### Video Transcript

In this video, we’ll learn how to
use integration to find the length of a curve. By this stage, you might be aware
of some of the key applications of integration, areas, volumes, and average
values. But did you know that integration
can also be used to find the length of a curve? In this video, we’ll derive the
formula for arc length using integration and then look at some of the key
applications of this formula.

Suppose that a curve 𝑐 is defined
by the equation 𝑦 equals 𝑓 of 𝑥, where 𝑓 is a continuous function over the close
interval 𝑎 to 𝑏.

That might look a little something
like this. We can then find an estimate to the
length of 𝑐 by dividing our interval into subintervals and then working out the
length of the straight line that joins the function at these intervals as shown. Now, imagine the number of
subintervals 𝑛 gets ever larger. What will happen to our
approximation? As the number of subintervals
increase, each straight line gets smaller, and our approximation will get closer and
closer to the exact value of the length of 𝑐. So we’re going to define the length
of the curve 𝐿 with an equation 𝑦 equals 𝑓 of 𝑥 as the limit of the length of
these inscribed polygons, assuming that limit exists.

The problem is this isn’t a hugely
useful formula for us. But it does allow us to derive an
integral formula for 𝐿 given that 𝑓 is continuous and differentiable over that
closed interval 𝑎 to 𝑏. But what we can do is work out the
length of the line segments by using the distance formula or the Pythagorean theorem
by defining Δ𝑦 subscript 𝑖 as the difference between 𝑦 subscript 𝑖 and 𝑦
subscript 𝑖 minus one, which is, of course, equal to the difference between 𝑓 𝑥
subscript 𝑖 minus 𝑓 of 𝑥 subscript 𝑖 minus one. We can say that the length of the
line segments are given by the square root of 𝑥 𝑖 minus 𝑥 𝑖 minus one squared
plus 𝑦 𝑖 minus 𝑦 𝑖 minus one square. And we can redefine that as the
square to of Δ𝑥 𝑖 squared plus Δ𝑦 𝑖 squared.

Then by applying the mean value
theorem to our function 𝑓 on the close interval 𝑥 𝑖 minus one 𝑥 𝑖, we find that
there’s this number 𝑥 𝑖 star between 𝑥 𝑖 minus one and 𝑥 𝑖 such that 𝑓 of 𝑥
𝑖 minus 𝑓 of 𝑥 𝑖 minus one is equal to the derivative of the function evaluated
at this number 𝑥 star 𝑖 times 𝑥 𝑖 minus 𝑥 𝑖 minus one. And then we redefine this using the
notation from before. We find that Δ𝑦 𝑖 is equal to 𝑓
prime of 𝑥 star 𝑖 times Δ𝑥. We replaced Δ𝑦 𝑖 in our
expression for the length of the line segments. And then we factor by the square
root of Δ𝑥 squared. And we find that the length of our
line segments is equal to Δ𝑥 times the square root of one plus 𝑓 prime of 𝑥 star
𝑖 squared.

We substitute this into our
original limit. And we find that 𝐿 is equal to the
limit as 𝑥 tends to infinity of the sum between 𝑖 equals one and 𝑛 of Δ𝑥 times
the square root of one plus 𝑓 prime of 𝑥 star 𝑖 squared. By definition, we can say that this
is equal to the integral between 𝑎 and 𝑏 of the square root of one of 𝑓 prime of
𝑥 squared with respect to 𝑥. And we’ve obtained the ultimate
formula.

This says that if 𝑓 prime is
continuous on the close interval 𝑎 to 𝑏. Then the length of the curve given
by 𝑦 equals 𝑓 of 𝑥, given that 𝑥 is greater than or equal to zero and less than
or equal to 𝑏, is 𝐿 equals the definite integral evaluated between 𝑎 and 𝑏 of
the square root of one plus 𝑓 prime of 𝑥 squared with respect to 𝑥. Using Libenie’s notation, we can
write this as the integral between 𝑎 and 𝑏 of the square root of one plus d𝑦 by
d𝑥 squared evaluated with respect to 𝑥. We’re now going to have a look at
the application of this formula.

Calculate the arc length of the
curve of 𝑦 equals the square root of four minus 𝑥 squared between 𝑥 equals zero
and 𝑥 equals two, giving your answer to five decimal places.

Using Libenie’s notation, the
formula for the arc length of a curve is given by the definite integral evaluated
between 𝑎 and 𝑏 of the square root of one plus d𝑦 by d𝑥 squared with respect to
𝑥. We know that 𝑦 is equal to the
square root of four minus 𝑥 squared, so we’ll begin by working out d𝑦 by d𝑥. If we write 𝑦 as four minus 𝑥
squared to the power of one-half, then we can use the general power rule to find the
derivative of this function with respect to 𝑥. It’s a half times four minus 𝑥
squared to the power of negative half multiplied by the derivative of the function
that sits inside the brackets. That’s negative two 𝑥.

