If $x = at^2$, $y = 2at$, then find $\dfrac{d^2y}{dx^2}$.

> [!TLDR] > To find the second derivative $\frac{d^2y}{dx^2}$ for parametric equations, we first find $\frac{dy}{dx}$ using the chain rule, and then differentiate this result with respect to $x$ again, applying the chain rule a second time. The final result is $\boxed{-\frac{1}{2at^3}}$. When dealing with parametric equations like $x = f(t)$ and $y = g(t)$, we cannot directly find $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$. Instead, we use the chain rule to relate the derivatives with respect to $t$. The core idea is that if $y$ is a function of $t$, and $t$ is a function of $x$, then $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$. Since we usually have $x$ as a function of $t$, we can write $\frac{dt}{dx} = \frac{1}{dx/dt}$. This leads to the fundamental formula for the first derivative in parametric form. > [!FORMULA] > For parametric equations $x = f(t)$ and $y = g(t)$, the first derivative $\frac{dy}{dx}$ is given by: > $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} $$ For the second derivative, $\frac{d^2y}{dx^2}$, we need to differentiate $\frac{dy}{dx}$ with respect to $x$. Since $\frac{dy}{dx}$ will typically be an expression in terms of $t$, we again apply the chain rule: > [!FORMULA] > The second derivative $\frac{d^2y}{dx^2}$ for parametric equations is given by: > $$ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt} $$ > This formula is crucial and often where students make mistakes. Let's apply this step-by-step to the given problem. 1. **Find the first derivatives with respect to $t$ for $x$ and $y$.** We are given $x = at^2$ and $y = 2at$. Differentiating $x$ with respect to $t$: $$ \frac{dx}{dt} = \frac{d}{dt}(at^2) = a \cdot (2t) = 2at $$ Differentiating $y$ with respect to $t$: $$ \frac{dy}{dt} = \frac{d}{dt}(2at) = 2a \cdot (1) = 2a $$ 2. **Calculate the first derivative $\frac{dy}{

If x = at², y = 2at, then find (d²y)/(dx²).

Question

x = at^2

CBSECBSE Class XII Board 2020Subjective· 2mImportance★★★★★

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Concept understanding — Parametric Second Derivative

Parametric Second Derivative

When a curve is given parametrically as $x = x(t)$ , $y = y(t)$ , its slope is

\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{y'(t)}{x'(t)},\qquad x'(t)\neq 0.

The second derivative $\dfrac{d^2y}{dx^2}$ measures how fast that slope changes — the concavity of the path. The catch is that $\frac{dy}{dx}$ comes out as a function of $t$ , but we need its rate of change with respect to $x$ .

The key idea

Differentiate the slope with respect to $t$ , then convert that $t$ -derivative into an $x$ -derivative by dividing by $dx/dt$ (chain rule):

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

Carrying this out with the quotient rule gives a compact closed form:

\frac{d^2y}{dx^2}=\frac{x'(t)\,y''(t)-y'(t)\,x''(t)}{\big[x'(t)\big]^3}.

Watch out

Do not write $\dfrac{d^2y}{dx^2}=\dfrac{d^2y/dt^2}{d^2x/dt^2}$ . The parametric second derivative is not the ratio of the second $t$ -derivatives — that tempting shortcut is wrong.

Worked illustration

For the cycloid $x=t-\sin t$ , $y=1-\cos t$ :

First derivatives: $\dfrac{dx}{dt}=1-\cos t$ , $\dfrac{dy}{dt}=\sin t$ , so $\dfrac{dy}{dx}=\dfrac{\sin t}{1-\cos t}$ .
Differentiate $\dfrac{dy}{dx}$ with respect to $t$ , then divide by $\dfrac{dx}{dt}=1-\cos t$ , which simplifies to $\frac{d^2y}{dx^2}=-\frac{1}{(1-\cos t)^2}.$ …

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To find the second derivative $\frac{d^2y}{dx^2}$ for parametric equations, we first find $\frac{dy}{dx}$ using the chain rule, and then differentiate this result with respect to $x$ again, applying the chain rule a second time. The final result is $\boxed{-\frac{1}{2at^3}}$ .

When dealing with parametric equations like $x = f(t)$ and $y = g(t)$ , we cannot directly find $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$ . Instead, we use the chain rule to relate the derivatives with respect to $t$ .

The core idea is that if $y$ is a function of $t$ , and $t$ is a function of $x$ , then $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$ . Since we usually have $x$ as a function of $t$ , we can write $\frac{dt}{dx} = \frac{1}{dx/dt}$ . This leads to the fundamental formula for the first derivative in parametric form.

x = f(t)

For the second derivative, $\frac{d^2y}{dx^2}$ , we need to differentiate $\frac{dy}{dx}$ with respect to $x$ . Since $\frac{dy}{dx}$ will typically be an expression in terms of $t$ , we again apply the chain rule:

\frac{d^2y}{dx^2}

Let's apply this step-by-step to the given problem.

