What is a parametric equation?
Usually a curve is written as y in terms of x, like y = x². In parametric form we write both x and y using a third variable, the parameter t:
x = f(t), y = g(t).
Pick a value of t, work out x and y, plot the point. Join the points in order of t and you get the curve. The order also gives the curve a direction (orientation): which way the point moves as t increases.
Example: x = t, y = t² for −2 ≤ t ≤ 2 gives the points (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4): a parabola.
Changing to Cartesian form
To find the ordinary (Cartesian) equation, eliminate t:
- Make t the subject of one equation and substitute. x = 2t, y = 4t − t² → t = x/2 → y = 2x − x²/4.
- Use an identity for trig forms. x = 3cos t, y = 3sin t → x² + y² = 9(cos²t + sin²t) = 9, a circle of radius 3.
Useful standard forms: circle x = a cos t, y = a sin t; ellipse x = a cos t, y = b sin t; parabola x = at², y = 2at. Always note the allowed range of x and y (for example x = t² can never be negative).
Differentiating parametric equations
First derivative: dy/dx = (dy/dt) ÷ (dx/dt), as long as dx/dt ≠ 0.
- dy/dt = 0 (and dx/dt ≠ 0): horizontal tangent.
- dx/dt = 0 (and dy/dt ≠ 0): vertical tangent.
Second derivative: d²y/dx² = [d/dt (dy/dx)] ÷ (dx/dt). A common error is to divide d²y/dt² by d²x/dt²: that is wrong.
Tangent at a point: find the point from t, find the slope m = dy/dx at that t, then use y − y₁ = m(x − x₁). The normal has slope −1/m.
Motion and arc length
When t is time, the position is r(t) = (x(t), y(t)).
- Velocity v = (dx/dt, dy/dt).
- Speed = |v| = √((dx/dt)² + (dy/dt)²).
- Acceleration a = (d²x/dt², d²y/dt²).
Arc length (distance travelled along the curve) from t = a to t = b:
L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt.
For the circle x = 3cos t, y = 3sin t the speed is always 3, so from 0 to 2π the length is 3 × 2π = 6π, the circumference. Polar curves r = f(θ) are a special case: x = r cos θ, y = r sin θ.
Try it: draw a parametric curve by hand
Make a table for x = t + 1, y = 2t − 1 with t = 0, 1, 2, 3. Plot the points and join them. Predict: what shape do you get? (A straight line; eliminate t to find y = 2x − 3.) Then open the free-play step and slide t for the parabola and the ball to check your predictions.
Key formulas and definitions
- x = f(t), y = g(t): each t gives one point (x, y)
- dy/dx = (dy/dt) ÷ (dx/dt), dx/dt ≠ 0
- d²y/dx² = [d/dt (dy/dx)] ÷ (dx/dt)
- Circle: x = a cos t, y = a sin t; ellipse: x = a cos t, y = b sin t
- Speed = √((dx/dt)² + (dy/dt)²)
- Arc length L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt
Worked examples
1. x = 2t − 1, y = t + 3. Find the Cartesian equation.
From y: t = y − 3. Substitute: x = 2(y − 3) − 1 = 2y − 7. So x = 2y − 7, or y = (x + 7)/2, a straight line.
2. x = 4cos t, y = 2sin t. Find the Cartesian equation.
cos t = x/4 and sin t = y/2. Since cos²t + sin²t = 1: x²/16 + y²/4 = 1. An ellipse with semi-axes 4 and 2.
3. x = t², y = t³. Find dy/dx at t = 2.
dx/dt = 2t, dy/dt = 3t². dy/dx = 3t² ÷ 2t = 3t/2. At t = 2: dy/dx = 3.
4. For x = t², y = t³ find d²y/dx².
dy/dx = 3t/2. d/dt(3t/2) = 3/2. Divide by dx/dt = 2t: d²y/dx² = (3/2) ÷ (2t) = 3/(4t).
5. Find the tangent to x = t², y = 2t at t = 1.
Point: (1, 2). dx/dt = 2t = 2, dy/dt = 2, slope = 1. Tangent: y − 2 = 1(x − 1), so y = x + 1.
6. A particle moves with x = 3t, y = 4t (metres, seconds). Find its speed and the distance travelled from t = 0 to t = 2.
dx/dt = 3, dy/dt = 4. Speed = √(9 + 16) = 5 m/s. Arc length = ∫₀² 5 dt = 10 m.
7. Find the arc length of x = cos t, y = sin t from t = 0 to t = π/2.
dx/dt = −sin t, dy/dt = cos t. √(sin²t + cos²t) = 1. L = ∫₀^(π/2) 1 dt = π/2 ≈ 1.57 (a quarter of a unit circle).
Common mistakes
- Writing dy/dx = (dx/dt) ÷ (dy/dt). The y-rate goes on top: dy/dx = ẏ ÷ ẋ.
- Finding d²y/dx² as (d²y/dt²) ÷ (d²x/dt²). Correct: differentiate dy/dx with respect to t, then divide by dx/dt.
- Forgetting the range of x and y after eliminating t (x = t² gives only x ≥ 0).
- Using the speed formula without the square root, or forgetting to square both rates.