Parametric descriptions of curves | Year 12

01 / LEARN

The idea, step by step

One parameter controls both coordinates

A parametric curve is specified by $x = f (t)$ and $y = g (t)$ for the same value of t. To find a point, substitute one t into both formulas. The parameter may represent time, or simply be a convenient variable.

Eliminating t can produce a Cartesian equation, but that equation alone may lose the direction of travel, repeated visits and restrictions on the traced portion. Keep the parameter interval with the model.

Establish orientation

Calculate points in increasing parameter order, then plot them in that order. A vertical line can be parameterised by fixing x and varying y, even though it is not a single-valued function y of x.

When eliminating t by squaring or taking roots, check for extra branches. Substitute allowed parameter values back into the original equations to establish the correct range.

A common trap

A Cartesian equation completely describes the motion.

Instead: It can omit orientation, speed, repetitions and the parameter range.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

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

$t = x - 1$ , so substitute this into y.
The parameter endpoints give x=0 and 3.
x increases with t, so traversal is left to right.

y = {(x - 1)}^{2}

0 \leq x \leq 3

, from

(0, 1)

through

(1, 0)

(3, 4)

Example 2

Worked example

Parameterise the segment from $(2, - 1)$ to $(5, 3)$ .

The endpoint displacement is (3, 4).
Add t times this displacement to the starting point.
At t=0 and 1 the required endpoints are obtained.

x = 2 + 3 t

y = - 1 + 4 t

0 \leq t \leq 1

Check your understanding

Eliminate t from $x = 2 t$ , $y = t + 3$ .

Show answer

$y = x / 2 + 3$

Use t=x/2 in the y equation.

03 / WORKSHEET

Your turn

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

Worksheet PDF 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
$t = (x - 1) / 3$ .
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.
$d x / d t = - 1$ .
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).

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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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