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Pure · Differentiation

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Parametric differentiation

Finding dy/dx for a curve given parametrically as dy/dx = (dy/dt) / (dx/dt), using it for tangents and normals, and the second derivative of a parametric curve.

Pure · Differentiation

Parametric differentiation

Finding dy/dx for a curve given parametrically as dy/dx = (dy/dt) / (dx/dt), using it for tangents and normals, and the second derivative of a parametric curve.

Why it works

A parametric curve gives xx and yy each as a function of the parameter tt, not of each other. To find the gradient dydx\frac{dy}{dx} you need the rate of change of yy against xx — but all you can differentiate directly is each coordinate against tt. The chain rule bridges the gap: dydx=dydt×dtdx=  dydt  dxdt.\frac{dy}{dx} = \frac{dy}{dt}\times\frac{dt}{dx} = \frac{\;\frac{dy}{dt}\;}{\frac{dx}{dt}}. Differentiate yy with respect to tt, differentiate xx with respect to tt, and divide. The dtdt's "cancel" in the Leibniz picture, which is exactly why it works.

Take x=t2, y=t3x = t^2,\ y = t^3. Then dxdt=2t\frac{dx}{dt} = 2t and dydt=3t2\frac{dy}{dt} = 3t^2, so dydx=3t22t=3t2.\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}. The gradient comes out in terms of tt — which is natural, since each point of the curve is labelled by its tt-value. To get the gradient at a specific point, find the tt for that point and substitute. (Where dxdt=0\frac{dx}{dt} = 0 the tangent is vertical and dydx\frac{dy}{dx} is undefined — here that is t=0t = 0, the cusp at the origin.)

Tangents and normals then work as always: gradient m=dydxm = \frac{dy}{dx} at the point's tt-value, the point itself (x(t),y(t))(x(t), y(t)), and y−y1=m(x−x1)y - y_1 = m(x - x_1) for the tangent (or gradient −1m-\frac1m for the normal). You almost never need the Cartesian equation.

The second derivative. A classic trap: d2ydx2\frac{d^2y}{dx^2} is not d2y/dt2d2x/dt2\frac{d^2y/dt^2}{d^2x/dt^2}. The second derivative is the rate of change of dydx\frac{dy}{dx} with respect to xx, so apply the same bridge again — differentiate dydx\frac{dy}{dx} with respect to tt, then divide by dxdt\frac{dx}{dt}: d2ydx2=ddt ⁣(dydx)dxdt.\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}.