Parametric Forms

Sometimes $x$ and $y$ are not linked directly at all — each is tied separately to a third variable, the parameter: $x = f(t)$, $y = g(t)$. The circle is the classic case: $x = a\cos\theta$, $y = a\sin\theta$ traces the whole circle as $\theta$ runs, with no single formula $y = h(x)$ needed.

The chain rule links the three variables: $\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$, and solving for $\frac{dy}{dx}$:

$$\boxed{\;\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)} \qquad \left(\text{provided } \frac{dx}{dt} \neq 0\right)\;}$$

The routine: differentiate each coordinate with respect to the parameter, then divide — $\frac{dy}{dt}$ on top, $\frac{dx}{dt}$ underneath. The answer naturally comes out in terms of the parameter, and that is perfectly acceptable.

Two cautions worth marks:

1. Right way up. The quotient is $\frac{dy/dt}{dx/dt}$, never the reciprocal — a flipped fraction is the most common parametric error.
2. Do not differentiate $y$ w.r.t. $x$ piecewise by eliminating $t$ unless elimination is genuinely easier; the division route is almost always shorter, and for curves like the cycloid, elimination is barely possible.

The standard curves

The half-angle collapse in the cycloid row — $\frac{\sin\theta}{1 + \cos\theta} = \tan\frac{\theta}{2}$ — is a simplification examiners expect on sight; its partner $\frac{\sin\theta}{1 - \cos\theta} = \cot\frac{\theta}{2}$ appears just as often.

Solved Examples

Example 1 — The circle

Find $\dfrac{dy}{dx}$ if $x = a\cos\theta$, $y = a\sin\theta$.

Step 1 — differentiate each w.r.t. $\theta$: $\frac{dx}{d\theta} = -a\sin\theta$, $\frac{dy}{d\theta} = a\cos\theta$.

Step 2 — divide: $$\frac{dy}{dx} = \frac{a\cos\theta}{-a\sin\theta} = -\cot\theta$$

Answer: $-\cot\theta$ — geometrically sensible: at $\theta = \frac{\pi}{2}$ (top of the circle) the tangent is horizontal, and indeed $-\cot\frac{\pi}{2} = 0$.

Example 2 — The parabola

Find $\dfrac{dy}{dx}$ if $x = at^2$, $y = 2at$.

Step 1 — differentiate: $\frac{dx}{dt} = 2at$, $\frac{dy}{dt} = 2a$.

Step 2 — divide: $\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$.

Answer: $\dfrac{1}{t}$ — this parametrisation of $y^2 = 4ax$ powers half the conic-section problems in coordinate geometry.

Example 3 — The cycloid and the half-angle collapse

Find $\dfrac{dy}{dx}$ if $x = a(\theta + \sin\theta)$, $y = a(1 - \cos\theta)$.

Step 1 — differentiate: $\frac{dx}{d\theta} = a(1 + \cos\theta)$, $\frac{dy}{d\theta} = a\sin\theta$.

Step 2 — divide: $\frac{dy}{dx} = \dfrac{\sin\theta}{1 + \cos\theta}$.

Step 3 — half-angle clean-up: $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ and $1 + \cos\theta = 2\cos^2\frac{\theta}{2}$, so the quotient is $\tan\frac{\theta}{2}$.

Answer: $\tan\dfrac{\theta}{2}$ — leaving the answer unsimplified loses the presentation mark.

Example 4 — The astroid, twice over

Find $\dfrac{dy}{dx}$ if $x^{2/3} + y^{2/3} = a^{2/3}$.

Step 1 — parametrise: $x = a\cos^3\theta$, $y = a\sin^3\theta$ satisfies the relation (check: $a^{2/3}(\cos^2 + \sin^2) = a^{2/3}$).

Step 2 — differentiate and divide: $$\frac{dy}{dx} = \frac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta} = -\tan\theta$$

Step 3 — translate back if asked: $\tan\theta = \left(\frac{y}{x}\right)^{1/3}$, so $\frac{dy}{dx} = -\sqrt[3]{\frac{y}{x}}$.

Answer: $-\tan\theta$, equivalently $-\sqrt[3]{y/x}$ — the same problem solved implicitly takes twice the algebra.

Example 5 — A scaled cycloid with the other half-angle

Find $\dfrac{dy}{dx}$ if $y = 12(1 - \cos t)$, $x = 10(t - \sin t)$.

Step 1 — differentiate: $\frac{dx}{dt} = 10(1 - \cos t)$, $\frac{dy}{dt} = 12\sin t$.

Step 2 — divide: $\frac{dy}{dx} = \frac{12\sin t}{10(1 - \cos t)}$.

Step 3 — half-angle (the $1 - \cos$ twin): $\sin t = 2\sin\frac{t}{2}\cos\frac{t}{2}$ and $1 - \cos t = 2\sin^2\frac{t}{2}$: $$\frac{dy}{dx} = \frac{6}{5}\cot\frac{t}{2}$$

Answer: $\dfrac{6}{5}\cot\dfrac{t}{2}$ — note $1 + \cos$ collapses to a tangent, $1 - \cos$ to a cotangent.

Example 6 — When the answer is a constant

Find $\dfrac{dy}{dx}$ if $x = a\cos\theta$, $y = b\cos\theta$.

Step 1 — differentiate and divide: $\frac{dy}{dx} = \frac{-b\sin\theta}{-a\sin\theta} = \frac{b}{a}$.

Answer: the constant $\dfrac{b}{a}$ — as it must be: eliminating $\theta$ gives $y = \frac{b}{a}x$, a straight line through the origin.