Polynomially varying coordinates | Year 12

01 / LEARN

Understand the method.

THE CORE RULE

r^{'} (t) = (\begin{array}{cccccccccc} x^{'} (t) \\ y^{'} (t) \end{array})

Differentiate each coordinate. A turning point of one coordinate need not be a stop; rest requires every coordinate rate to be zero.

Keep all branches of a polynomial path

Polynomial coordinates can produce curved paths, reversals and loops. If $x = t$ , direct substitution eliminates the parameter; if $x = t^{2}$ , the sign of $t$ may still matter.

Use the original equations and parameter interval to check branches. Squaring a coordinate relationship can hide its sign and does not by itself establish the direction of travel.

\begin{array}{c} x = t^{2}, y = t^{3} \\ ⟹ y^{2} = x^{3} \end{array}

A coordinate can turn while the particle keeps moving

A maximum or minimum height is a turning point of the vertical coordinate. Find candidates using $y^{'} (t) = 0$ , then confirm the change using a sign test or a second derivative.

Check the other coordinate rate before claiming the particle stops. At a maximum height, it can still move horizontally. For an absolute maximum on a closed time interval, also check its endpoints.

\begin{array}{c} x^{'} (t) \neq 0, y^{'} (t) = 0 \\ ⟹ r^{'} (t) \neq 0 \end{array}

The path x=t, y=4t-t squared for zero to four seconds, with endpoints (0,0) and (4,0), maximum (2,4), and a horizontal rightward velocity vector at the maximum.

Test complete positions and complete rates

A revisited point requires every coordinate to agree at two distinct parameter values. Repeated horizontal coordinates alone are not enough.

At rest means every coordinate rate is zero at the same instant. A repeated point does not imply a stop there, and a stop does not mean the particle stays there afterwards.

\begin{array}{c} r (t_{1}) = r (t_{2}) \\ t_{1} \neq t_{2} \end{array}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find a maximum height without inventing a stop

A particle has $x = t$ and $y = 4 t - t^{2}$ metres for $0 \leq t \leq 4$ seconds. Find its greatest height and decide whether it stops there.

Worked solution

1
Differentiate the vertical coordinate.
$\begin{aligned} y^{'} (t) & = 4 - 2 t \\ 4 - 2 t & = 0 \\ t & = 2 \end{aligned}$
An interior maximum height must have zero vertical rate.
2
Confirm the maximum and compare endpoint heights.
$\begin{aligned} y^{′′} (t) & = - 2 < 0 \\ y (0) & = 0 \\ y (2) & = 4 \\ y (4) & = 0 \end{aligned}$
The curve is concave down, and the interior height exceeds both endpoint heights.
3
Check both coordinate rates at the maximum.
$\begin{aligned} r^{'} (2) & = (\begin{array}{cccccccccc} 1 \\ 4 - 2 (2) \end{array}) \\ = (\begin{array}{cccccccccc} 1 \\ 0 \end{array}) m/s \end{aligned}$
The horizontal rate remains nonzero. The particle continues moving horizontally at its greatest height.

Answer

The greatest height is $4$ metres at $t = 2$ seconds, at $(2, 4)$ . The particle does not stop.

EXAMPLE 02

Retain both branches after elimination

Eliminate $t$ from $x = t^{2}$ , $y = t^{3}$ for $- 1 \leq t \leq 1$ . State the restrictions and the traversal order.

Worked solution

1
Make equal even powers of the parameter.
$\begin{aligned} y^{2} & = t^{6} \\ = {(t^{2})}^{3} \\ = x^{3} \end{aligned}$
This relation does not require choosing a square-root sign.
2
Use the allowed parameter values to establish the branches.
$\begin{aligned} 0 & \leq x \leq 1 \\ y & = \pm x^{3 / 2} \end{aligned}$
Negative $t$ gives negative $y$ and positive $t$ gives positive $y$ . Both branches occur, meeting at the origin.
3
Evaluate the endpoints and the central parameter value.
$\begin{aligned} t = - 1 & : (1, - 1) \\ t = 0 & : (0, 0) \\ t = 1 & : (1, 1) \end{aligned}$
Increasing parameter order gives the lower branch towards the origin, then the upper branch away from it.

