Q.Find $\frac{dy}{dx}$ in the following: $x = a \cos \theta, y = b \cos \theta$

The key idea is Implicit Differentiation (or parametric differentiation, since both $x$ and $y$ are given in terms of $\theta$).

Step 1: Differentiate each parametric equation with respect to $\theta$:

Step 2: Use the chain rule for parametric curves:

Step 3: Cancel $-\sin \theta$ (provided $\sin \theta \neq 0$):

When you see two equations like $x = a\cos\theta$ and $y = b\cos\theta$, the natural instinct might be to eliminate $\theta$ first. But that’s unnecessary here — and would actually obscure the simplicity. Both $x$ and $y$ are already functions of the same parameter $\theta$, so we can differentiate each with respect to $\theta$ and then take their ratio. This is the essence of parametric differentiation.

The key idea: if $x$ and $y$ are both given in terms of a third variable (the parameter), then

provided $dx/d\theta \neq 0$. This works because the chain rule lets us cancel $d\theta$ like a fraction — but only when both derivatives exist and the denominator is non-zero.

Let’s apply it step by step.

1. Differentiate $x$ with respect to $\theta$. $x = a\cos\theta$ The derivative of $\cos\theta$ is $-\sin\theta$, so

2. Differentiate $y$ with respect to $\theta$. $y = b\cos\theta$ Similarly,

3. Take the ratio.

4. Simplify. The $-\sin\theta$ cancels (provided $\sin\theta \neq 0$), leaving

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

Parametric answers built from $\sin,\cos$ of the parameter very often simplify with a double-angle or half-angle identity ($\sin 2\theta = 2\sin\theta\cos\theta$, $1-\cos\theta = 2\sin^2\frac{\theta}{2}$, $1+\cos\theta = 2\cos^2\frac{\theta}{2}$, etc.) — always look for one before leaving the answer as a raw ratio.

Applying to this problem: with $x=a\cos\theta,\ y=b\cos\theta$, both derivatives with respect to $\theta$ pick up the same factor $-\sin\theta$, so it cancels in the ratio, leaving the constant $\dfrac{b}{a}$ — a signal that the curve is really the straight line $y=\frac{b}{a}x$, whose slope never depends on $\theta$.

Common Mistakes

Mistake 1: Trying to eliminate $\theta$ before differentiating.

Why it's wrong: the elimination is unnecessary extra work here — both $x$ and $y$ already reduce to the same trig function of $\theta$, so the parametric-ratio method gives the slope in one line. Correct approach: differentiate $x$ and $y$ with respect to $\theta$ directly and divide.

Mistake 2: Cancelling $-\sin\theta$ without noting it must be nonzero.

Why it's wrong: at $\theta = 0,\pi$ the parametric derivatives are both $0$, so the ratio $\frac{b}{a}$ formally still holds only away from those points where $dx/d\theta=0$. Correct approach: state the cancellation is valid for $\sin\theta \neq 0$.