### Video Transcript

In this video, we’ll learn how to
use integration to find the arc length of a curve defined by parametric equations of
the form 𝑥 equals 𝑓 of 𝑡 and 𝑦 equals 𝑔 of 𝑡. We’ll begin by recalling the
formula for the arc length of a curve defined as 𝑦 is equal to some function of
𝑥. Then, we’ll look at how we can
generalize this formula for parametrically defined curves and consider a number of
examples of the process.

Given an equation 𝑦 in terms of 𝑥
and 𝑥-values greater than or equal to 𝑎 and less than or equal to 𝑏, the arc
length 𝐿 is given by the definite integral between 𝑎 and 𝑏 of the square root of
one plus d𝑦 by d𝑥 squared with respect to 𝑥. We want to find a way to use this
formula for curves defined parametrically. Remember, these are of the form 𝑥
equals 𝑓 of 𝑡 and 𝑦 equals 𝑔 of 𝑡. We also know that in this case, d𝑦
by d𝑥 is equal to d𝑦 by d𝑡 over d𝑥 by d𝑡, which can be written as 𝑔 prime of
𝑡 over 𝑓 prime of 𝑡.

Now, using the fact that d𝑥 by d𝑡
is equal to 𝑓 prime of 𝑡, we’re going to equivalently say that d𝑥 equals 𝑓 prime
of 𝑡 d𝑡. Redefining our limits so they’re in
terms of 𝑡 and we find that we can rewrite the arc length as the definite integral
between 𝛼 and 𝛽 of the square root of one plus d𝑦 by d𝑡 over d𝑥 by d𝑡 squared
𝑓 prime of 𝑡 d𝑡. We separate the components of our
fraction and rewrite 𝑓 prime of 𝑡 as d𝑥 by d𝑡. Now, this looks really nasty. But we’re next going to add the
fractions inside of our root by creating a common denominator of d𝑥 by d𝑡
squared.

So inside of our root, we have d𝑥
by d𝑡 all squared over d𝑥 by d𝑡 all squared plus d𝑦 by d𝑡 all squared over d𝑥
by d𝑡 all squared. We can then factor out that
denominator and notice we need to use the absolute value symbols as we need this to
be positive to be able to do so. And we see that the arc length is
equal to the definite integral between 𝛼 and 𝛽 of one over the absolute value of
d𝑥 by d𝑡 times the square root of d𝑥 by d𝑡 squared plus d𝑦 by d𝑡 squared times
d𝑥 by d𝑡 d𝑡. Now, in fact, if we assume that the
curve is traced out from left to right, we can lose the absolute value symbol. And then, we notice that one over
d𝑥 by d𝑡 times d𝑥 by d𝑡 is simply one. And so, we’re left with the given
formula for the arc length of a curve between the limits of 𝑡 equals 𝛼 and 𝑡
equals 𝛽.

Now, it is important to note that
the Cartesian form of the arc length formula is only valid for 𝑦 equals 𝑓 of 𝑥
when 𝑓 prime is continuous on the closed interval 𝑎 to 𝑏. So using this to make our
parametric version has a similar consequence on what parametric equations we can use
this formula on. In fact, if we start with the
parametric equations 𝑥 equals 𝑓 of 𝑡 and 𝑦 equals 𝑔 of 𝑡, then 𝑓 prime and 𝑔
prime must be continuous on the closed interval from 𝛼 to 𝛽. It’s also worth remembering that
with parametric equations the curve can loop over itself. This might lead to answers, which
are longer than the actual arc length, in which case we would need to determine a
range for 𝑡, for which the arc is traced out exactly once.

So let’s have a look at how we can
apply this formula.

Express the length of the curve
with parametric equations 𝑥 equals 𝑡 squared minus 𝑡 and 𝑦 equals 𝑡 to the
fourth power, where 𝑡 is greater than or equal to one and less than or equal to
four, as an integral.

We recall that the formula for the
arc length 𝐿 of a curve defined parametrically between the limits of 𝑡 equals 𝛼
and 𝑡 equals 𝛽 is the definite integral between 𝛼 and 𝛽 of the square root of
d𝑥 by d𝑡 squared plus d𝑦 by d𝑡 squared with respect to 𝑡. Now, our curve is defined
parametrically by 𝑥 equals 𝑡 squared minus 𝑡 and 𝑦 equals 𝑡 to the fourth
power. And we want to find this arc length
between the limits of 𝑡 equals one and 𝑡 equals four. So we let 𝛼 be equal to one, 𝛽 be
equal to four. And we see we’re going to need to
differentiate 𝑥 and 𝑦 with respect to 𝑡.

Now, to differentiate a polynomial
term, we simply multiply the entire term by the exponent and then reduce the
exponent by one. So the derivative of 𝑡 squared is
two 𝑡. And when we differentiate negative
𝑡, we get negative one. d𝑥 by d𝑡 is, therefore, two 𝑡 minus one. And this satisfies the criteria
that the derivative of this function is continuous. d𝑦 by d𝑡 is the first
derivative of 𝑡 to the fourth power. That’s four 𝑡 cubed, which is also
a continuous function. Now, we noticed that we’re going to
have to square these in our formula for the arc length. So let’s work out d𝑥 by d𝑡
squared and d𝑦 by d𝑡 squared before substituting into the formula.

