AQA A Level Maths: Pure复习笔记9.2.1 Parametric Differentiation

Parametric Differentiation

How do I find dy/dx from parametric equations?

  • Ensure you are familiar with Parametric Equations – Basics first


  • This method uses the chain rule and the reciprocal property of derivatives
    • Questions usually involve finding gradients, tangents and normals
  • The chain rule is always needed when there are three variables or more – see Connected Rates of Change

How do I find gradients, tangents and normals from parametric equations?

  • To find a gradient …
    • STEP 1: Find dx/dt and dy/dt
    • STEP 2: Find dy/dx in terms of t

Using either dy/dx = dy/dt ÷ dx/dt

or dy/dx = dy/dt × dt/dx where dt/dx = 1 ÷ dx/dt

    • STEP 3: Find the value of t at the required point
    • STEP 4: Substitute this value of t into dy/dx to find the gradient 9.2.1-Notes-para_grad
  • to then go on to find the equation of a tangent …
    • STEP 5: Find the y coordinate
    • STEP 6: Use the gradient and point to find the equation of the tangent


  • To find a normal...
    • STEP 7: Use perpendicular lines property to find the gradient of the normal m1 × m2 = -1
    • STEP 8: Use gradient and point to find the equation of the normal y - y1 = m(x - x1)

What else may I be asked to do?

  • Questions may require use of tangents and normals as per the coordiante geometry sections
    • Find points of intersection between a tangent/normal and x/y axes
    • Find areas of basic shapes enclosed by axes and/or tangents/normal
  • Find stationary points (dy/dx = 0)


  • You may also be asked about horizontal and vertical tangents
    • At horizontal (parallel to the x-axis) tangents, dy/dt = 0
    • At vertical (parallel to y-axis) tangents, dx/dt = 0


Just for fun …

  • Try plotting the graph from the question below using graphing software
  • Plenty of free online tools do this – for example Desmos and Geogebra
  • Try changing the domain of t to -π/3 ≤ t ≤ π/3

Worked Example