Polynomially varying coordinates | Year 12

01 / LEARN

The idea, step by step

Polynomial coordinates can make a curved path

For x=t and $y = 4 t - t^{2}$ , elimination gives $y = 4 x - x^{2}$ . The parameter also tells us which direction the curve is traversed, because x increases with t.

If x is not one-to-one, a single formula y (x) may not describe the whole path. Use parameter values and coordinate ranges rather than choosing a square-root sign without justification.

Coordinate extrema do not imply rest

A y turning point needs dy/dt=0 with an appropriate sign change. The x velocity can remain nonzero, so the particle may keep moving horizontally at its maximum height.

To prove a point is revisited, check both coordinates at two parameter values. Repeated x values alone are not enough.

A common trap

Zero vertical velocity means the particle stops.

Instead: All velocity components must vanish simultaneously.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

For x=t and $y = 4 t - t^{2}$ on $0 \leq t \leq 4$ , find maximum y and decide whether the particle stops there.

$y^{'} = 4 - 2 t = 0$ at t=2 and $y^{''} = - 2$ .
$y (2) = 4$ .
$x^{'} = 1$ , so velocity there is (1, 0).

Maximum 4 at t=2; it does not stop.

Example 2

Worked example

Eliminate t from $x = t^{2}$ , $y = t^{3}$ , $- 1 \leq t \leq 1$ .

$y^{2} = t^{6} = {(t^{2})}^{3}$ .
Negative t gives negative y and positive t gives positive y.
The trace goes from (1, -1) through (0, 0) to (1, 1).

y^{2} = x^{3}

0 \leq x \leq 1

, with both signs of y.

Check your understanding

Does $r = (t, t^{2})$ stop at t=0?

Show answer

No; velocity is (1, 0).

Differentiate both coordinates.

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

Eliminate t from $x = t + 2$ , $y = t^{2} - 1$ .
2 marks
Find maximum y for $x = t$ , $y = 6 t - t^{2}$ , $0 \leq t \leq 6$ .
2 marks
Eliminate t from $x = t^{2} - 1$ , $y = 2 t$ .
2 marks
Find velocity and acceleration for $(t^{2}, 3 t - t^{3})$ .
2 marks
For $(t^{2}, t^{2} - 2 t)$ , find the point where vertical velocity is zero.
2 marks
For $(t^{2}, t^{3} - t)$ , show t=-1 and 1 give the same point.
2 marks

Read the worked solutions

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

Eliminate t from $x = t + 2$ , $y = t^{2} - 1$ .
2 marks
$t = x - 2$ .
Substitute into y.
$y = {(x - 2)}^{2} - 1$
Find maximum y for $x = t$ , $y = 6 t - t^{2}$ , $0 \leq t \leq 6$ .
2 marks
$y^{'} = 6 - 2 t = 0$ gives 3.
The endpoints have y=0; the stationary value is 9.
9 at t=3.
Eliminate t from $x = t^{2} - 1$ , $y = 2 t$ .
2 marks
Use t=y/2.
Substitute into x.
$x = y^{2} / 4 - 1$
Find velocity and acceleration for $(t^{2}, 3 t - t^{3})$ .
2 marks
Differentiate each coordinate once.
Differentiate again.
$(2 t, 3 - 3 t^{2})$ and $(2, - 6 t)$ .
For $(t^{2}, t^{2} - 2 t)$ , find the point where vertical velocity is zero.
2 marks
$d y / d t = 2 t - 2$ .
Substitute t=1 into both coordinates.
At t=1, point (1, -1).
For $(t^{2}, t^{3} - t)$ , show t=-1 and 1 give the same point.
2 marks
The squared coordinates are both 1.
Both y coordinates are 0.
Both give (1, 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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