By distributing the parentheses, we
find that two 𝑡 minus one all squared is four 𝑡 squared minus four 𝑡 plus
one. And four 𝑡 cubed all squared is
16𝑡 to the sixth power. Our final step is to substitute
into the formula for the arc length. And we find that 𝐿 is equal to the
definite integral between one and four of the square root of four 𝑡 squared minus
four 𝑡 plus one plus 16𝑡 to the sixth power d𝑡. We might choose to rewrite the
expression inside our root in descending powers of 𝑡. And when we do, we find that the
arc length of the curve defined by our parametric equations for 𝑡 is greater than
or equal to one and less than or equal to four is the integral shown.

In our next example, we’re going to
look at how to actually evaluate one of these expressions.

Find the length of the curve with
parametric equations 𝑥 equals three cos 𝑡 minus cos three 𝑡 and 𝑦 equals three
sin 𝑡 minus sin three 𝑡, where 𝑡 is greater than or equal to zero and less than
or equal to 𝜋.

We recall that the formula we used
to find the arc length of a curve defined parametrically for values of 𝑡 from 𝛼 to
𝛽 is the definite integral between 𝛼 and 𝛽 of the square root of d𝑥 by d𝑡
squared plus d𝑦 by d𝑡 squared with respect to 𝑡. In this case, 𝑥 is equal to three
cos 𝑡 minus cos three 𝑡 and 𝑦 is equal to three sin 𝑡 minus sin three 𝑡. And we’re interested in the length
of the curve between 𝑡 is greater than or equal to zero and less than or equal to
𝜋.

So we’ll let 𝛼 be equal to zero
and 𝛽 be equal to 𝜋. We’re also going to need to work
out d𝑥 by d𝑡 and d𝑦 by d𝑡. And so, since we’re working with
trigonometric expressions, we recall the derivative of cos of 𝑎𝑡 and sin of
𝑎𝑡. They are negative 𝑎 sin of 𝑎𝑡
and 𝑎 cos 𝑎𝑡, respectively, for real constant values of 𝑎. This means d𝑥 by d𝑡 is negative
three sin 𝑡 minus negative three sin three 𝑡. And of course, that becomes plus
three sin three 𝑡. Similarly, d𝑦 by d𝑡 is three cos
𝑡 minus three cos three 𝑡.

Before we substitute into the
formula, we’re actually going to square these and find their sum. Negative three sin 𝑡 plus three
sin three 𝑡 all squared is nine sin squared 𝑡 minus 18 sin 𝑡 sin three 𝑡 plus
nine sin square three 𝑡. Then, three cos 𝑡 minus three cos
three 𝑡 all squared is nine cos squared 𝑡 minus 18 cos 𝑡 cos three 𝑡 plus nine
cos squared three 𝑡. At this stage, we recall the
trigonometric identity sin squared 𝑡 plus cos squared 𝑡 equals one. And we see that we have nine sin
squared 𝑡 plus nine cos squared 𝑡. Well, that must be equal to
nine. Similarly, we have nine sin squared
three 𝑡 plus nine cos squared three 𝑡, which is also equal to nine. And we also have negative 18 times
sin 𝑡 sin three 𝑡 plus cos 𝑡 cos three 𝑡. All I’ve done here is factored the
negative 18 out.

Next, we’re going to use the
trigonometric identity cos of 𝐴 minus 𝐵 is equal to cos 𝐴 cos 𝐵 plus sin 𝐴 sin
𝐵. And this means that sin 𝑡 sin
three 𝑡 plus cos 𝑡 cos three 𝑡 must be equal to cos of three 𝑡 minus 𝑡, which
is, of course, simply cos of two 𝑡. So this becomes 18 minus 18 cos of
two 𝑡. And so, we find that the arc length
is equal to the definite integral between zero and 𝜋 of the square root of 18 minus
18 cos of two 𝑡 d𝑡.

Let’s clear some space and evaluate
this integral. Now, in fact, the integral of the
square root of 18 minus 18 cos of two 𝑡 still isn’t particularly nice to
calculate. And so, we go back to the fact that
cos of two 𝑡 is equal to two cos squared 𝑡 minus one. We replace cos of two 𝑡 with this
expression and then distribute the parentheses. And our integrand is now equal to
the square root of 36 minus 36 cos squared 𝑡. We take out the common factor of 36
and then rearrange the identity sin squared 𝑡 plus cos squared 𝑡 equals one. So that one minus cos squared 𝑡 is
equal to sin squared 𝑡. So our integrand is six times the
square root of sin squared 𝑡, which is, of course, simply six sin 𝑡.

