Video: Arc Length by Integration

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

14:26

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 π‘₯, where 𝑓 is a continuous function over the close interval π‘Ž 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 π‘₯ 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 π‘Ž 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 subject 𝑖 minus one, which is, of course, equal to the difference between 𝑓 π‘₯ subscript 𝑖 minus 𝑓 of π‘₯ subscript 𝑖 minus one. We can say that the length of the line segments are given by the square root of π‘₯ 𝑖 minus π‘₯ 𝑖 minus one squared plus 𝑦 𝑖 minus 𝑦 𝑖 minus one square. And we can redefine that as the square to of Ξ”π‘₯ 𝑖 squared plus Δ𝑦 𝑖 squared.

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

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

This says that if 𝑓 prime is continuous on the close interval π‘Ž to 𝑏. Then the length of the curve given by 𝑦 equals 𝑓 of π‘₯, given that π‘₯ is greater than or equal to zero and less than or equal to 𝑏, is 𝐿 equals the definite integral evaluated between π‘Ž and 𝑏 of the square root of one plus 𝑓 prime of π‘₯ squared with respect to π‘₯. Using Libenie’s notation, we can write this as the integral between π‘Ž and 𝑏 of the square root of one plus d𝑦 by dπ‘₯ squared evaluated with respect to π‘₯. 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 π‘₯ squared between π‘₯ equals zero and π‘₯ 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 π‘Ž and 𝑏 of the square root of one plus d𝑦 by dπ‘₯ squared with respect to π‘₯. We know that 𝑦 is equal to the square root of four minus π‘₯ squared, so we’ll begin by working out d𝑦 by dπ‘₯. If we write 𝑦 as four minus π‘₯ 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 π‘₯. It’s a half times four minus π‘₯ squared to the power of negative half multiplied by the derivative of the function that sits inside the brackets. That’s negative two π‘₯.

Dividing through by two and rewriting d𝑦 by dπ‘₯, we obtain negative π‘₯ over the square root of four minus π‘₯ squared. We let π‘Ž 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 π‘₯ over the square of four minus π‘₯ squared squared with respect to π‘₯. And this simplifies quite nicely. The integrand becomes the square root of one plus π‘₯ squared over four minus π‘₯ squared. We can simplify the expression inside the square root by multiplying both the numerator and denominator of the number one by four minus π‘₯ squared. And when we add that, we get four minus π‘₯ squared plus π‘₯ squared over four minus π‘₯ squared. Negative π‘₯ squared plus π‘₯ squared is zero. And so our integrand becomes the square root of four over four minus π‘₯ 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 π‘₯ 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 π‘₯ 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 π‘₯ is one over the square root of one minus π‘₯ 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 π‘₯ squared over four as π‘₯ over two squared. We choose 𝑒 then for our substitution to be equal to π‘₯ over two. Then d𝑒 by dπ‘₯ is equal to one-half. And we can say this is the equivalent to saying two d𝑒 equals dπ‘₯. We’re also going to need to change the limits on our integral. When π‘₯ is equal to two, 𝑒 is equal to two divided by two, which is one. And when π‘₯ is equal to zero, 𝑒 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 𝑒 squared times two d𝑒. 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 𝑒 squared. It’s simply arc sin of 𝑒. And we’re looking to evaluate two times arc sine of 𝑒 between one and zero. That’s two times arc sin of one minus arc sin of zero, which is simply πœ‹. 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 π‘₯ in terms of 𝑦. For a curve with equation π‘₯ 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 dπ‘₯ 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 π‘₯ between 𝑦 equals zero and 𝑦 equals four.

Remember, for a curve with equation π‘₯ 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 dπ‘₯ by d𝑦 squared d𝑦. Let’s compare this formula to our question.

Our function π‘₯ in terms of 𝑦 is π‘₯ equals 𝑦 squared. So we can see we’re going to need to differentiate π‘₯ with respect to 𝑦. The derivative of π‘₯ 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 dπ‘₯ squared dπ‘₯ if 𝑦 is equal to 𝑓 of π‘₯ over some closed interval π‘Ž to 𝑏. And d𝑠 equals the square root of one plus dπ‘₯ by d𝑦 squared d𝑦 if π‘₯ 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 π‘₯ from a point given.

Find the arc length function for the curve 𝑦 equals four π‘₯ 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 π‘₯ is given by 𝐿 equals the integral d𝑠, where d𝑠 is equal to the square root of one plus d𝑦 by dπ‘₯ squared dπ‘₯. In our case, 𝑦 is equal to four π‘₯ to the power of three over two. So we’ll begin by simply working out what d𝑦 by dπ‘₯ is. Remember to differentiate the function of this sort, we multiply by the power and then reduce that power by one. So d𝑦 by dπ‘₯ is equal to three over two times four π‘₯ to the power of one-half. Four divided by two is two. So we see that d𝑦 by dπ‘₯ is equal to six π‘₯ 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 π‘₯ to the power of one-half squared dπ‘₯.

Well, this simplifies quite nicely; we end up with the square root of one plus 36π‘₯. And this means that 𝐿 is going to be equal to the integral of the square root of one plus 36π‘₯ evaluated with respect to π‘₯. But what are our limits? Well, we’re working in terms of π‘₯. And we know we have a starting point at π‘₯ equals one. So that’s the lower limit. We’re looking for a general function, so the upper limit is simply going to be π‘₯. And we can see that the argument function will be given by the integral evaluated between one and π‘₯ of the square root of one plus 36π‘₯ dπ‘₯. 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 𝑒 be equal to the function inside the square root; that’s one plus 36π‘₯. We can say that the derivative of 𝑒 with respect π‘₯ is, therefore, 36. And we can also equivalently say that one over 36 d𝑒 equals dπ‘₯. 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 π‘₯, so when π‘₯ is equal to π‘₯, 𝑒 is just one plus 36π‘₯. And the lower limit is one, so when π‘₯ is equal to one, 𝑒 is one plus 36 times one which is 37. We now replace one plus 36 with 𝑒. I’ve changed the square root to be the power of one-half. And we replace dπ‘₯ with 36 d𝑒. So all that’s left is to evaluate this integral between the limits given.

The integral of 𝑒 to the power of one-half is 𝑒 to the power of three over two divided by three over two. So this means that the integral of 𝑒 to the power of one-half over 36 is equal to 𝑒 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 𝑒 to the power of three over two. Evaluated between 37 and one plus 36π‘₯. That’s one over 54 times one plus 36π‘₯ 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 π‘₯ 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 π‘₯ equals one and any other point by substituting that value of π‘₯ 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 π‘₯ and π‘₯ 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 dπ‘₯ squared dπ‘₯ if 𝑦 is a function in π‘₯ over some close interval π‘Ž to 𝑏. And it’s equal to the square root of one plus dπ‘₯ d𝑦 squared d𝑦 if π‘₯ 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 π‘₯ or 𝑦, depending on the context.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.