Find the first derivatives with respect to $t$ for $x$ and $y$ . We are given $x = at^2$ and $y = 2at$ . Differentiating $x$ with respect to $t$ :

\frac{dx}{dt} = \frac{d}{dt}(at^2) = a \cdot (2t) = 2at

Differentiating $y$ with respect to $t$:

\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a \cdot (1) = 2a

Calculate the first derivative $\frac{dy}{dx}$ . Using the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ :

\frac{dy}{dx} = \frac{2a}{2at}

Assuming $t \neq 0$, we can simplify this expression:

\frac{dy}{dx} = \frac{1}{t}

Calculate the second derivative $\frac{d^2y}{dx^2}$ . Now we need to differentiate $\frac{dy}{dx}$ (which is $\frac{1}{t}$ ) with respect to $x$ . Since $\frac{1}{t}$ is a function of $t$ , we must use the chain rule: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$x = a\cos\theta$
(a) $-\cot\theta$
(b) $\tan\theta$
(c) $a\tan\theta$
(d) None of these
›Reveal solutionSolution

For a parametric curve, $\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta}$ .

$x = a\cos\theta \Rightarrow \dfrac{dx}{d\theta} = -a\sin\theta$

$y = a\sin\theta \Rightarrow \dfrac{dy}{d\theta} = a\cos\theta$

…

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CBSE 2026Set ANNUAL1 mark
$Q.If x=a sec \theta and y=b tan \theta, then find \frac{dy}{dx}.$
›Reveal solutionSolution
$\dfrac{dy}{dx}=\dfrac{b}{a}\csc\theta$ .

Concept. For a curve given parametrically as $x=f(\theta),\ y=g(\theta)$ , $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$ .

Steps.

$x=a\sec\theta\Rightarrow \dfrac{dx}{d\theta}=a\sec\theta\tan\theta$ .

$y=b\tan\theta\Rightarrow \dfrac{dy}{d\theta}=b\sec^2\theta$ .

$\dfrac{dy}{dx}=\dfrac{b\sec^2\theta}{a\sec\theta\tan\theta}=\dfrac{b\sec\theta}{a\tan\theta}$ . …
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CBSE 2025Set ANNUAL1 markMCQ
$x = a \sec\theta$
(a) $\dfrac{b}{a} \sec\theta$
(b) $\dfrac{b}{a} \text{cosec}\,\theta$
(c) $\dfrac{b}{a} \cot\theta$
(d) none of these
›Reveal solutionSolution

Differentiate x and y separately with respect to the parameter θ, then divide.

$x = a\sec\theta \Rightarrow \dfrac{dx}{d\theta} = a\sec\theta\tan\theta$

$y = b\tan\theta \Rightarrow \dfrac{dy}{d\theta} = b\sec^2\theta$

…

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CBSE 2024Set ANNUAL1 mark
$\dfrac{dy}{dx}$
›Reveal solutionSolution

Use parametric differentiation: $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ .

Given $x = 2at^2$ , $y = at^4$ .

$\frac{dx}{dt} = 4at, \qquad \frac{dy}{dt} = 4at^3$

So: …

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CBSE 2022Set ANNUAL1 markMCQ
$x = a\cos\theta$
(a) $\tan\theta$
(b) $-\cot\theta$
(c) $-\tan\theta$
(d) $\sec^2\theta$
›Reveal solutionSolution

Parametric differentiation gives $\frac{a\cos\theta}{-a\sin\theta}=-\cot\theta$ .

$\frac{dx}{d\theta}=-a\sin\theta$ , $\frac{dy}{d\theta}=a\cos\theta$ .

…

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CBSE 2022Set ANNUAL1 markMCQ
$x = at^2$
(a) $t$
(b) $\frac{1}{t}$
(c) $t^2$
(d) None of these
›Reveal solutionSolution

Parametric: $\frac{dy/dt}{dx/dt}=\frac{2a}{2at}=\frac1t$ .

$\frac{dx}{dt}=2at$ , $\frac{dy}{dt}=2a$ .

…

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CBSE 2018Set ANNUAL1 markMCQ
$x = t^{2}$
(a) $\dfrac{3}{2}$
(b) $\dfrac{3}{4t}$
(c) $\dfrac{3}{2t}$
(d) $\dfrac{3t}{2}$
›Reveal solutionSolution

For parametric curves, first find dy/dx as a function of t, then differentiate that again with respect to t and divide by dx/dt once more.

$x=t^2 \Rightarrow \dfrac{dx}{dt}=2t$ ; $y=t^3 \Rightarrow \dfrac{dy}{dt}=3t^2$

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3t^2}{2t} = \dfrac{3t}{2}$ …

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CBSE 2016Set ANNUAL1 markMCQ
$x=t^2$
(a) $\dfrac{4t}{3}$
(b) $\dfrac{2t}{3}$
(c) $\dfrac{3}{2t}$
(d) $\dfrac{3}{4t}$
›Reveal solutionSolution

second derivative of a parametric curve

$x=t^2\Rightarrow \dfrac{dx}{dt}=2t$ ; $\ y=t^3\Rightarrow \dfrac{dy}{dt}=3t^2$ .

$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3t^2}{2t}=\dfrac{3t}{2}$

Differentiate again with respect to $x$ (using $\dfrac{d}{dx}=\dfrac{1}{dx/dt}\dfrac{d}{dt}$ ):

…

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Q.If x=at2x = at^2x=at2, y=2aty = 2aty=2at, then find d2ydx2\dfrac{d^2y}{dx^2}dx2d2y​.

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Q.If $x = at^2$ , $y = 2at$ , then find $\dfrac{d^2y}{dx^2}$ .

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