Parametric Differentiation (AQA A-Level Mathematics): Flashcards

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Parametric Differentiation
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Chain rule for parametric: dydt×dtdx=?\frac{dy}{dt} \times \frac{dt}{dx} = ?

dydx\frac{dy}{dx}

Parametric differentiation definition

Finding derivatives where xx and yy are functions of parameter tt

dydx\frac{dy}{dx} in parametric form

dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

Second derivative formula for parametric equations

d2ydx2=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}

Properties calculated using parametric differentiation

Slope, concavity, and other curve properties

Gradient of normal to a curve

−1m-\frac{1}{m} where mm is gradient of tangent

Tangent line equation form

y−y1=m(x−x1)y - y_1 = m(x - x_1)

First step in parametric differentiation

Differentiate x=f(t)x=f(t) and y=g(t)y=g(t) w.r.t. tt

Convert parametric to Cartesian method

Eliminate parameter tt from equations

dtdx\frac{dt}{dx} in terms of dxdt\frac{dx}{dt}

1dxdt\frac{1}{\frac{dx}{dt}} (reciprocal)

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