Arc length in two and three dimensions | Year 12

01 / LEARN

Understand the method.

THE CORE RULE

\begin{aligned} L & = \int_{a}^{b} ∣ r^{'} (t) ∣ d t \\ = \int_{a}^{b} \sqrt{\begin{array}{c} {(x^{'})}^{2} + {(y^{'})}^{2} \\ + {(z^{'})}^{2} \end{array}} d t \end{aligned}

For $a < b$ , integrate the nonnegative speed. Omit the third squared component for a two-dimensional path.

Take the magnitude before integrating

Over a short time interval $Δ t > 0$ , the displacement is approximately $r^{'} (t) Δ t$ . Its length is approximately $∣ r^{'} (t) ∣ Δ t$ . Adding these small lengths gives the arc-length integral.

The order matters: first differentiate, then take the magnitude, then integrate. With position in metres and time in seconds, speed is in m/s and the resulting length is in metres.

\begin{aligned} L & = \int_{a}^{b} s (t) d t \\ s (t) & = ∣ r^{'} (t) ∣ \end{aligned}

Simplify speed and check its sign

Square the derivative components, add them and simplify before integrating. A trigonometric identity may make speed constant; a polynomial expression may allow a simple substitution.

Remember $\sqrt{t^{2}} = ∣ t ∣$ . Remove the absolute value only after checking the parameter interval. If an antiderivative is outside the course techniques, set up the correct definite integral and use appropriate numerical technology.

\begin{array}{c} \sqrt{4 t^{2} (1 + t^{2})} \\ = 2 ∣ t ∣ \sqrt{1 + t^{2}} \end{array}

Count every part of the journey

Arc length counts the whole route, including repeated or retraced sections. Displacement is the vector from start to finish; its magnitude is direct endpoint separation.

For a smooth path, length is at least the endpoint separation. Returning to the starting point can give zero displacement while the distance travelled remains positive.

L \geq ∣ r (b) - r (a) ∣

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Simplify a curved path’s speed

Find the length of $r (t) = (t^{2}, \frac{2}{3} t^{3})$ for $0 \leq t \leq 1$ , with coordinates in metres.

Worked solution

1
Differentiate and take the magnitude.
$\begin{aligned} r^{'} (t) & = (2 t, 2 t^{2}) \\ ∣ r^{'} (t) ∣ & = \sqrt{4 t^{2} + 4 t^{4}} \\ = 2 ∣ t ∣ \sqrt{1 + t^{2}} \end{aligned}$
Factor $4 t^{2}$ out of the square root, retaining the absolute value at this stage.
2
Use the interval to remove the absolute value.
$L = \int_{0}^{1} 2 t \sqrt{1 + t^{2}} d t$
Here $t \geq 0$ , so $∣ t ∣ = t$ . The magnitude is nonnegative throughout the interval.
3
Substitute $u = 1 + t^{2}$ and change the limits.
$\begin{aligned} d u & = 2 t d t \\ t = 0 & \Rightarrow u = 1 \\ t = 1 & \Rightarrow u = 2 \\ L & = \int_{1}^{2} \sqrt{u} d u \end{aligned}$
The factor $2 t d t$ is exactly the derivative of the expression inside the root.
4
Evaluate the definite integral.
$\begin{aligned} L & = {[\frac{2}{3} u^{3 / 2}]}_{1}^{2} \\ = \frac{2}{3} (2 \sqrt{2} - 1) \end{aligned}$
Subtract the lower-bound value after evaluating the antiderivative at both bounds.

Answer

The curve length is $\frac{2}{3} (2 \sqrt{2} - 1)$ metres.

EXAMPLE 02

Separate two turns from zero displacement

A particle follows $r (t) = (cos t, sin t)$ metres for $0 \leq t \leq 4 π$ seconds. Find its distance travelled, displacement and endpoint separation.

