Arc Length

Section 1.8 Arc Length

Motivating Questions

How can a definite integral be used to measure the length of a curve in 2- or 3-space?

Given a space curve, there are two natural geometric questions one might ask: how long is the curve and how much does it bend? In this section and the optional next section, we answer both questions by developing techniques for measuring the length of a space curve as well as its curvature.

🔗

Preview Activity 1.8.1.

🔗

In earlier investigations, we have used integration to calculate quantities such as area, volume, mass, and work. We are now interested in determining the length of a space curve.

🔗

Consider the smooth curve in 3-space defined by the vector-valued function

r,

where

r (t) = ⟨ x (t), y (t), z (t) ⟩ = ⟨ \cos (t), \sin (t), t ⟩

🔗

for

t

in the interval

[0, 2 π] .

Pictures of the graph of

r

are shown in Figure 1.8.1. We will use the integration process to calculate the length of this curve. In this situation we partition the interval

[0, 2 π]

into

n

subintervals of equal length and let

0 = t_{0} < t_{1} < t_{2} < \dots < t_{n} = b

be the endpoints of the subintervals. We then approximate the length of the curve on each subinterval with some related quantity that we can compute. In this case, we approximate the length of the curve on each subinterval with the length of the segment connecting the endpoints. Figure 1.8.1 illustrates the process in three different instances using increasing values of

n .

Figure 1.8.1. Approximating the length of the curve with $n = 3,$ $n = 6,$ and $n = 9 .$

Write a formula for the length of the line segment that connects the endpoints of the curve on the $i$ th subinterval $[t_{i - 1}, t_{i}] .$ (This length is our approximation of the length of the curve on this interval.)
Use your formula in part (a) to write a sum that adds all of the approximations to the lengths on each subinterval.
What do we need to do with the sum in part (b) in order to obtain the exact value of the length of the graph of $r (t)$ on the interval $[0, 2 π] ?$

🔗

Subsection 1.8.1 Arc Length

🔗

Consider a smooth curve in 3-space that is parametrically described by the vector-valued function

r

defined by

r (t) = ⟨ x (t), y (t), z (t) ⟩ .

Preview Activity 1.8.1 shows that to approximate the length of the curve defined by

r (t)

as the values of

t

run over an interval

[a, b],

we partition the interval

[a, b]

into

n

subintervals of equal length

Δ t,

with

a = t_{0} < t_{1} < \dots < t_{n} = b

as the endpoints of the subintervals. On each subinterval, we approximate the length of the curve by the length of the line segment connecting the endpoints. The points on the curve corresponding to

t = t_{i - 1}

and

t = t_{i}

are

(x (t_{i - 1}), y (t_{i - 1}), z (t_{i - 1}))

and

(x (t_{i}), y (t_{i}), z (t_{i})),

respectively, so the length of the line segment connecting these points is

\sqrt{(x (t_{i}) - x (t_{i - 1}))^{2} + (y (t_{i}) - y (t_{i - 1}))^{2} + (z (t_{i}) - z (t_{i - 1}))^{2}} .

🔗

Now we add all of these approximations together to obtain an approximation to the length

L

of the curve:

L \approx \sum_{i = 1}^{n} \sqrt{(x (t_{i}) - x (t_{i - 1}))^{2} + (y (t_{i}) - y (t_{i - 1}))^{2} + (z (t_{i}) - z (t_{i - 1}))^{2}} .

🔗

We now want to take the limit of this sum as

n

goes to infinity, but in its present form it might be difficult to see how. We first introduce

Δ t

by multiplying by

\frac{Δ t}{Δ t},

and see that

\begin{aligned} L & \approx \sum_{i = 1}^{n} \sqrt{(x (t_{i}) - x (t_{i - 1}))^{2} + (y (t_{i}) - y (t_{i - 1}))^{2} + (z (t_{i}) - z (t_{i - 1}))^{2}} \\ = \sum_{i = 1}^{n} \sqrt{(x (t_{i}) - x (t_{i - 1}))^{2} + (y (t_{i}) - y (t_{i - 1}))^{2} + (z (t_{i}) - z (t_{i - 1}))^{2}} \frac{Δ t}{Δ t} \\ = \sum_{i = 1}^{n} \sqrt{(x (t_{i}) - x (t_{i - 1}))^{2} + (y (t_{i}) - y (t_{i - 1}))^{2} + (z (t_{i}) - z (t_{i - 1}))^{2}} \frac{Δ t}{\sqrt{(Δ t)^{2}}} \end{aligned}

🔗

To get the difference quotients under the radical, we use properties of the square root function to see further that

\begin{aligned} L & \approx \sum_{i = 1}^{n} \sqrt{[(x (t_{i}) - x (t_{i - 1}))^{2} + (y (t_{i}) - y (t_{i - 1}))^{2} + (z (t_{i}) - z (t_{i - 1})^{2}] \frac{1}{(Δ t)^{2}}} Δ t \\ = \sum_{i = 1}^{n} \sqrt{{(\frac{x (t_{i}) - x (t_{i - 1})}{Δ t})}^{2} + {(\frac{y (t_{i}) - y (t_{i - 1})}{Δ t})}^{2} + {(\frac{z (t_{i}) - z (t_{i - 1})}{Δ t})}^{2}} Δ t . \end{aligned}

🔗

Recall that as

n \to \infty

we also have

Δ t \to 0 .

