Parametric Differentiation (Cambridge International A-Level Mathematics)

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

  • $\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$
  • $\frac{dt}{dx}=1\div\frac{dx}{dt}$

• Equivalently, $\frac{dy}{dx}=\frac{dy}{dt}\div\frac{dx}{dt}$

• $\frac{dy}{dx}$ will be in terms of t – this is fine

  • 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

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)

How do I solve problems using parametric differentiation?

• Questions may require use of tangents and normals as per the coordinate 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/normals

• 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

Worked Example