Dividing through by two and
rewriting d𝑦 by d𝑥, we obtain negative 𝑥 over the square root of four minus 𝑥
squared. We let 𝑎 be equal to zero and 𝑏
be equal to two. And this means the length of the
curve that we’re interested in is equal to the definite integral between zero and
two of the square root of one plus negative 𝑥 over the square of four minus 𝑥
squared squared with respect to 𝑥. And this simplifies quite
nicely. The integrand becomes the square
root of one plus 𝑥 squared over four minus 𝑥 squared. We can simplify the expression
inside the square root by multiplying both the numerator and denominator of the
number one by four minus 𝑥 squared. And when we add that, we get four
minus 𝑥 squared plus 𝑥 squared over four minus 𝑥 squared. Negative 𝑥 squared plus 𝑥 squared
is zero. And so our integrand becomes the
square root of four over four minus 𝑥 squared.

This can be simplified even
further. If we take out our factor of four,
we see that the integrand could be written as the square root of four times one over
the square of four minus 𝑥 squared. The square to four is two. And we’re allowed state constant
factors outside of the integral. So we have that 𝐿 is equal to two
times the definite integral evaluated between zero and two of one over the square of
four minus 𝑥 squared. Now, this might look really
nasty. But we can evaluate this integral
by using a substitution. We recall that the derivative of
arc sine of 𝑥 is one over the square root of one minus 𝑥 squared. So we rewrite our integrand
slowly.

This time we take out our factor
four on the denominator. And we see that two times one over
the square of four cancels. We can also write 𝑥 squared over
four as 𝑥 over two squared. We choose 𝑢 then for our
substitution to be equal to 𝑥 over two. Then d𝑢 by d𝑥 is equal to
one-half. And we can say this is the
equivalent to saying two d𝑢 equals d𝑥. We’re also going to need to change
the limits on our integral. When 𝑥 is equal to two, 𝑢 is
equal to two divided by two, which is one. And when 𝑥 is equal to zero, 𝑢 is
zero divided by two, which is zero.

So we now see that 𝐿 is equal to
the definite integral between zero and one of one over the square root of one minus
𝑢 squared times two d𝑢. Once again, we’ll take out our
common factor of two. And we can now evaluate the
integral of one over the square root of one minus 𝑢 squared. It’s simply arc sin of 𝑢. And we’re looking to evaluate two
times arc sine of 𝑢 between one and zero. That’s two times arc sin of one
minus arc sin of zero, which is simply 𝜋. And we see that correct to five
decimal places, the arc length of the curve 𝐿 is 3.14159.

We’re now going to look at how to
find the arc length of a curve defined as 𝑥 in terms of 𝑦. For a curve with equation 𝑥 equals
𝑔 of 𝑦, where 𝑔 of 𝑦 is continuous and has a continuous derivative on the close
interval 𝑐 to 𝑑, the length of the curve between 𝑦 equals 𝑐 and 𝑦 equals 𝑑 is
given by the definite integral evaluated between 𝑐 and 𝑑 of the square root of one
plus d𝑥 by d𝑦 squared d𝑦. The application of this formula is
just the same as the application of the previous formula we used. Let’s see what this might look
like.

Find the length of the length of
the arc of the curve 𝑦 squared equals 𝑥 between 𝑦 equals zero and 𝑦 equals
four.

Remember, for a curve with equation
𝑥 equals 𝑔 of 𝑦, where 𝑔 of 𝑦 is continuous and has a continuous derivative on
the close interval 𝑐 to 𝑑, the arc length of the curve between 𝑦 equals 𝑐 and 𝑦
equals 𝑑 is given by the definite integral evaluated between 𝑐 and 𝑑 of the
square root of one plus d𝑥 by d𝑦 squared d𝑦. Let’s compare this formula to our
question.

Our function 𝑥 in terms of 𝑦 is
𝑥 equals 𝑦 squared. So we can see we’re going to need
to differentiate 𝑥 with respect to 𝑦. The derivative of 𝑥 with respect
to 𝑦 is two 𝑦. We’re also told that we need to
find the length of the arc between 𝑦 equals zero and 𝑦 equals four. So we’re going to say that 𝑐 is
equal to zero and 𝑑 is equal to four. Substituting everything we know
into this formula and we obtain 𝐿 to be equal to the definite integral between zero
and four of the square root of one plus two 𝑦 squared, evaluated with respect to
𝑦. Well, two 𝑦 all squared is the
same as four 𝑦 squared. So to find the length of the arc
we’re interested in, we need to evaluate the definite integral of the square root of
one plus four 𝑦 squared between the limits of zero and four. We’re going to use our graphical
calculator to do so. And when we do, we obtain that the
length of the arc is equal to 16.81863 which is equal to 16.8 units correct to three
significant figures.