Since

\begin{aligned} x^{'} (t) & = lim_{Δ t \to 0} \frac{x (t_{i}) - x (t_{i - 1})}{Δ t}, \\ y^{'} (t) & = lim_{Δ t \to 0} \frac{y (t_{i}) - y (t_{i - 1})}{Δ t}, and \\ z^{'} (t) & lim_{Δ t \to 0} \frac{z (t_{i}) - z (t_{i - 1})}{Δ t}, \end{aligned}

🔗

we see that

lim_{n \to \infty} \sum_{i = 1}^{n} \sqrt{{(\frac{x (t_{i}) - x (t_{i - 1})}{Δ t})}^{2} + {(\frac{y (t_{i}) - y (t_{i - 1})}{Δ t})}^{2} + {(\frac{z (t_{i}) - z (t_{i - 1})}{Δ t})}^{2}} Δ t

🔗

is equal to

\int_{a}^{b} \sqrt{(x^{'} (t))^{2} + (y^{'} (t))^{2} + (z^{'} (t))^{2}} d t .

🔗

Noting further that

| r^{'} (t) | = \sqrt{(x^{'} (t))^{2} + (y^{'} (t))^{2} + (z^{'} (t))^{2}},

🔗

we can rewrite our arclength formula in a more succinct form as follows.

🔗

The length of a curve.

🔗

r (t)

defines a smooth curve

C

on an interval

[a, b],

then the length

L

C

is given by

\begin{matrix} (1.8.1) & L = \int_{a}^{b} | r^{'} (t) | d t . \end{matrix}

🔗

Note that formula (1.8.1) applies to curves in any dimensional space. Moreover, this formula has a natural interpretation: if

r (t)

records the position of a moving object, then

r^{'} (t)

is the object’s velocity and

| r^{'} (t) |

its speed. Formula (1.8.1) says that we simply integrate the speed of an object traveling over the curve to find the distance traveled by the object, which is the same as the length of the curve, just as in one-variable calculus.

🔗

Activity 1.8.2.

🔗

Here we calculate the arc length of two familiar curves.

Use Equation (1.8.1) to calculate the circumference of a circle of radius $r .$
Find the exact length of the spiral defined by $r (t) = ⟨ \cos (t), \sin (t), t ⟩$ on the interval $[0, 2 π] .$

🔗

We can adapt the arc length formula to curves in 2-space that define

y

as a function of

x

as the following activity shows.

🔗

Activity 1.8.3.

🔗

Let

y = f (x)

define a smooth curve in 2-space. Parameterize this curve and use Equation (1.8.1) to show that the length of the curve defined by

f

on an interval

[a, b]

\int_{a}^{b} \sqrt{1 + [f^{'} (t)]^{2}} d t .

🔗

Subsection 1.8.2 Summary

🔗

The integration process shows that the length $L$ of a smooth curve defined by $r (t)$ on an interval $[a, b]$ is

$L = \int_{a}^{b} | r^{'} (t) | d t .$

🔗

Exercises 1.8.3 Exercises

🔗

The WeBWorK problems are written by many different authors. Some authors use parentheses when writing vectors, e.g.,

(x (t), y (t), z (t))

instead of angle brackets

⟨ x (t), y (t), z (t) ⟩ .

Please keep this in mind when working WeBWorK exercises.

🔗

1.

🔗

Find the length of the curve

x = - (3 + 3 t), y = - (1 + 4 t), z = 4 - 4 t,

🔗

for

4 \leq t \leq 5 .

🔗

length =

🔗

(Think of second way that you could calculate this length, too, and see that you get the same result.)

🔗

2.

🔗

Consider the curve

r = (e^{- 2 t} \cos (6 t), e^{- 2 t} \sin (6 t), e^{- 2 t}) .

🔗

Compute the arclength function

s (t) :

(with initial point

t = 0

🔗

3.

🔗

Find the length of the given curve:

r (t) = (3 t, - 4 \sin t, - 4 \cos t)

🔗

where

- 1 \leq t \leq 1 .

🔗

4.

🔗

Consider the path

r (t) = (12 t, 6 t^{2}, 6 \ln t)

defined for

t > 0 .

🔗

Find the length of the curve between the points

(12, 6, 0)

and

(24, 24, 6 \ln (2)) .

🔗

5.

🔗

Consider the single variable function defined by

y = 4 x^{2} - x^{3} .

Find a parameterization of the form $r (t) = ⟨ x (t), y (t) ⟩$ that traces the curve $y = 4 x^{2} - x^{3}$ on the interval from $x = - 3$ to $x = 3 .$
Write a definite integral that, if evaluated, gives the exact length of the given curve from $x = - 3$ to $x = 3 .$ Why is the integral difficult to evaluate exactly?

Prev Top Next