Vector-valued position functions | Year 12

01 / LEARN

Understand the method.

THE CORE RULE

Δ r = r (b) - r (a)

Displacement is final position minus initial position. Its magnitude is endpoint separation, which need not equal distance travelled.

Position locates the particle relative to an origin

A position vector lists coordinates measured from a chosen origin. In three dimensions, the coefficients of $i$ , $j$ and $k$ are the horizontal, second and third coordinates.

The magnitude $∣ r (t) ∣$ is the distance from the origin at that instant. A constant third coordinate means the path remains in a plane parallel to the $x y$ plane.

\begin{aligned} r (t) & = x (t) i \\ + y (t) j \\ + z (t) k \\ = (\begin{array}{cccccccccc} x (t) \\ y (t) \\ z (t) \end{array}) \end{aligned}

Subtract positions to find displacement

Displacement points from the initial position to the final position. Subtract earlier from later in every component, keeping the same coordinate order and units.

Its magnitude measures the direct endpoint separation. Changing the origin changes both position vectors by the same amount, so it does not change their difference.

∣ Δ r ∣ = \sqrt{\begin{array}{c} {(Δ x)}^{2} + {(Δ y)}^{2} \\ + {(Δ z)}^{2} \end{array}}

Projection into the xy plane of the path r=(t squared,2t,3) from t=1 to t=2. The curved path links (1,2) to (4,4); a straight chord shows displacement (3,2,0). The third coordinate is constant at z=3 metres.

Separate travelled distance from endpoint separation

Distance travelled follows the actual route. It is at least the endpoint separation, and is greater if the route curves or doubles back. Returning to the start gives zero displacement after a nonzero journey.

Keep the meaning of the parameter explicit. If $t$ is seconds and positions are metres, derivatives have units m/s. A derivative with respect to a dimensionless curve parameter is not automatically a physical velocity.

\begin{array}{c} distance travelled \\ \geq ∣ r (b) - r (a) ∣ \end{array}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Evaluate positions and subtract in three dimensions

For $r (t) = (t^{2}, 2 t, 3)$ metres with $t$ in seconds, find the position at $t = 2$ , the displacement from $t = 1$ to $t = 2$ , and its magnitude.

Worked solution

1
Evaluate the complete position at both times.
$\begin{aligned} r (1) & = (1, 2, 3) m \\ r (2) & = (4, 4, 3) m \end{aligned}$
The constant third coordinate stays equal to $3$ metres; it is not removed when evaluating a position.
2
Subtract the initial position from the final position.
$\begin{aligned} Δ r & = (\begin{array}{cccccccccc} 4 - 1 \\ 4 - 2 \\ 3 - 3 \end{array}) \\ = (\begin{array}{cccccccccc} 3 \\ 2 \\ 0 \end{array}) m \end{aligned}$
The zero third displacement means there is no change in that coordinate, even though its position coordinate is nonzero.
3
Take the magnitude of the displacement.
$\begin{aligned} ∣ Δ r ∣ & = \sqrt{3^{2} + 2^{2} + 0^{2}} \\ = \sqrt{13} m \end{aligned}$
This is the separation of the two positions. By comparison, $∣ r (2) ∣ = \sqrt{41}$ metres is the final distance from the origin.

Answer

Position $(4, 4, 3)$ metres; displacement $(3, 2, 0)$ metres; endpoint separation $\sqrt{13}$ metres.

EXAMPLE 02

Return to the start after a nonzero journey

A particle has position $r (t) = (cos t, sin t)$ metres for $0 \leq t \leq 2 π$ seconds. Find its displacement, endpoint separation and distance travelled.

Worked solution

1
Evaluate the two endpoint positions.
$\begin{aligned} r (0) & = (1, 0) \\ r (2 π) & = (1, 0) \end{aligned}$
The sine and cosine values repeat after a full revolution.
2
Find the displacement and its magnitude.
$\begin{aligned} Δ r & = (\begin{array}{cccccccccc} 1 \\ 0 \end{array}) - (\begin{array}{cccccccccc} 1 \\ 0 \end{array}) \\ = (\begin{array}{cccccccccc} 0 \\ 0 \end{array}) m \end{aligned}$
The endpoints coincide, so both the displacement vector and the endpoint separation are zero.
3
Identify the actual route.
$\begin{aligned} x^{2} + y^{2} & = 1 \\ d & = 2 π (1) \\ = 2 π m \end{aligned}$
As $t$ increases from $0$ to $2 π$ , the particle makes one complete circuit of the unit circle, whose circumference is $2 π$ metres.

Answer

Displacement $(0, 0)$ metres; endpoint separation $0$ metres; distance travelled $2 π$ metres.

EXAMPLE 03

Describe the same drone journey from a new origin

A drone follows $r (t) = (1 + 2 t, 4 - t, 2 + t)$ metres for $0 \leq t \leq 3$ seconds. Find its displacement. A new origin is at old coordinates $(1, 1, 1)$ metres; show that the displacement is unchanged.

