Parametric Differentiation

MATTechniques of Differentiation

When $x$ and $y$ are each given in terms of a parameter $t$ or $\theta$ , the derivative $\dfrac{dy}{dx}$ is found as the ratio of their derivatives with respect to the parameter. Second derivatives require an extra division by $\dfrac{dx}{d\theta}$ .

ISC commonly sets these on cycloids, ellipses and astroid-type curves and asks for evaluation at specific parameter values.

First parametric derivative

\dfrac{dy}{dx} = \dfrac{\,dy/d\theta\,}{\,dx/d\theta\,}

Valid where

\dfrac{dx}{d\theta}\ne 0

;

x=x(\theta)

y=y(\theta)

Second parametric derivative

\dfrac{d^2y}{dx^2} = \dfrac{d}{d\theta}\!\left(\dfrac{dy}{dx}\right)\cdot \dfrac{1}{\,dx/d\theta\,}

Differentiate the first derivative w.r.t.

\theta

, then divide again by

\dfrac{dx}{d\theta}

— do not divide by

\dfrac{d^2x}{d\theta^2}

Cycloid derivative

x=a(\theta-\sin\theta),\; y=a(1-\cos\theta)\;\Rightarrow\;\dfrac{dy}{dx}=\cot\dfrac{\theta}{2}

From

\dfrac{a\sin\theta}{a(1-\cos\theta)}

using

\sin\theta=2\sin\tfrac\theta2\cos\tfrac\theta2

1-\cos\theta=2\sin^2\tfrac\theta2

Ellipse derivative

x=a\cos\theta,\; y=b\sin\theta \;\Rightarrow\; \dfrac{dy}{dx}=-\dfrac{b}{a}\cot\theta

From

\dfrac{b\cos\theta}{-a\sin\theta}

; for

x=a\cos^3\theta,\,y=a\sin^3\theta

the result is

-\tan\theta

First differentiate $x$ and $y$ separately with respect to the parameter, then take their ratio.
For the second derivative, differentiate $\dfrac{dy}{dx}$ (a function of $\theta$ ) with respect to $\theta$ and divide once more by $\dfrac{dx}{d\theta}$ .
Use half-angle identities to simplify cycloid results to $\cot\dfrac{\theta}{2}$ .
For $x=a\cos^3\theta,\,y=a\sin^3\theta$ (astroid), $\dfrac{dy}{dx}=-\tan\theta$ , so at $\theta=\tfrac\pi4$ it equals $-1$ .
When asked to evaluate at a particular $\theta$ , simplify the general expression fully before substituting.
$\dfrac{dx}{d\theta}=0$ marks points (e.g. vertical tangents) where $\dfrac{dy}{dx}$ is undefined.
Keep the parameter throughout; only the final numeric answer should be parameter-free if a value is requested.

Where the marks go

Computing the second derivative as $\dfrac{d^2y/d\theta^2}{d^2x/d\theta^2}$ — this is wrong; you must divide $\dfrac{d}{d\theta}\!\left(\dfrac{dy}{dx}\right)$ by $\dfrac{dx}{d\theta}$ .
Forgetting the extra factor $\dfrac{1}{dx/d\theta}$ when finding $\dfrac{d^2y}{dx^2}$ .
Substituting the parameter value too early, before simplifying with half-angle or double-angle identities.
Sign errors from $\dfrac{dx}{d\theta}=-a\sin\theta$ in trig parametrisations.

How the board asks it

Numericalfirst parametric derivative as ratio of parameter derivatives
If $x = a(\theta - \sin\theta)$ and $y = a(1 - \cos\theta)$ , find $\dfrac{dy}{dx}$ and show that it simplifies to $\cot\dfrac{\theta}{2}$ .
Numericalsimplify the general expression, then substitute the parameter
If $x = a\cos^3\theta$ and $y = a\sin^3\theta$ , find the value of $\dfrac{dy}{dx}$ at $\theta = \dfrac{\pi}{4}$ .
Numericaldifferentiate dy/dx with respect to the parameter and divide again by $dx/d\theta$
If $x = a\cos\theta$ and $y = b\sin\theta$ , find $\dfrac{d^2y}{dx^2}$ in terms of the parameter $\theta$ .
Derive / provefirst parametric derivative with trigonometric simplification
If $x = a\left(\cos t + \log\tan\dfrac{t}{2}\right)$ and $y = a\sin t$ , prove that $\dfrac{dy}{dx} = \tan t$ .
Numericalevaluate the parametric dy/dx at the given parameter value
For the curve $x = a(\theta + \sin\theta)$ , $y = a(1 - \cos\theta)$ , find the slope of the tangent at $\theta = \dfrac{\pi}{2}$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.