When we integrate six sin 𝑡, we
get negative six cos 𝑡. So the arc length is equal to
negative six cos 𝑡 evaluated between those limits. That’s negative six cos of 𝜋 minus
negative six cos of zero, which is equal to 12. And so, we found that the arc
length of the curve that we’re interested in is 12 units. As you might expect, not only does
this process work for curves defined by trigonometric equations, but also those
defined by exponential and logarithmic ones.

Find the length of the curve with
parametric equations 𝑥 equals 𝑒 to the power of 𝑡 minus 𝑡 and 𝑦 equals four 𝑒
to the power of 𝑡 over two, where 𝑡 is greater than or equal to zero and less than
or equal to two.

We know that the formula we used to
find the arc length of curves defined parametrically from values of 𝑡 greater than
or equal to 𝛼 and less than or equal to 𝛽 is the definite integral between 𝛼 and
𝛽 of the square root of d𝑥 by d𝑡 squared plus d𝑦 by d𝑡 squared with respect to
𝑡. Now, in this case, we’re interested
in the length of the curve, where 𝑡 is greater than or equal to zero and less than
or equal to 𝑡. So we’ll let 𝛼 to be equal to zero
and 𝛽 be equal to two. Then, our parametric equations are
𝑥 equals 𝑒 to the power of 𝑡 minus 𝑡 and 𝑦 equals four 𝑒 to the power of 𝑡
over two.

It’s quite clear that we’re going
to need to work out d𝑥 by d𝑡 and d𝑦 by d𝑡. And so, we firstly recall that the
derivative of 𝑒 to the power of 𝑡 is 𝑒 to the power of 𝑡. The derivative of negative 𝑡 is
one. So d𝑥 by d𝑡 is 𝑒 to the power of
𝑡 minus one. Now, we’re going to use the chain
rule to differentiate 𝑦 with respect to 𝑡. We let 𝑢 be equal to 𝑡 over
two. So that d𝑢 by d𝑡 is equal to
one-half. Then, d𝑦 by d𝑡 is d𝑦 by d𝑢
times d𝑢 by d𝑡. Now, 𝑦 is equal to four 𝑒 to the
power of 𝑢. So d𝑦 by d𝑡 is four 𝑒 to the
power of 𝑢 times a half, which is two 𝑒 to the power of 𝑢. But of course, we want d𝑦 by d𝑡
in terms of 𝑡. So we replace 𝑢 with 𝑡 over
two. And we find that d𝑦 by d𝑡 equals
two 𝑒 to the power of 𝑡 over two.

Now, in fact, for our arc length
formula, we need d𝑥 by d𝑡 squared and d𝑦 by d𝑡 squared. So we’re going to square each of
our expressions. When we do, we find that 𝑒 to the
power of 𝑡 minus one squared is 𝑒 to the power of two 𝑡 minus two 𝑒 to the power
of 𝑡 plus one. And d𝑦 by d𝑡 squared is simply
four 𝑒 to the power of 𝑡. Now, we substitute everything we
know into our formula for the arc length. And we get the definite integral
between zero and two of the square root of 𝑒 to the power of two 𝑡 minus two 𝑒 to
the power of 𝑡 plus one plus four 𝑒 to the power of 𝑡 with respect to 𝑡.

We notice that negative two 𝑒 to
the power of 𝑡 plus four 𝑒 to the power of 𝑡 is two 𝑒 to the power of 𝑡. And that’s great because we see we
can factor 𝑒 to the power of two 𝑡 plus two 𝑒 to the power of 𝑡 plus one a
little like we would with a quadratic. We get 𝑒 to the power of 𝑡 plus
one times 𝑒 to the power of 𝑡 plus one or 𝑒 to the power of 𝑡 plus one
squared. And of course, the square root of
𝑒 to the power of 𝑡 plus one squared is just 𝑒 to the power of 𝑡 plus one. When we integrate 𝑒 to the power
of 𝑡, we get 𝑒 to the power of 𝑡. And the integral of one is 𝑡.

This means the arc length is 𝑒
squared plus two minus 𝑒 to the power of of zero plus zero. And, of course, 𝑒 to the power of
zero is one. So this simplifies to 𝑒 to the
power of two plus one. And the length of the curve with
our parametric equations for values of 𝑡 from zero to two is 𝑒 to the power of two
plus one units.

In this video, we’ve learned that
for a curve defined parametrically by 𝑥 equals 𝑓 of 𝑡 and 𝑦 equals 𝑔 of 𝑡, the
arc length of the curve for values of 𝑡 greater than or equal to 𝛼 and less than
or equal to 𝛽 is given by the definite integral between 𝛼 and 𝛽 of the square
root of d𝑥 by d𝑡 squared plus d𝑦 by d𝑡 squared with respect to 𝑡. In this case, 𝑓 prime and 𝑔 prime
must be continuous functions on the closed interval 𝛼 to 𝛽. We saw that it’s also worth
remembering that with parametric equations, the curve can loop over itself. And in which case, we do need to
determine a range for 𝑡, for which the arc is traced out exactly once.