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9.7PureStretch

Parametric Differentiation

When a curve is given parametrically, the gradient dy/dx is found by dividing dy/dt by dx/dt. This is a direct application of the chain rule.

24 min Differentiation
Edexcel A level Maths: 9.7 Parametric DifferentiationCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Find dy/dx from parametric equations
  • Use dy/dx = (dy/dt)/(dx/dt)
  • Find tangents and normals
  • Locate stationary points
1

The method

Differentiate xx and yy each with respect to tt, then dydx=dy/dtdx/dt\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}.

dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}
Parametric gradient.
Example 1For x=t2x = t^2, y=t3y = t^3, find dydx\dfrac{dy}{dx}.
1dxdt=2t\tfrac{dx}{dt} = 2t, dydt=3t2\tfrac{dy}{dt} = 3t^2.
2dydx=3t22t=3t2\tfrac{dy}{dx} = \tfrac{3t^2}{2t} = \tfrac{3t}{2}.
Answer
3t2\dfrac{3t}{2}

Tip — Leave the gradient in terms of t unless a specific point is given.

2

Stationary points

Stationary points occur where dydt=0\dfrac{dy}{dt} = 0 (and dxdt≠0\dfrac{dx}{dt} \ne 0). Solve for tt, then find the coordinates.

Formula recap

dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}
Parametric gradient.
dydt=0⇒stationary\frac{dy}{dt} = 0 \Rightarrow \text{stationary}
Turning points.

Common mistakes to avoid

Multiplying dy/dt by dx/dt.
Divide: dy/dx = (dy/dt)/(dx/dt).
Forgetting to substitute t to find a numerical gradient.
Plug the given t into dy/dx.

Key takeaways

  • dy/dx = (dy/dt)/(dx/dt).
  • Leave in terms of t, or substitute a value.
  • Stationary where dy/dt = 0.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

  1. For parametric curves, dy/dx =

    • dy/dtdx/dt\frac{dy/dt}{dx/dt}
    • dx/dtdy/dt\frac{dx/dt}{dy/dt}
    • dydt⋅dxdt\frac{dy}{dt}\cdot\frac{dx}{dt}
    • dydt\frac{dy}{dt}
    Show answer and explanation

    Answer: dy/dtdx/dt\frac{dy/dt}{dx/dt}

    Divide.

  2. For x=t2x = t^2, y=t3y = t^3, dy/dx =

    • 3t2\frac{3t}{2}
    • 23t\frac{2}{3t}
    • 3t23t^2
    • 3t22\frac{3t^2}{2}
    Show answer and explanation

    Answer: 3t2\frac{3t}{2}

    3t²/2t.

  3. Stationary points occur where:

    • dydt=0\frac{dy}{dt} = 0
    • dxdt=0\frac{dx}{dt} = 0
    • t=0t = 0
    • y=0y = 0
    Show answer and explanation

    Answer: dydt=0\frac{dy}{dt} = 0

    Gradient zero.

  4. If x=2tx = 2t, y=t2y = t^2, dy/dx =

    • tt
    • 2t2t
    • 1t\frac{1}{t}
    • t2t^2
    Show answer and explanation

    Answer: tt

    2t/2.

  5. Parametric differentiation is an application of the:

    • chain rule
    • product rule
    • quotient rule
    • binomial theorem
    Show answer and explanation

    Answer: chain rule

    dy/dx = dy/dt · dt/dx.

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