Worked solution

1
Find speed, including its magnitude.
$\begin{aligned} v (t) & = (\begin{array}{cccccccccc} - sin t \\ cos t \end{array}) \\ s (t) & = \sqrt{{sin}^{2} t + {cos}^{2} t} \\ = 1 \end{aligned}$
The particle moves at $1$ m/s around a radius- $1$ circle.
2
Integrate speed over the full interval.
$\begin{aligned} L & = \int_{0}^{4 π} 1 d t \\ = {[t]}_{0}^{4 π} \\ = 4 π \end{aligned}$ m.
The interval covers two complete turns. Both turns add to the distance.
3
Subtract the endpoint positions.
$\begin{aligned} r (4 π) & = (1, 0) \\ = r (0) \\ Δ r & = (\begin{array}{cccccccccc} 1 \\ 0 \end{array}) - (\begin{array}{cccccccccc} 1 \\ 0 \end{array}) \\ = (0, 0) \\ ∣ Δ r ∣ & = 0 \end{aligned}$
The particle returns to its start. Neither displacement nor its magnitude records the intervening journey.

Answer

Distance $4 π$ m; displacement $(0, 0)$ m; endpoint separation $0$ m.

EXAMPLE 03

Measure a cable around a cylinder

A cable follows $r (t) = (2 cos t, 2 sin t, 3 t)$ metres for $0 \leq t \leq π$ . Find the cable length. Compare it with direct endpoint separation.

A half-turn helical cable on a radius-2 cylinder rises from (2,0,0) to (-2,0,3 pi). The curved route is distinguished from its dashed direct endpoint chord.

Worked solution

1
Differentiate the three coordinates.
$r^{'} (t) = (\begin{array}{cccccccccc} - 2 sin t \\ 2 cos t \\ 3 \end{array})$
The cable makes a half-turn around a radius- $2$ cylinder while rising by $3 π$ m.
2
Simplify the magnitude and integrate.
$\begin{array}{c} ∣ r^{'} (t) ∣^{2} \\ \begin{aligned} = 4 {sin}^{2} t + 4 {cos}^{2} t \\ + 9 \\ = 4 + 9 \end{aligned} \\ ∣ r^{'} (t) ∣ = \sqrt{13} \\ \begin{aligned} L & = \int_{0}^{π} \sqrt{13} d t \\ = π \sqrt{13} \end{aligned} \end{array}$
The magnitude is constant. Integrate it over the parameter interval of length $π$ .
3
Calculate the endpoint displacement.
$\begin{aligned} r (0) & = (2, 0, 0) \\ r (π) & = (- 2, 0, 3 π) \\ Δ r & = (- 4, 0, 3 π) \\ ∣ Δ r ∣ & = \sqrt{16 + 9^{π 2}} \end{aligned}$
The dashed straight chord joins the same endpoints but cuts across the curved route.

Answer

Cable length $π \sqrt{13}$ m, greater than the endpoint separation $\sqrt{16 + 9^{π 2}}$ m. The curved route requires the arc length.

03 / YOUR TURN

Try it yourself.

Calculate, explain, then apply.

CHECK 01

Integrate a constant speed

Find the length of $r (t) = (t, 2 t, 2 t)$ for $0 \leq t \leq 3$ , with coordinates in metres.

Show answer & reasoning

$9$ m.

The derivative is $(1, 2, 2)$ , with magnitude $\sqrt{1 + 4 + 4} = 3$ . Therefore $L = \int_{0}^{3} 3 d t = 9$ m. This straight path also has endpoint separation $9$ m.

CHECK 02

Count a retraced section twice

For $r (t) = (t^{2}, 0)$ metres on $- 1 \leq t \leq 1$ , find the distance travelled and displacement. Explain why the speed is not $2 t$ throughout.

Show answer & reasoning

Distance $2$ m; displacement $(0, 0)$ m. Speed is $2 ∣ t ∣$ , not $2 t$ .

Velocity is $(2 t, 0)$ , so speed is $\sqrt{4 t^{2}} = 2 ∣ t ∣$ . The particle goes from $(1, 0)$ to $(0, 0)$ and back to $(1, 0)$ . Thus $L = \int_{- 1}^{0} (- 2 t) d t + \int_{0}^{1} 2 t d t = 1 + 1 = 2$ m.

