# Lesson Video: Arc Length of Parametric Curves Mathematics • Higher Education

In this video, we will learn how to use integration to find the arc length of a parametrically defined curve.

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### 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.