We saw that the processes in the
two examples we’ve looked at are almost identical. We can, therefore, redefine and
formalize. So we have a single formula. Arc length is given by 𝐿 equals
the integral d𝑠 where d𝑠 is equal to the square root of one plus d𝑦 by d𝑥
squared d𝑥 if 𝑦 is equal to 𝑓 of 𝑥 over some closed interval 𝑎 to 𝑏. And d𝑠 equals the square root of
one plus d𝑥 by d𝑦 squared d𝑦 if 𝑥 is equal to 𝑔 of 𝑦 for some closed interval
𝑐 to 𝑑. In our final example, we’re going
to look how to define an arc length function 𝑠 of 𝑥 from a point given.

Find the arc length function for
the curve 𝑦 equals four 𝑥 to the power of three over two, taking the point with
coordinates one, four as the starting point.

Remember, the arc length formula
for a function 𝑦 equals 𝑓 of 𝑥 is given by 𝐿 equals the integral d𝑠, where d𝑠
is equal to the square root of one plus d𝑦 by d𝑥 squared d𝑥. In our case, 𝑦 is equal to four 𝑥
to the power of three over two. So we’ll begin by simply working
out what d𝑦 by d𝑥 is. Remember to differentiate the
function of this sort, we multiply by the power and then reduce that power by
one. So d𝑦 by d𝑥 is equal to three
over two times four 𝑥 to the power of one-half. Four divided by two is two. So we see that d𝑦 by d𝑥 is equal
to six 𝑥 to the power of one-half. And this means we can substitute
what we know so far into our formula for d𝑠. We get the square root of one plus
six 𝑥 to the power of one-half squared d𝑥.

Well, this simplifies quite nicely;
we end up with the square root of one plus 36𝑥. And this means that 𝐿 is going to
be equal to the integral of the square root of one plus 36𝑥 evaluated with respect
to 𝑥. But what are our limits? Well, we’re working in terms of
𝑥. And we know we have a starting
point at 𝑥 equals one. So that’s the lower limit. We’re looking for a general
function, so the upper limit is simply going to be 𝑥. And we can see that the argument
function will be given by the integral evaluated between one and 𝑥 of the square
root of one plus 36𝑥 d𝑥. So how do we evaluate this
Integral? We cannot rely on calculator, and,
in fact, we’re going to need to use a substitution.

We have a composite function. We’re going to let 𝑢 be equal to
the function inside the square root; that’s one plus 36𝑥. We can say that the derivative of
𝑢 with respect 𝑥 is, therefore, 36. And we can also equivalently say
that one over 36 d𝑢 equals d𝑥. Before we can make any substitution
though, we’re going to need to change the limits. We use our substitution to do
so. And the upper limit is 𝑥, so when
𝑥 is equal to 𝑥, 𝑢 is just one plus 36𝑥. And the lower limit is one, so when
𝑥 is equal to one, 𝑢 is one plus 36 times one which is 37. We now replace one plus 36 with
𝑢. I’ve changed the square root to be
the power of one-half. And we replace d𝑥 with 36 d𝑢. So all that’s left is to evaluate
this integral between the limits given.

The integral of 𝑢 to the power of
one-half is 𝑢 to the power of three over two divided by three over two. So this means that the integral of
𝑢 to the power of one-half over 36 is equal to 𝑢 to the power of three over two
over 36 divided by three over two. And, of course, dividing by three
over two is the same as multiplying by two-thirds. We can simplify this a little. And we see that our arc length
function is now one over 54 times 𝑢 to the power of three over two. Evaluated between 37 and one plus
36𝑥. That’s one over 54 times one plus
36𝑥 to the power of three over two minus one over 54 times 37 to the power of three
over two. And we’re done.

We found the arc length function
for the curve 𝑦 equals four 𝑥 to the power of three over two, taking the point
with coordinates one, four as the starting point. It’s useful to know that now we
have this arc length function. We can find the length of the curve
between the point 𝑥 equals one and any other point by substituting that value of 𝑥
into this formula.

In this video, we’ve seen that we
can use integration to help us find the arc length of curves given in the form 𝑦
equals 𝑓 of 𝑥 and 𝑥 equals 𝑔 of 𝑦. We saw that the arc length formula
is given by 𝐿 is equal to the integral d𝑠, where d𝑠 is equal to the square root
of one plus d𝑦 by d𝑥 squared d𝑥 if 𝑦 is a function in 𝑥 over some close
interval 𝑎 to 𝑏. And it’s equal to the square root
of one plus d𝑥 d𝑦 squared d𝑦 if 𝑥 is a function in 𝑦 over some closed interval
𝑐 to 𝑑. We also saw that we can use this
formula to find the general equation for the arc length for a function from a given
starting point by changing the upper limit of our integral to 𝑥 or 𝑦, depending on
the context.