Worked solution

1
Evaluate positions and subtract.
$\begin{aligned} r (0) & = (1, 4, 2) \\ r (3) & = (7, 1, 5) \\ Δ r & = (6, - 3, 3) m \end{aligned}$
The displacement measures the change between positions rather than either position on its own.
2
Subtract the new origin coordinates from both positions.
$\begin{aligned} q (0) & = (0, 3, 1) \\ q (3) & = (6, 0, 4) \end{aligned}$
Write $q (t)$ for position relative to the new origin. Its coordinates are the old coordinates minus the position of that origin.
3
Compare the new displacement and separation.
$\begin{aligned} Δ q & = (\begin{array}{cccccccccc} 6 \\ 0 \\ 4 \end{array}) - (\begin{array}{cccccccccc} 0 \\ 3 \\ 1 \end{array}) \\ = (\begin{array}{cccccccccc} 6 \\ - 3 \\ 3 \end{array}) m \\ ∣ Δ r ∣ & = \sqrt{36 + 9 + 9} \\ = 3 \sqrt{6} m \end{aligned}$
The same origin shift cancels in the subtraction. Here the constant coordinate rates give a straight journey, so its travelled distance also equals $3 \sqrt{6}$ metres.

Answer

The displacement is $(6, - 3, 3)$ metres in both coordinate systems. Its magnitude is $3 \sqrt{6}$ metres.

03 / YOUR TURN

Try it yourself.

Calculate, explain, then apply.

CHECK 01

Calculate a displacement

Find the displacement from $t = 0$ to $t = 2$ for $r (t) = (1 + t, t^{2}, 4 - t)$ metres.

Show answer & reasoning

$(2, 4, - 2)$ metres.

The positions are $r (0) = (1, 0, 4)$ and $r (2) = (3, 4, 2)$ . Final minus initial gives $(3 - 1, 4 - 0, 2 - 4)$ .

CHECK 02

Interpret a position magnitude

A particle is at $(3, 4)$ metres. A learner says it has therefore travelled $5$ metres. What does the value $5$ actually measure, and what information is missing?

Show answer & reasoning

It is $5$ metres from the chosen origin. Its travelled distance needs the starting point and route.

$\sqrt{3^{2} + 4^{2}} = 5$ is the magnitude of the current position vector. The particle could have started elsewhere or taken a longer route to this position.

CHECK 03

Apply displacement to a survey route

A survey robot travels along straight segments from $(0, 0)$ to $(3, 4)$ and then to $(6, 0)$ metres. Find its displacement, endpoint separation and total distance travelled.

Show answer & reasoning

Displacement $(6, 0)$ metres; separation $6$ metres; distance travelled $10$ metres.

Subtract the initial position from the final position for displacement. Each segment has length $\sqrt{3^{2} + 4^{2}} = 5$ metres, so the actual route is $5 + 5 = 10$ metres long.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Evaluate $r (3)$ for $(2 t, t^{2} - 1, 5)$ .
2 marks
Find displacement from $0$ to $1$ for $(t, t^{2}, t^{3})$ .
2 marks
Find distance from the origin for $(3 cos t, 3 sin t, 2)$ .
2 marks
Find endpoint separation between $(1, 2, 3)$ and $(4, 6, 3)$ .
2 marks
Identify the Cartesian path of $(t^{2}, 2 t)$ .
2 marks
Give a position function for the line through $(1, 0, 2)$ with direction $(2, - 1, 3)$ .
2 marks

Read the worked solutions

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

Evaluate $r (3)$ for $(2 t, t^{2} - 1, 5)$ .
2 marks
Substitute $3$ into each coordinate.
The third coordinate stays $5$ .
$(6, 8, 5)$
Find displacement from $0$ to $1$ for $(t, t^{2}, t^{3})$ .
2 marks
The initial vector is zero.
The final vector is $(1, 1, 1)$ .
$(1, 1, 1)$
Find distance from the origin for $(3 cos t, 3 sin t, 2)$ .
2 marks
Magnitude squared is $9 ({cos}^{2} t + {sin}^{2} t) + 4$ .
This equals $13$ .
$\sqrt{13}$
Find endpoint separation between $(1, 2, 3)$ and $(4, 6, 3)$ .
2 marks
The displacement is $(3, 4, 0)$ .
Its magnitude is $5$ .
$5$ units.
Identify the Cartesian path of $(t^{2}, 2 t)$ .
2 marks
$t = \frac{y}{2}$ .
Substitute into $x$ .
$y^{2} = 4 x$ .
Give a position function for the line through $(1, 0, 2)$ with direction $(2, - 1, 3)$ .
2 marks
At zero the given point is obtained.
The parameter moves along the required direction.
$r = (1, 0, 2) + t (2, - 1, 3)$ .

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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.

Position locates the particle relative to an origin

Subtract positions to find displacement

Separate travelled distance from endpoint separation

See the method in action.

Evaluate positions and subtract in three dimensions

Evaluate the complete position at both times.

Subtract the initial position from the final position.

Take the magnitude of the displacement.

Return to the start after a nonzero journey

Evaluate the two endpoint positions.

Find the displacement and its magnitude.

Identify the actual route.

Describe the same drone journey from a new origin

Evaluate positions and subtract.

Subtract the new origin coordinates from both positions.

Compare the new displacement and separation.

Try it yourself.

Calculate a displacement

Interpret a position magnitude

Apply displacement to a survey route

Keep practising