Arc length in two and three dimensions | Year 12

01 / LEARN

The idea, step by step

Integrate small distances

A small parameter increment gives displacement approximately $r^{'} (t) Δ t$ . Its magnitude is approximately $∣ r^{'} (t) ∣ Δ t$ . Adding these lengths and taking a limit gives $L = \int_{a}^{b} ∣ r^{'} (t) ∣ d t$ .

For two coordinates use $\sqrt{x^{' 2} + y^{' 2}}$ ; for three include $z^{' 2}$ . The nonnegative integrand counts distance along the curve, not signed displacement.

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

Simplify the speed first

Differentiate, square components, add and simplify before integrating. Trigonometric identities can make the speed constant. Polynomial models may yield a perfect square or a simple substitution.

When the antiderivative is outside course techniques, give the correct definite integral and use appropriate numerical technology. Retracing still adds distance; do not replace arc length with endpoint separation.

A common trap

Integrating velocity gives curve length.

Instead: Vector integration gives displacement; length requires the magnitude before integration.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Find length of $(3 t, 4 t)$ on $0 \leq t \leq 2$ .

The derivative is (3, 4), speed 5.
Integrate 5 from 0 to 2.
The resulting length is 10.

10 units.

Example 2

Worked example

Find length of $(2 cos t, 2 sin t, 3 t)$ on $0 \leq t \leq π$ .

The derivative is $(- 2 sin t, 2 cos t, 3)$ .
Squared speed is 4+9.
Integrate constant speed $\sqrt{13}$ over interval length pi.

π \sqrt{13}

units.

Check your understanding

Find length of $(t, 2 t, 2 t)$ on $0 \leq t \leq 3$ .

Show answer

The speed is 3 throughout the interval.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF 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 $0 \leq t \leq π / 2$ .
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 $0 \leq t \leq π / 2$ .
2 marks
The speed is 3.
Multiply by the parameter interval length.
$3 π / 2$
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 squared, 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 pi.
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 topic 6.3–6.5: polynomial and trigonometric parameterisations, related coordinate rates, motion, tangent directions and arc length. This is school Vector calculus, not vector-field calculus.

Official curriculum source ↗

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