# 9.2.1 Parametric Differentiation

A Level Only

#### 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
• dy/dx = dy/dt × dt/dx
• dt/dx = 1 ÷ dx/dt
• Equivalently, dy/dx = dy/dt ÷ dx/dt
• dy/dx will be in terms of t – this is fine
• Questions usually involve finding gradients, tangents and normals
• 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 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)

• 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
A Level Only

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