Q.Find $\frac{dy}{dx}$ in the following: $x = a (\theta - \sin \theta), y = a (1 + \cos \theta)$

The key idea is that both $x$ and $y$ are given as functions of a common parameter $\theta$, so we use parametric differentiation:

Step 1: Differentiate $x$ with respect to $\theta$:

Step 2: Differentiate $y$ with respect to $\theta$:

Step 3: Form the ratio:

$\frac{dy}{dx} = \frac{-a \sin \theta}{a (1 - \cos \theta)} = \frac{-\sin \theta}{1 - \cos \theta}$ …

The problem gives $x$ and $y$ as functions of a parameter $\theta$, not directly as $y$ in terms of $x$. This is a classic parametric differentiation setup. Instead of trying to eliminate $\theta$ (which would be messy here), we use the chain rule in reverse:

provided $dx/d\theta \neq 0$. This works because both $x$ and $y$ are functions of $\theta$, so their derivatives with respect to $\theta$ are straightforward.

Let’s go step by step.

1. Differentiate $x$ with respect to $\theta$. $x = a(\theta - \sin\theta)$. The derivative of $\theta$ is $1$, and the derivative of $-\sin\theta$ is $-\cos\theta$. So:

2. Differentiate $y$ with respect to $\theta$. $y = a(1 + \cos\theta)$. The derivative of $1$ is $0$, and the derivative of $\cos\theta$ is $-\sin\theta$. So:

3. Apply the parametric formula.

The $a$ cancels neatly.

4. Simplify using trigonometric identities. The expression $-\frac{\sin\theta}{1 - \cos\theta}$ can be rewritten. Recall the half-angle identities:

Substitute these in: …

Method: Differentiating a Parametric Curve

This method applies whenever a curve is given through a third variable (a parameter — commonly $t$ or $\theta$) instead of $y$ written directly as a function of $x$.

Steps

Step 1: Recognise the parametric form

If you are given $x = f(\text{param})$ and $y = g(\text{param})$ instead of $y = h(x)$, do not try to eliminate the parameter first — it is often messy or impossible. Differentiate each equation separately with respect to the parameter instead.

Step 2: Differentiate $x$ and $y$ with respect to the parameter

Use the ordinary rules (product rule, chain rule, standard derivatives) to find $\dfrac{dx}{d\theta}$ (or $\dfrac{dx}{dt}$) and $\dfrac{dy}{d\theta}$ (or $\dfrac{dy}{dt}$).

Step 3: Divide — the parametric-derivative formula

This is justified by the chain rule: $\dfrac{dy}{d\theta} = \dfrac{dy}{dx}\cdot\dfrac{dx}{d\theta}$, so dividing recovers $\dfrac{dy}{dx}$.

Step 4: Simplify with trigonometric identities where possible …

Common Mistakes

Mistake 1: Using the wrong half-angle identity for $1-\cos\theta$.

Why it's wrong: $1-\cos\theta=2\sin^2\frac{\theta}{2}$, while $1+\cos\theta=2\cos^2\frac{\theta}{2}$ — swapping them (a very common mix-up) gives $-\tan\frac{\theta}{2}$ instead of $-\cot\frac{\theta}{2}$. Correct approach: memorise which identity pairs with which sign and double-check by picking a test value of $\theta$.

Mistake 2: Losing the negative sign from $\frac{dy}{d\theta}=-a\sin\theta$. …