Q.If $x = 3\sin t - \sin 3t$, $y = 3\cos t - \cos 3t$, find $\dfrac{dy}{dx}$ at $t = \dfrac{\pi}{3}$.

Concept: Implicit Differentiation (parametric form)

We have $x$ and $y$ given as functions of $t$. The derivative $\frac{dy}{dx}$ is found using

First, differentiate each with respect to $t$:

Thus

At $t = \frac{\pi}{3}$: …

The core idea here is parametric differentiation. When $x$ and $y$ are both given in terms of a third variable (here $t$), you cannot directly write $y$ as a function of $x$. Instead, you find the slope $\frac{dy}{dx}$ by dividing the rate of change of $y$ with respect to $t$ by the rate of change of $x$ with respect to $t$:

This works because the $dt$ terms cancel, just like fractions. The trick is to compute each derivative carefully, then simplify before plugging in the value of $t$.

Let’s work through it step by step.

1. Differentiate $x$ with respect to $t$. $x = 3\sin t - \sin 3t$ Using standard derivatives: $\frac{d}{dt}(\sin at) = a\cos at$

2. Differentiate $y$ with respect to $t$. $y = 3\cos t - \cos 3t$ Using $\frac{d}{dt}(\cos at) = -a\sin at$

3. Form the ratio for $\frac{dy}{dx}$.

4. Simplify using trigonometric identities.

   Use the sum-to-product formulas:

   • $\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$
   • $\cos B - \cos A = 2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$ (note the order)

   For numerator: $\sin 3t - \sin t = 2\cos\frac{3t+t}{2}\sin\frac{3t-t}{2} = 2\cos 2t \sin t$

   For denominator: $\cos t - \cos 3t = 2\sin\frac{t+3t}{2}\sin\frac{3t-t}{2} = 2\sin 2t \sin t$ …

Method: Simplifying Parametric Derivatives with Sum-to-Product Identities

Use this when differentiating $x(t)$ and $y(t)$ produces a difference of two sine or cosine terms with different multiples of the parameter (e.g. $\sin 3t - \sin t$) — a sum-to-product identity converts this into a product, which is much easier to simplify and cancel.

Steps

Step 1: Differentiate $x$ and $y$ term by term

Use $\dfrac{d}{dt}\sin(kt) = k\cos(kt)$ and $\dfrac{d}{dt}\cos(kt) = -k\sin(kt)$ for each term, then factor out any common numerical coefficient.

Step 2: Rewrite each difference using a sum-to-product formula

Apply the correct one to the numerator-type expression and to the denominator-type expression, matching $A$ and $B$ to the actual multiples of $t$ present.

Step 3: Cancel the shared factor …

Common Mistakes

Mistake 1: Misremembering the sign convention in the sum-to-product identity

Why it's wrong: for $\cos t-\cos3t$, writing it as $-2\sin\left(\tfrac{t+3t}{2}\right)\sin\left(\tfrac{3t-t}{2}\right)$ (the formula for $\cos A-\cos B$ with $A<B$) instead of correctly ordering it as $2\sin\left(\tfrac{t+3t}{2}\right)\sin\left(\tfrac{3t-t}{2}\right)$ flips the sign of the whole denominator. Correct approach: carefully track which angle is $A$ and which is $B$ in $\cos A-\cos B=2\sin\frac{A+B}{2}\sin\frac{B-A}{2}$ rather than applying the formula on autopilot.

Mistake 2: Cancelling the common factor $\sin t$ without checking it's nonzero …