Q.If $x = a\sin 2t\,(1 + \cos 2t)$ and $y = b\cos 2t\,(1 - \cos 2t)$, show that $\left.\dfrac{dy}{dx}\right|_{t = \frac{\pi}{4}} = \dfrac{b}{a}$.

Concept: Parametric Differentiation – find $\frac{dy}{dx}$ by computing $\frac{dy}{dt}$ and $\frac{dx}{dt}$ separately, then divide.

Step 1 – Differentiate $x$ with respect to $t$

$x = a\sin 2t\,(1 + \cos 2t)$

Using product rule:

$\frac{dx}{dt} = a[2\cos 2t\,(1+\cos 2t) + \sin 2t\,(-2\sin 2t)]$

$= 2a[\cos 2t + \cos^2 2t - \sin^2 2t]$

Since $\cos^2 2t - \sin^2 2t = \cos 4t$, we get

$\frac{dx}{dt} = 2a(\cos 2t + \cos 4t)$.

Step 2 – Differentiate $y$ with respect to $t$

$y = b\cos 2t\,(1 - \cos 2t)$

$\frac{dy}{dt} = b[-2\sin 2t\,(1-\cos 2t) + \cos 2t\,(2\sin 2t)]$

$= 2b[-\sin 2t + \sin 2t\cos 2t + \sin 2t\cos 2t]$

$= 2b(-\sin 2t + 2\sin 2t\cos 2t)$

$= 2b(-\sin 2t + \sin 4t)$.

Step 3 – Form $\frac{dy}{dx}$ and evaluate at $t = \frac{\pi}{4}$ …

For a curve given parametrically, $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ (provided $\dfrac{dx}{dt}\neq 0$).

Differentiate $x=a\sin 2t\,(1+\cos 2t)$.

Differentiate $y=b\cos 2t\,(1-\cos 2t)=b(\cos 2t-\cos^2 2t)$.

Evaluate at $t=\dfrac{\pi}{4}$, so $2t=\dfrac{\pi}{2}$ and $4t=\pi$: …

Method: Evaluating a Parametric Derivative at a Specific Parameter Value

Use this whenever a question asks for $\dfrac{dy}{dx}$ at one particular value of the parameter (e.g. $t=\tfrac{\pi}{4}$) rather than as a general expression in $t$.

Steps

Step 1: Differentiate $x$ and $y$ with respect to the parameter in general form first

Do NOT substitute the given value of $t$ yet. Differentiate fully, using the Product Rule where needed, and simplify using trig identities (e.g. double-angle formulas) to get $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ in their cleanest possible form.

Step 2: Form the general ratio

Keeping this as a general expression (rather than evaluating term-by-term too early) avoids arithmetic slips.

Step 3: Substitute the given parameter value LAST …

Common Mistakes

Mistake 1: Substituting $t=\pi/4$ before differentiating

Why it's wrong: plugging in the specific value of $t$ into $x$ and $y$ first turns them into constants, so there is nothing left to differentiate — the derivative must come from the general expressions. Correct approach: differentiate $x(t)$ and $y(t)$ symbolically first, and only substitute $t=\pi/4$ into the resulting $dx/dt$ and $dy/dt$ at the very end.

Mistake 2: Missing the chain-rule factor when differentiating $\cos^2 2t$

Why it's wrong: since $y=b\cos2t-b\cos^2 2t$, differentiating the second term as $2\cos2t\cdot(-\sin2t)$ without also multiplying by the derivative of the inner $2t$ (an extra factor of $2$) undercounts the rate of change. Correct approach: $\dfrac{d}{dt}\cos^2 2t=2\cos2t\cdot(-2\sin2t)=-4\sin2t\cos2t=-2\sin4t$. …