Answer

$y^{2} = x^{3}$ with $0 \leq x \leq 1$ , including both signs of $y$ , traced from $(1, - 1)$ through $(0, 0)$ to $(1, 1)$ .

EXAMPLE 03

Check whether a plotting head returns and stops

A plotting head follows $x = {(t - 1)}^{2}$ , $y = {(t - 1)}^{3} - (t - 1)$ metres for $0 \leq t \leq 2$ seconds. Show that it returns to its starting point, and decide whether it stops anywhere.

Worked solution

1
Compare the endpoint positions.
$\begin{aligned} r (0) & = (1, - 1 + 1) \\ = (1, 0) \\ r (2) & = (1, 1 - 1) \\ = (1, 0) \end{aligned}$
Both coordinates agree at the two distinct times, so the starting point is revisited.
2
Differentiate each coordinate.
$r^{'} (t) = (\begin{array}{cccccccccc} 2 (t - 1) \\ 3 {(t - 1)}^{2} - 1 \end{array})$
The chain rule applies to each power of $t - 1$ .
3
Check the only time when the horizontal rate is zero.
$\begin{array}{c} x^{'} (t) = 0 ⟹ t = 1 \\ y^{'} (1) = - 1 \end{array}$
At any other time the horizontal rate is nonzero; at $t = 1$ the vertical rate is nonzero. The rates never vanish together.

Answer

It returns to $(1, 0)$ metres at $t = 2$ seconds, but it is never at rest during the journey.

03 / YOUR TURN

Try it yourself.

Calculate, explain, then apply.

CHECK 01

Calculate the coordinate rates

For $x = t^{2} + 1$ and $y = 3 t - t^{2}$ metres, with $t$ in seconds, find the position and velocity at $t = 2$ .

Show answer & reasoning

Position $(5, 2)$ metres and velocity $(4, - 1)$ m/s.

Substitution gives $(2^{2} + 1, 3 (2) - 2^{2}) = (5, 2)$ . Differentiate before substituting: $r^{'} (t) = (2 t, 3 - 2 t)$ .

CHECK 02

Interpret an instant of rest

For $r (t) = (t^{2}, t^{3})$ metres with $t$ in seconds, is the particle at rest at $t = 0$ ? Does that mean it remains at the origin?

Show answer & reasoning

It is at rest at that instant, but it does not remain at the origin.

$r^{'} (0) = (0, 0)$ , so both coordinate rates vanish. For any $t > 0$ , the position $(t^{2}, t^{3})$ differs from $(0, 0)$ ; instantaneous rest is not permanent rest.

CHECK 03

Apply a maximum-height test

A drone is modelled by $x = 2 t$ , $y = 2 + 6 t - t^{2}$ metres for $0 \leq t \leq 6$ seconds. Find its greatest height and say whether it stops there.

Show answer & reasoning

Greatest height $11$ metres at $t = 3$ seconds; it keeps moving with velocity $(2, 0)$ m/s.

$y^{'} = 6 - 2 t = 0$ at $t = 3$ and $y^{′′} = - 2$ . The height is $2 + 18 - 9 = 11$ , above both endpoint heights of $2$ metres. The nonzero horizontal rate prevents a stop.

Keep practising

6 worksheet questions · 12 marks

Open worksheet 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 $t = \frac{y}{2}$ .
Substitute into $x$ .
$x = \frac{y^{2}}{4} - 1$
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
$\frac{d y}{d t} = 2 t - 2$ .
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 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.

Keep all branches of a polynomial path

A coordinate can turn while the particle keeps moving

Test complete positions and complete rates

See the method in action.

Find a maximum height without inventing a stop

Differentiate the vertical coordinate.

Confirm the maximum and compare endpoint heights.

Check both coordinate rates at the maximum.

Retain both branches after elimination

Make equal even powers of the parameter.

Use the allowed parameter values to establish the branches.

Evaluate the endpoints and the central parameter value.

Check whether a plotting head returns and stops

Compare the endpoint positions.

Differentiate each coordinate.

Check the only time when the horizontal rate is zero.

Try it yourself.

Calculate the coordinate rates

Interpret an instant of rest

Apply a maximum-height test

Keep practising