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 Video by Zeeshan Zamurred Differentiation

Edexcel A level Maths: 9.7 Parametric DifferentiationWatch the full walkthrough before the notes below.

What you'll be able to do

The method

Differentiate $x$ and $y$ each with respect to $t$ , then $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ .

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}

Parametric gradient.

\frac{d x}{d t} = 2 t

\frac{d y}{d t} = 3 t^{2}

\frac{d y}{d x} = \frac{3 t ^{2}}{2 t} = \frac{3 t}{2}

Answer

\frac{3 t}{2}

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

Stationary points occur where $\frac{d y}{d t} = 0$ (and $\frac{d x}{d t} \neq = 0$ ). Solve for $t$ , then find the coordinates.

Formula recap

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}

Parametric gradient.

\frac{d y}{d t} = 0 \Rightarrow 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

Test yourself

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