Parametric descriptions of curves | Year 12

01 / LEARN

Understand the method.

THE CORE RULE

\begin{aligned} r (t) & = (\begin{array}{cccccccccc} x (t) \\ y (t) \end{array}) \\ a & \leq t \leq b \end{aligned}

One parameter value gives one complete point. Keep its interval and increasing-parameter direction with the path.

One parameter chooses both coordinates

A parametric description gives the coordinates as functions of the same parameter. Substitute one value of $t$ into both functions to find a point; do not mix coordinates from different values.

The parameter may be time, but it can also be a dimensionless way of moving along a curve. Plot a few values in increasing parameter order to establish the direction. A vertical line is valid too: $x = 2$ , $y = t$ fixes one coordinate while the other varies, even though it is not a function $y = f (x)$ .

\begin{aligned} x & = t + 1 \\ y & = t^{2} \\ - 1 & \leq t \leq 2 \end{aligned}

The restricted parabola x=t+1, y=t squared, from t=-1 to t=2. Marked points (0,1), (1,0) and (3,4), with arrows showing increasing t from left to right.

Eliminate the parameter, then keep its restrictions

Eliminate the parameter by expressing it using one coordinate and substituting into the other equation. This finds a Cartesian relationship between the coordinates.

A Cartesian equation can include more points than the parameter interval permits. Keep the traced portion and check any branches introduced by squaring or taking square roots.

\begin{array}{c} t = x - 1 \\ ⟹ y = {(x - 1)}^{2} \end{array}

Keep the information the Cartesian equation loses

Use the parameter endpoints to locate the start and finish. Intermediate values reveal turns and repeated visits that the Cartesian equation alone cannot show.

To parameterise a directed segment, start at $A$ and add a fraction of its displacement towards $B$ . Restrict that fraction to $0 \leq u \leq 1$ .

r (u) = A + u (B - A)

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Describe the part of a parabola that is traced

For $x = t + 1$ , $y = t^{2}$ and $- 1 \leq t \leq 2$ , find the Cartesian equation, traced portion and direction.

Worked solution

1
Solve the first equation for the parameter.
$t = x - 1$
This expression can replace every occurrence of $t$ in the other coordinate.
2
Substitute into the second equation.
$y = {(x - 1)}^{2}$
Every point generated by the parameter satisfies this Cartesian equation.
3
Evaluate endpoints and one intermediate point.
$\begin{aligned} t = - 1 & : (0, 1) \\ t = 0 & : (1, 0) \\ t = 2 & : (3, 4) \end{aligned}$
The coordinate $x = t + 1$ increases throughout the interval, so $0 \leq x \leq 3$ and the curve is traced left to right.

Answer

$y = {(x - 1)}^{2}$ , restricted to $0 \leq x \leq 3$ , from $(0, 1)$ through $(1, 0)$ to $(3, 4)$ .

EXAMPLE 02

Build a directed segment

Parameterise the segment from $A = (2, - 1)$ to $B = (5, 3)$ , starting at $A$ when $u = 0$ and finishing at $B$ when $u = 1$ .

Worked solution

1
Find the displacement from the start to the finish.
$\begin{aligned} B - A & = (\begin{array}{cccccccccc} 5 - 2 \\ 3 - (- 1) \end{array}) \\ = (\begin{array}{cccccccccc} 3 \\ 4 \end{array}) \end{aligned}$
Subtract the starting coordinates from the finishing coordinates in the same order.
2
Add a parameter multiple to the starting point.
$r (u) = (\begin{array}{cccccccccc} 2 \\ - 1 \end{array}) + u (\begin{array}{cccccccccc} 3 \\ 4 \end{array})$
As $u$ increases from $0$ to $1$ , the multiple of the displacement increases from none to all of it.
3
Check both endpoint values.
$\begin{aligned} r (0) & = (2, - 1) \\ r (1) & = (5, 3) \end{aligned}$
The interval is essential: unrestricted real values would describe the whole line.

Answer

$x = 2 + 3 u$ , $y = - 1 + 4 u$ , with $0 \leq u \leq 1$ .

EXAMPLE 03

Read an outward and return journey

A small robot has coordinates $x = {(t - 1)}^{2}$ , $y = {(t - 1)}^{2}$ metres for $0 \leq t \leq 2$ seconds. Describe the path and explain why $y = x$ does not tell the whole story.

Worked solution

1
Compare the coordinate formulas.
$y = x, 0 \leq x \leq 1$
The squared quantity lies between $0$ and $1$ , so the geometric path is a segment of the diagonal line.
2
Evaluate the start, middle and finish.
$\begin{aligned} r (0) & = (1, 1) \\ r (1) & = (0, 0) \\ r (2) & = (1, 1) \end{aligned}$
The squared quantity decreases until $t = 1$ and then increases. The order shows the reversal.
3
Compare the parametric and Cartesian information.
$r (0) = r (2)$
The robot returns to its initial position after tracing the same segment in opposite directions. The equation $y = x$ records neither the reversal nor the timing.

