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Parametric Motion: Direction, Speed & Orientation

You’ll be able to

A path plus a schedule

A parametric description gives x(t) and y(t) separately, so it records not just where the curve goes but when each point is reached and which way it is traveled. That extra information — the orientation — is exactly what a rectangular equation throws away. Two very different motions can share one path, and only the parametrization distinguishes them.

Parametrizing a circle and an ellipse
circle radius r: x = r cos t, y = r sin t · ellipse: x = a cos t, y = b sin t
Both are traced counterclockwise starting at (r, 0) or (a, 0). Swapping sine and cosine, or negating one, reverses or rotates the starting point.
Worked example

A particle moves with x(t) = 3 cos t and y(t) = 3 sin t for 0 ≤ t ≤ 2π. Find the path, the direction of travel, and the position at t = π/2.

  1. 1.Eliminate t using the Pythagorean identity: x² + y² = 9cos²t + 9sin²t = 9(cos²t + sin²t) = 9.
  2. 2.So the path is the circle of radius 3 centered at the origin.
  3. 3.At t = 0: (3, 0). At t = π/2: (3cos(π/2), 3sin(π/2)) = (0, 3).
  4. 4.Moving from (3, 0) to (0, 3) is counterclockwise.
  5. 5.The full interval 0 to 2π traces exactly one revolution.
Answer: The path is x² + y² = 9, traced counterclockwise once, starting and ending at (3, 0). At t = π/2 the particle is at (0, 3). Checking two consecutive t values is the reliable way to establish orientation — the rectangular equation alone cannot tell you.

Eliminating the parameter

Two techniques cover most cases. If one equation is easy to invert, solve for t and substitute: from x = t + 1 we get t = x − 1, and substituting into y = t² gives y = (x − 1)². If the pair involves sine and cosine, use a Pythagorean identity instead: from x = 2cos t and y = 3sin t, form (x/2)² + (y/3)² = cos²t + sin²t = 1. Either way, note the resulting domain — the parameter interval may trace only part of the full rectangular curve.

Watch out

Eliminating the parameter can enlarge the curve. x = t², y = t traces only the right half of the parabola x = y², since t² is never negative. The rectangular equation describes the whole parabola, so the parametric domain must be reported alongside it.

Checkpoint

For x = 2 cos t and y = 2 sin t with 0 ≤ t ≤ π, what portion of the circle is traced?

Many parametrizations, one path

The path x² + y² = 9 can be traced by (3cos t, 3sin t) counterclockwise at one revolution per 2π; by (3cos 2t, 3sin 2t) counterclockwise at twice the speed; by (3cos(−t), 3sin(−t)) = (3cos t, −3sin t) clockwise; or by (3sin t, 3cos t) clockwise starting from the top. The path is a set of points; the parametrization is one way of moving along it. Questions about speed and direction are questions about the parametrization, not the path.

Worked example

A projectile has x(t) = 40t and y(t) = 30t − 4.9t², with t in seconds and distances in meters. Find the rectangular path and the horizontal distance when it returns to y = 0.

  1. 1.From x = 40t, solve t = x/40.
  2. 2.Substitute: y = 30(x/40) − 4.9(x/40)² = 0.75x − 4.9x²/1600.
  3. 3.So y = 0.75x − 0.0030625x², a downward parabola. ✓ (The x² coefficient is negative.)
  4. 4.For the landing point, use the parametric form: y = 0 gives t(30 − 4.9t) = 0, so t = 0 or t = 30/4.9 ≈ 6.122 s.
  5. 5.x at that time: 40(6.122) ≈ 244.9 m.
Answer: The path is the parabola y = 0.75x − 0.0030625x², and the projectile lands about 244.9 m downrange after about 6.12 s. The parametric form is far better suited to the timing question — solving y = 0 in t takes one line, while the rectangular form obscures the time entirely.
Checkpoint

Which parametrization traces the circle x² + y² = 1 clockwise starting at (1, 0)?

Answer the 2 checkpoints as you read.

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