Find $\dfrac{dy}{dx}$ of the function expressed in parametric form: $x = 3\cos\theta - 2\cos^3\theta,\ y = 3\sin\theta - 2\sin^3\theta$.

Find (dy)/(dx) of the function expressed in parame – Class 12 Mathematics

NCERT Exemplar · Q46

\dfrac{dy}{dx}

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For parametric equations, $\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta}$ . After differentiating and cancelling the common factor $\cos 2\theta$ , $\dfrac{dy}{dx} = \cot\theta$ .

Given $x = 3\cos\theta - 2\cos^3\theta$ and $y = 3\sin\theta - 2\sin^3\theta$ , differentiate each with respect to $\theta$ .

Differentiate $x$ (chain rule on $\cos^3\theta$ gives $3\cos^2\theta\cdot(-\sin\theta)$ ):

\frac{dx}{d\theta} = -3\sin\theta + 6\cos^2\theta\,\sin\theta = 3\sin\theta\,(2\cos^2\theta - 1) = 3\sin\theta\,\cos 2\theta.

Differentiate $y$ :

$\frac{dy}{d\theta} = 3\cos\theta - 6\sin^2\theta\,\cos\theta = 3\cos\theta\,(1 - 2\sin^2\theta) = 3\cos\theta\,\cos 2\theta.$ …

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

CA Foundation 2026Set may-20261 markMCQ
$x(m) = am^2$
›Reveal solutionSolution

Parametric differentiation: $\dfrac{dy}{dx}=\dfrac{dy/dm}{dx/dm}=\dfrac{-2a/m^3}{2am}=-\dfrac{1}{m^4}$ .

Step 1 — Differentiate $x$ with respect to $m$
$x=am^2\ \Rightarrow\ \frac{dx}{dm}=2am$
Step 2 — Differentiate $y$ with respect to $m$
$y=\frac{a}{m^2}=am^{-2}\ \Rightarrow\ \frac{dy}{dm}=-2am^{-3}=-\frac{2a}{m^3}$
Step 3 — Form the ratio

$\frac{dy}{dx}=\frac{dy/dm}{dx/dm}=\frac{-2a/m^3}{2am}=\frac{-2a}{m^3}\cdot\frac{1}{2am}=-\frac{1}{m^4}$ …
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GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$x = a(\theta - \sin\theta)$
(a) $-\cot\left(\dfrac{\theta}{2}\right)$
(b) $\tan\left(\dfrac{\theta}{2}\right)$
(c) $\cot\left(\dfrac{\theta}{2}\right)$
(d) $-\tan\left(\dfrac{\theta}{2}\right)$
›Reveal solutionSolution

For a parametric curve, $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$ ; the half-angle identities simplify the result.

$\dfrac{dx}{d\theta}=a(1-\cos\theta)$ , $\dfrac{dy}{d\theta}=a\sin\theta$ .

…

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CA Foundation 2025Set jan-20251 markMCQ
$x = at^2$
›Reveal solutionSolution

Parametric derivative: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{a(3t^2-1)}{2at} = \dfrac{3t^2-1}{2t}$ .

Step 1 — Differentiate $y$ with respect to $t$
$y = a(t^3 - t) \;\Rightarrow\; \frac{dy}{dt} = a(3t^2 - 1)$
Step 2 — Differentiate $x$ with respect to $t$
$x = at^2 \;\Rightarrow\; \frac{dx}{dt} = 2at$
Step 3 — Divide to get $dy/dx$
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a(3t^2 - 1)}{2at} = \frac{3t^2 - 1}{2t}$
Why the other options are wrong: (B) mis-differentiates $t^3-t$ as $3t^2-t$ ; (C) drops the 2 in the denominator; (D) has a sign error ( $+1$ instead of $-1$ ). …
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CA Foundation 2025Set may-20251 markMCQ
$\frac{dy}{dx}$
›Reveal solutionSolution

$\dfrac{dy}{dt}=x$ , $\dfrac{dx}{dt}=y$ , so $\dfrac{dy}{dx}=\dfrac{x}{y}$ .

Step 1 — Differentiate x and y with respect to the parameter t
$\frac{dx}{dt}=\frac{d}{dt}\!\left(\frac{e^t+e^{-t}}{2}\right)=\frac{e^t-e^{-t}}{2}=y$ $\frac{dy}{dt}=\frac{d}{dt}\!\left(\frac{e^t-e^{-t}}{2}\right)=\frac{e^t+e^{-t}}{2}=x$
Step 2 — Use parametric differentiation
$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{x}{y}$
Why the other options are wrong: (A) $\frac{y}{x}$ inverts the ratio (swaps numerator and denominator); (C) and (D) do not simplify from the two parametric derivatives. …
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CA Foundation 2024Set sep-20241 markMCQ
$x = t^2$
›Reveal solutionSolution

dy/dx = 3t/2; differentiating again and dividing by dx/dt gives d²y/dx² = 3/(4t).

Step 1 — First derivative (parametric)
$\dfrac{dx}{dt} = 2t, \qquad \dfrac{dy}{dt} = 3t^2$ $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3t^2}{2t} = \dfrac{3t}{2}$
Step 2 — Second derivative (parametric)

Differentiate dy/dx with respect to t, then divide by dx/dt:

$\dfrac{d^2y}{dx^2} = \dfrac{\dfrac{d}{dt}\!\left(\dfrac{3t}{2}\right)}{\dfrac{dx}{dt}} = \dfrac{3/2}{2t} = \dfrac{3}{4t}$ …
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GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$x = at^2, y = 2at$
(a) $\frac{1}{t}$
(b) $t$
(c) $-t$
(d) $a$
›Reveal solutionSolution

For a parametric curve, $\frac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ .

…

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Q.Find $\dfrac{dy}{dx}$ of the function expressed in parametric form: $x = 3\cos\theta - 2\cos^3\theta,\ y = 3\sin\theta - 2\sin^3\theta$ .

Concept understanding — Parametric Differentiation

Method: Parametric Differentiation with Power-of-Trig-Function Terms

Steps

Common Mistakes

Step 1 — Differentiate $x$ with respect to $m$

Step 2 — Differentiate $y$ with respect to $m$

Step 3 — Form the ratio

Step 1 — Differentiate $y$ with respect to $t$

Step 2 — Differentiate $x$ with respect to $t$

Step 3 — Divide to get $dy/dx$

Step 1 — Differentiate x and y with respect to the parameter t

Step 2 — Use parametric differentiation

Step 1 — First derivative (parametric)

Step 2 — Second derivative (parametric)

Q.Find dydx\dfrac{dy}{dx}dxdy​ of the function expressed in parametric form: x=3cos⁡θ−2cos⁡3θ, y=3sin⁡θ−2sin⁡3θx = 3\cos\theta - 2\cos^3\theta,\ y = 3\sin\theta - 2\sin^3\thetax=3cosθ−2cos3θ, y=3sinθ−2sin3θ.

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Q.Find $\dfrac{dy}{dx}$ of the function expressed in parametric form: $x = 3\cos\theta - 2\cos^3\theta,\ y = 3\sin\theta - 2\sin^3\theta$ .

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