Q.If $x = a (\cos t + t \sin t)$ and $y = a (\sin t - t \cos t)$, find $\frac{d^2y}{dx^2}$.

Concept: Parametric Second Derivative — we compute $\frac{dy}{dx}$ as $\frac{dy/dt}{dx/dt}$, then differentiate again with respect to $x$ using $\frac{d}{dx} = \frac{d/dt}{dx/dt}$.

Step 1: First derivatives

Step 2: First derivative

Step 3: Second derivative …

When you see a curve given in parametric form — $x$ and $y$ each as functions of a third variable $t$ — the first derivative $\frac{dy}{dx}$ is straightforward: it's $\frac{dy/dt}{dx/dt}$, provided $dx/dt \neq 0$. But the second derivative is trickier. You cannot just differentiate $\frac{dy}{dx}$ with respect to $x$ directly, because $\frac{dy}{dx}$ is itself a function of $t$, not $x$.

The key insight: treat $\frac{dy}{dx}$ as a function of $t$, then use the chain rule again. Since $\frac{d}{dx} = \frac{d/dt}{dx/dt}$, we have:

This is the parametric second derivative formula. It's not just a mechanical step — it's the chain rule applied twice: first to get $\frac{dy}{dx}$, then to differentiate that result with respect to $x$ via $t$.

Let's apply it to the given equations.

1. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$

We have:

Differentiate $x$ with respect to $t$:

Notice the $\sin t$ terms cancel — that's neat. Similarly for $y$:

The $\cos t$ terms cancel here too. So both derivatives are simple products.

2. Find $\frac{dy}{dx}$

Using the parametric formula:

This is beautifully simple — the $a$ and $t$ cancel, leaving just $\tan t$. …

Method: Second Derivative of a Parametric Curve

For a curve given by $x=f(t),\ y=g(t)$, the second derivative $\dfrac{d^2y}{dx^2}$ is not obtained by dividing $\dfrac{d^2y}{dt^2}$ by $\dfrac{d^2x}{dt^2}$ — that is a common trap. It must be built from the first derivative.

Steps

Step 1: Find $\dfrac{dy}{dx}$ as usual

Step 2: Recognise that $\dfrac{dy}{dx}$ is itself a function of $t$, not of $x$

To differentiate it again with respect to $x$, you must route through $t$ using the chain rule a second time:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{dx/dt}.$ …

Common Mistakes

Mistake 1: Differentiating $\dfrac{dy}{dx}$ directly with respect to $x$, ignoring that it's a function of $t$.

Why it's wrong: $\frac{dy}{dx}=\tan t$ cannot be differentiated with respect to $x$ using ordinary single-variable rules — $t$ is the actual independent variable here, so the chain rule through $t$ is mandatory. Correct approach: always route the second differentiation through $\frac{d}{dt}\left(\frac{dy}{dx}\right)\Big/\frac{dx}{dt}$.

Mistake 2: Forgetting the extra division by $dx/dt$ after differentiating $\tan t$. …