Answer

It moves from $(1, 1)$ to $(0, 0)$ during the first second, then returns to $(1, 1)$ during the second. The segment is traced twice.

03 / YOUR TURN

Try it yourself.

Calculate, explain, then apply.

CHECK 01

Calculate a Cartesian equation

Eliminate $t$ from $x = 3 t + 1$ and $y = 2 t - 4$ , where $t$ is real.

Show answer & reasoning

$y = \frac{2}{3} (x - 1) - 4$ .

Solve $t = \frac{x - 1}{3}$ , then substitute into $y$ . Both coordinates must use the same value of $t$ .

CHECK 02

Interpret the traced portion

For $x = t^{2}$ , $y = t$ and $- 1 \leq t \leq 1$ , is $y = \sqrt{x}$ the whole traced curve? Give the correct relationship and direction.

Show answer & reasoning

No. $x = y^{2}$ with $- 1 \leq y \leq 1$ , traced from $(1, - 1)$ through $(0, 0)$ to $(1, 1)$ .

Negative parameter values give negative $y$ . Selecting only the positive square root discards half the path; the parameter order also identifies its orientation.

CHECK 03

Apply a timed segment model

A camera moves at constant coordinate rates from $(- 1, 2)$ metres at $t = 0$ to $(5, - 1)$ metres at $t = 3$ seconds. Give its parametric equations and time interval.

Show answer & reasoning

$x = - 1 + 2 t$ , $y = 2 - t$ , for $0 \leq t \leq 3$ .

The displacement is $(6, - 3)$ metres over $3$ seconds, so the coordinate rates are $(2, - 1)$ m/s. Substituting $t = 0$ and $t = 3$ verifies both endpoints.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find the point at $t = 2$ for $x = t - 1$ , $y = t^{2} + 1$ .
2 marks
Eliminate $t$ from $x = 3 t + 1$ , $y = 2 t - 4$ .
2 marks
For $x = t^{2}$ , $y = t$ , $t \geq 0$ , give the Cartesian equation and restriction.
2 marks
Parameterise the segment from $(0, 4)$ to $(6, 1)$ .
2 marks
For $x = 4 - t$ , $y = t^{2}$ , $0 \leq t \leq 2$ , find the start and end points.
2 marks
Explain the repeated trace for $x = t^{2}$ , $y = t^{2}$ , $- 1 \leq t \leq 1$ .
2 marks

Read the worked solutions

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

Find the point at $t = 2$ for $x = t - 1$ , $y = t^{2} + 1$ .
2 marks
Substitute into both coordinates.
$x = 1$ , $y = 5$ .
$(1, 5)$
Eliminate $t$ from $x = 3 t + 1$ , $y = 2 t - 4$ .
2 marks
$t = \frac{x - 1}{3}$ .
Substitute into $y$ .
$y = \frac{2}{3} (x - 1) - 4$
For $x = t^{2}$ , $y = t$ , $t \geq 0$ , give the Cartesian equation and restriction.
2 marks
Use $t = y$ .
The parameter restriction selects the upper branch.
$x = y^{2}$ , $y \geq 0$ .
Parameterise the segment from $(0, 4)$ to $(6, 1)$ .
2 marks
The displacement is $(6, - 3)$ .
Add its $t$ multiple to the starting point.
$x = 6 t$ , $y = 4 - 3 t$ , $0 \leq t \leq 1$ .
For $x = 4 - t$ , $y = t^{2}$ , $0 \leq t \leq 2$ , find the start and end points.
2 marks
Evaluate the two endpoints.
$\frac{d x}{d t} = - 1$ .
From $(4, 0)$ to $(2, 4)$ , moving left.
Explain the repeated trace for $x = t^{2}$ , $y = t^{2}$ , $- 1 \leq t \leq 1$ .
2 marks
Opposite $t$ values have equal squares.
The Cartesian equation $y = x$ does not show the reversal.
It travels from $(1, 1)$ to $(0, 0)$ , then back to $(1, 1)$ .

Checking saved progress…

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.

One parameter chooses both coordinates

Eliminate the parameter, then keep its restrictions

Keep the information the Cartesian equation loses

See the method in action.

Describe the part of a parabola that is traced

Solve the first equation for the parameter.

Substitute into the second equation.

Evaluate endpoints and one intermediate point.

Build a directed segment

Find the displacement from the start to the finish.

Add a parameter multiple to the starting point.

Check both endpoint values.

Read an outward and return journey

Compare the coordinate formulas.

Evaluate the start, middle and finish.

Compare the parametric and Cartesian information.

Try it yourself.

Calculate a Cartesian equation

Interpret the traced portion

Apply a timed segment model

Keep practising