Welcome to a presentation on arc length.
If f of x is continuous and differentiable
on the closed interval from a to b,
then the arc length is equal to the definite integral
from a to b of the square root of one plus f,
prime of x squared, integrated with respect to x.
One thing we have to be careful about
when we apply this formula
is that this is the derivative of the function.
A common error is just to sub in the original function,
and of course that would not be correct.
And we can find the arc length in a similar way
if we have the function in the form of x equals g of y.
And of course, if we integrate with respects to y,
our limits of integration will be the respects to y.
Let's take a look at the idea behind
where this formula comes from.
If we wanted to determine the arc length
of this black function, on the interval from zero to three,
we could start with a single segment,
and the length of this segment would be an approximation
of the actual arc length.
However, as we increase the number of segments used,
as we see here, you can quickly see that,
as the number of segments increase,
becomes a better approximation,
and as the number of segments approaches infinity,
the length of those segments
will approach the actual arc length.
So, using this idea of segments
to approximate the arc length,
we can get a better understanding
of where the formula comes from.
And it's really based on the Pythagorean theorem,
so if we wanted to determine the arc length
of this blue function on the interval from one to two,
we could use this red segment as an approximation.
If we take this segment and form a right triangle,
this length here would be delta y,
and this length here would be delta x.
And what the hypotenuse equal delta s.
So, using the Pythagorean theorem,
we can state that delta s squared
must equal delta x squared
plus delta y squared.
And if we solve this for delta s,
we'd have delta s equals the principle square root
of delta x squared plus delta y squared.
And then what we're going to do is multiply the right side
of the equation by delta x over delta x.
Next, we're going to take this delta x in the denominator
and move it into the square root.
Remember, when we do that, it would become delta x squared.
Now we'll simplify this fraction here.
So, this would give us the length of this single segment
in red, but as we saw in the animation,
we wanted some more and more of these segments,
and as we do that, it will approach the actual arc length
from one to two.
So, now we'll change the notations slightly
to denote that we're going to be using more and more segments
to determine this arc length.
Now, on the left we'll have the sum of the segments
from i equals one to n, and it must equal the sum
of the right side of the equation as well.
Now, you probably know where we're going with this.
If we let the number of segments approach infinity,
on the left side we would have just the arc length,
and on the right side we would actually have the formula
for arc length, which is going to be the definite integral
on the interval from a to b.
Or in this case, it's from one to two,
of the square root of one plus...
This would be the derivative squared.
Integrating with respects to x.
So, hopefully this gives you a little bit
better understanding of where this formula comes from.
The main thing here
is that we're using the Pythagorean theorem
to derive this formula as the number of segments
used to approximate the arc length approaches infinity.
Let's go and take a look at an example.
Here we want to determine the arc length of this function
on the given interval.
So, here's the graph of the function.
And if the interval is from zero to four,
we're trying to find the arc length from here to here.
So, the first step is to find the derivative
of the given function.
So, the derivative would be equal to...
1/3 times 3/2,
x to the 3/2 minus one.
This is going to be 1/2, x to the 1/2.
We remember for the formula,
we do have to square this derivative,
so let's go ahead and do that.
Well, if we square 1/2, we'll have 1/4.
If we square x to the 1/2, we'll just have x.
So, let's see if we can find this arc length now.
The arc length will equal the definite integral
from zero to four,
and because I know we're going to have to integrate,
let's rewrite this in rational exponent form.
So, instead of the square root, we'll write it
to the 1/2 power, and one plus the derivative squared
is equal to one plus 1/4 x.
Now, this will require u substitution,
where u will be equal to one plus 1/4 x.
Which means that du will equal 1/4 dx.
Looking at the integral, we're looking for a substitution
for dx, so let's go ahead and solve this for dx
by multiplying both sides by four.
We have four du
equals dx.
Let's rewrite this in terms of u.
We have u to the 1/2,
and then dx is equal to four du,
so we have four out here and du here.
Well, this will be equal
to four times u to the 3/2
divided by 3/2.
So, let's go ahead and clean this up
and rewrite it in terms of x.
So we have the arc length would be equal to four times 2/3.
U to the 3/2 and u is one plus 1/4 x.
We need to evaluate this at the upper and lower limits.
So, we'd have 8/3.
We'll first replace x with four.
So, it looks like we'd have one plus...
1/4 times four would be one,
so we'll have one plus one to the 3/2 power.
And next to zero, we'll have one plus zero to the 3/2 power.
So 8/3 times two to the 3/2...
minus...
Well, one to the 3/2 is just one,
so this would be the arc length of this function
on the interval from zero to four.
And if the decimal is approximately equal to 4.876.
Okay, that's all we're going to have time for on this video.
We'll go ahead and take a look at one more example
in part two.
Thank you for watching.