies to two cosine theta.
So our numerator is two cosine theta,
our denominator stays the same.
And now we should be able to integrate this using
u substitution.
Remember the derivative of sine is cosine.
So if we let u equal two plus two sine theta,
then the u is going to equal two cosine d theta.
Notice our integrand does contain two cosine theta d theta,
and so we'll have the square root of u in the denominator.
So let's go ahead and write this in terms of u.
We'll leave out the limits of integration right now,
because these are in terms of theta.
And we are rewriting this in terms of u.
So all of this becomes d u,
and this will be one over the square root of u,
which will be the same as u to the negative one-half power.
Let's go ahead and integrate this.
We're going to have two times u to the one-half
divided by one-half.
Let's go ahead and convert this back to theta.
So dividing by one-half is the same as multiplying by two,
so we're going to have four times u to the one-half
is the same as a square root
and u is two plus two sine theta.
And we would evaluate this at the upper and lower
limits of integration.
So let's go ahead and do that on the next page.
So we'll first replace theta with pi over two.
We'll first replace theta with pi over two,
so we'll have two plus,
the sine of pi over two is one,
so we'll have two times one, that'll be two,
minus two plus sine of negative pi over two.
Well, sine of negative pi over two is negative one,
and two times negative one is negative two.
So we're going to have four times the square root of four
minus the square root of zero.
This is going to give us four times two,
which is equal to eight.
So let's go back to our original slide now.
This is the arc length of this curve from zero to two pi.
Remember, when we set this up, we did double it,
so this is the total arc length from zero to two pi,
or the entire curve.
Okay, I hope this explanation was helpful.
Thank you.