CHECK 03

Choose enough cable for a rising route

A guide cable must follow $r (t) = (3 cos t, 3 sin t, 4 t)$ metres for $0 \leq t \leq \frac{π}{2}$ . Ignoring fastening allowance, is an $8$ m reel long enough?

Show answer & reasoning

Yes. The required length is $\frac{5 π}{2} \approx 7.854$ m, leaving about $0.146$ m.

The derivative is $(- 3 sin t, 3 cos t, 4)$ with magnitude $\sqrt{9 + 16} = 5$ . Hence $L = \int_{0}^{\frac{π}{2}} 5 d t = 5 π / 2 < 8$ . Compare the reel with route length, not the shorter direct distance between endpoints.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find length of $(5 t, 12 t)$ on $0 \leq t \leq 1$ .
2 marks
Find length of $(3 cos t, 3 sin t)$ on $0 \leq t \leq \frac{π}{2}$ .
2 marks
Find length of $(cos t, sin t, 2 t)$ on $0 \leq t \leq 2 π$ .
2 marks
Set up length for $(t, t^{2})$ on $0 \leq t \leq 1$ without evaluating.
2 marks
Find length of $(t^{2}, \frac{2}{3} t^{3})$ on $0 \leq t \leq 1$ .
2 marks
For $(cos t, sin t)$ on $0 \leq t \leq 4 π$ , find distance and displacement.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find length of $(5 t, 12 t)$ on $0 \leq t \leq 1$ .
2 marks
Speed is $\sqrt{25 + 144}$ .
Integrate $13$ over an interval of length $1$ .
$13$
Find length of $(3 cos t, 3 sin t)$ on $0 \leq t \leq \frac{π}{2}$ .
2 marks
The speed is $3$ .
Multiply by the parameter interval length.
$\frac{3 π}{2}$
Find length of $(cos t, sin t, 2 t)$ on $0 \leq t \leq 2 π$ .
2 marks
Speed squared is $5$ .
Integrate constant speed $\sqrt{5}$ .
$2 π \sqrt{5}$
Set up length for $(t, t^{2})$ on $0 \leq t \leq 1$ without evaluating.
2 marks
The derivative is $(1, 2 t)$ .
Take its magnitude before integration.
$\int_{0}^{1} \sqrt{1 + 4 t^{2}} d t$
Find length of $(t^{2}, \frac{2}{3} t^{3})$ on $0 \leq t \leq 1$ .
2 marks
Speed is $2 t \sqrt{1 + t^{2}}$ for nonnegative $t$ .
Use $u = 1 + t^{2}$ , so the integral becomes $\int_{1}^{2} \sqrt{u} d u$ .
$\frac{2}{3} (2 \sqrt{2} - 1)$
For $(cos t, sin t)$ on $0 \leq t \leq 4 π$ , find distance and displacement.
2 marks
Speed $1$ is integrated over interval length $4 π$ .
The path makes two full turns back to its start.
Distance $4 π$ ; displacement $(0, 0)$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics school Vector calculus: polynomial and trigonometric parameterisations, related coordinate rates, Bézier paths, componentwise differentiation and integration, motion, tangent directions, circular motion and arc length. Reasoning and proofs use these school topics; vector-field calculus is outside this chapter.

Official curriculum source ↗

In this lesson

Understand the method.

Take the magnitude before integrating

Simplify speed and check its sign

Count every part of the journey

See the method in action.

Simplify a curved path’s speed

Differentiate and take the magnitude.

Use the interval to remove the absolute value.

Substitute u=1+t2 and change the limits.

Evaluate the definite integral.

Separate two turns from zero displacement

Find speed, including its magnitude.

Integrate speed over the full interval.

Subtract the endpoint positions.

Measure a cable around a cylinder

Differentiate the three coordinates.

Simplify the magnitude and integrate.

Calculate the endpoint displacement.

Try it yourself.

Integrate a constant speed

Count a retraced section twice

Choose enough cable for a rising route

Keep practising

Substitute $u = 1 + t^{2}$ and change the limits.