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Periodic parametric motion

Model circular and figure-of-eight motion using parametric equations. Find full periods, return times and circular distances, and distinguish revisiting a point from repeating the whole motion.

Before you startRadians, trigonometric periods and parametric circles

01 / A repeat must satisfy both coordinates

One coordinate returning is not enough.

For circular motion x = 3 cos t, y = 3 sin t, a full turn returns both coordinates to their starting values. But other paths can revisit their start midway through a larger pattern.

A return time solves the two starting-coordinate equations. A full period repeats the motion from every time.

Compare the circle and figure of eight. At half a cycle, the figure of eight is back at its starting point but still has its other loop to trace.

Return or full repeat?Explore

PathAngular factor ωProgress through a full cycle

x = 3 cos t; y = 3 sin t.

At t = 0 s: (3, 0) m.

Full period: 2π seconds.

The first return to the starting point is after the full period.

Circle radius 3 m; constant speed 3 m/s.

Grey shows a whole cycle and blue shows the path so far. The ring marks the starting point. A return to that point need not finish the entire pattern.

02 / Read radius, centre and angular rate

Keep the angle inside the sine and cosine separate from time.

Interpret x = 4 cos(πt/6), y = 4 sin(πt/6), with metres and seconds.Worked example

x² + y² = 16

The path is a circle of radius 4 m centred at the origin.

Angle θ = πt/6 radians

The angular rate is π/6 rad/s.

At t = 0: (4, 0)

The point begins on the positive x-axis.

For small positive t, y increases

It moves anticlockwise.

Adding fixed values to the coordinate rules shifts the centre without changing the radius or time for a full turn.

03 / Find the time for a full circular cycle

The angle must increase by 2π radians.

Find the period of x = 4 cos(πt/6), y = 4 sin(πt/6).Worked example

(π/6)T = 2π

A complete revolution uses a 2π angle change.

T = 12 s

This is the smallest positive full period.

For x = a + r cos(ωt + φ), y = b + r sin(ωt + φ): T = 2π/|ω|, when r > 0 and ω ≠ 0.

The phase φ changes the starting point. A negative ω reverses direction but still gives a positive period. If r = 0 or ω = 0, the point is stationary and there is no unique smallest positive period.

04 / Use circular arc length for circular motion

Distance travelled uses the total swept angle, including extra revolutions.

How far does the radius-4 circle model travel in 30 s?Worked example

Swept angle = (π/6) × 30 = 5π radians

This is 2.5 full turns.

Distance = rθ = 4 × 5π = 20π m

Do not reduce the angle modulo 2π when finding distance.

Constant speed = 4 × π/6 = 2π/3 m/s

Equivalently circumference divided by period.

Final position (−4, 0); displacement magnitude 8 m

Distance and displacement are different.

For an ellipse or a figure of eight, rθ is not a valid general path-length formula. Those curves do not have one fixed distance from their centre.

05 / Solve position conditions in the stated time interval

Keep the time factor when converting angle solutions back to seconds.

For the radius-4, period-12 model, find y = 2 during 0 ≤ t < 12.Worked example

sin(πt/6) = 1/2

Divide the y-rule by 4.

πt/6 = π/6 or 5π/6

These are the complete angle solutions in [0, 2π).

t = 1 or 5 s

Convert each angle back to time.

Points (2√3, 2) and (−2√3, 2)

The same height occurs at two different positions.

Include or exclude the final time according to the question. A full-turn endpoint repeats the initial position, even though it is a different time.

06 / Trace a path with different coordinate frequencies

Two oscillations can produce a self-intersecting path.

Consider x = 3 sin t, y = 2 sin 2t, with t in seconds and coordinates in metres. Over 0 ≤ t ≤ 2π, the x-coordinate completes one sine cycle while y completes two.

Find some key positions.Worked example

t = 0: (0, 0)

Starting point.

t = π/4: (3√2/2, 2); t = π/2: (3, 0)

The first part of the right loop.

t = 3π/4: (3√2/2, −2); t = π: (0, 0)

The right loop returns to the origin.

π < t < 2π gives x < 0

The remaining interval traces the left loop.

The origin occurs at more than one parameter value. It is a crossing of the path, not evidence that the coordinates are invalid.

Watch the two loops meet at the same point

Pause, replay or seek freely. The notes explain the same idea and stay in view.

07 / Find the first positive return to the starting point

Solve both coordinates together and choose the earliest valid time.

When does x = 3 sin t, y = 2 sin 2t first return to (0, 0)?Worked example

sin t = 0 requires t = nπ

Check x first.

sin 2t = 0 at all those times

The same parameters also satisfy y = 0.

First positive return: t = π

The point has completed the right loop only.

Full period: 2π

After π the x-coordinate changes sign, so the entire motion has not repeated.

At t = π/4 the point is (3√2/2, 2). At t = π/4 + π it is (−3√2/2, 2). This counterexample proves that π is not a period, despite the origin return.

08 / Find a common period of both coordinate functions

The least positive common repeat is the period of the pair.

Find the full period of x = 2 cos 3t, y = sin 2t.Worked example

x has period 2π/3; y has period π

Both amplitudes are non-zero.

Require 3T = 2πm and 2T = 2πn

For positive integers m and n.

The smallest solution is m = 3, n = 2

Thus T = 2π.

This finds a full period, not necessarily the first return from every chosen starting point. Special intersections can be revisited sooner. Do not simply choose the larger of two periods unless it is actually a multiple of the other.

09 / Keep time units consistent

Frequency changes the timing; amplitudes change the geometry.

Compare x = 3 sin(2t), y = 2 sin(4t) with the earlier figure of eight.Worked example

Let θ = 2t

The coordinate pair is (3 sin θ, 2 sin 2θ).

The same geometric path is traced twice as fast

All corresponding positions occur after half the time.

Full period π s; first positive origin return π/2 s

Rescale both quantities, not only one.

If time is given in minutes but speed is required in metres per second, convert the elapsed time. A period of 2 minutes is 120 seconds. Do not label a rad/min coefficient as rad/s.

10 / Separate average speed from average velocity

A closed trip can cover distance while having zero net displacement.

A point makes one full revolution of a radius-5 m circle in 20 s at constant angular rate.Worked example

Distance travelled = 10π m

One circumference.

Average speed = 10π/20 = π/2 m/s

Distance divided by elapsed time.

Net displacement = (0, 0)

The end position equals the start.

x = 4 cos t, y = 2 sin t has period 2π. May you calculate its full path length as 2π × 4?

Hint

Is its distance from the centre constant?

Worked solution

No. The path is an ellipse, with distances 4 and 2 from the centre at its axis endpoints. The circle circumference formula does not apply.

12 / Recap

A repeated position can be part of a longer cycle.

For circular motion, use angular change 2π to find a full turn.
Use absolute angular rate for positive speed and period.
Find point returns by solving both coordinate conditions.
A full period repeats the pair from every time; a special point may return earlier.
Use rθ only for circular arc distance and retain whole extra turns.
Convert time units and distinguish distance from displacement.

Section 1 of 12 · A repeat must satisfy both coordinates