All subjects · Mathematics · WMA14 (P4)

Parametric differentiation

When a curve is given parametrically, you find its gradient without converting to Cartesian form, using the chain rule.

The rule

dy/dx = (dy/dt) ÷ (dx/dt).

Differentiate x and y with respect to t separately, then divide. The answer is in terms of t.

Tangents and normals

1. Substitute the given t to find the point (x, y).

2. Substitute t into dy/dx for the gradient m.

3. Tangent: y - y1 = m(x - x1). Normal: gradient -1/m.

Stationary points and vertical tangents

Stationary points: dy/dt = 0 (with dx/dt not 0). Vertical tangents: dx/dt = 0 (with dy/dt not 0). Solve for t, then find the coordinates.

Worked example

x = t2, y = t3 - 3t. Find the equation of the tangent at t = 2.

  1. dx/dt = 2t, dy/dt = 3t2 - 3
  2. dy/dx = (3t2 - 3)/(2t) = 9/4 at t = 2
  3. At t = 2: x = 4, y = 8 - 6 = 2

Answer: y - 2 = (9/4)(x - 4), or 9x - 4y - 28 = 0

Key idea: dy/dx = (dy/dt)/(dx/dt). Find the point and gradient from the same t, then use y - y1 = m(x - x1).

Check yourself

  1. x = 3t, y = t2. Find dy/dx.
    • 3/(2t)
    • 2t/3
    • 6t
    • 2t
    Show answer

    2t/3. dy/dt = 2t and dx/dt = 3, so dy/dx = 2t/3.

  2. x = t2 + 1, y = 4t. Find the gradient of the curve at t = 2.
    Show answer

    1. dy/dx = 4/(2t) = 2/t. At t = 2: 1.

  3. x = 2cos t, y = 3sin t. Find the gradient at t = π/4.
    Show answer

    -1.5. dy/dx = 3cos t/(-2sin t). At t = π/4, cos t = sin t, so dy/dx = -3/2 = -1.5.

  4. x = t2, y = t3. Find the gradient of the normal at t = 1. Give your answer as a fraction in its simplest form.
    Show answer

    -2/3. dy/dx = 3t2/(2t) = 3t/2 = 3/2 at t = 1. The normal gradient is -2/3.

  5. x = t2, y = t3 - 12t. Find the y-coordinate of the stationary point with t > 0.
    Show answer

    -16. dy/dt = 3t2 - 12 = 0 gives t = 2. Then y = 8 - 24 = -16.

  6. x = t2, y = 2t. Find where the tangent at t = 3 crosses the y-axis. Give the y-coordinate.
    Show answer

    3. dy/dx = 2/(2t) = 1/t = 1/3. The point is (9, 6). Tangent: y - 6 = (1/3)(x - 9). At x = 0: y = 6 - 3 = 3.

  7. Where does a parametric curve have a vertical tangent?
    • Where dx/dt = 0 and dy/dt is not 0
    • Where dy/dt = 0 and dx/dt is not 0
    • Where t = 0
    • Where x = 0
    Show answer

    Where dx/dt = 0 and dy/dt is not 0. dy/dx = (dy/dt)/(dx/dt) becomes infinitely steep when dx/dt = 0.

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Last updated 7 November 2026. Spotted a mistake? Tell us.