CUET UG 2025 — Mathematics Calculus

For parametric equations where both $x$x and $y$y depend on parameter $t$t:

$$x = 4t$$x=4t

$$y = \frac{4}{t}$$y=t4​

The first derivative is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$dxdy​=dx/dtdy/dt​

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$$\frac{dx}{dt} = 4$$dtdx​=4

$$y = 4t^{-1}$$y=4t−1

$$\frac{dy}{dt} = -4t^{-2}$$dtdy​=−4t−2

$$\frac{dy}{dt} = -\frac{4}{t^2}$$dtdy​=−t24​

Therefore:

$$\frac{dy}{dx} = \frac{-4/t^2}{4}$$dxdy​=4−4/t2​

$$\frac{dy}{dx} = -\frac{1}{t^2}$$dxdy​=−t21​

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For the second derivative: $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \times \frac{dt}{dx}$dx2d2y​=dtd​(dxdy​)×dxdt​

$$\frac{dy}{dx} = -t^{-2}$$dxdy​=−t−2

$$\frac{d}{dt}\left(-t^{-2}\right) = 2t^{-3}$$dtd​(−t−2)=2t−3

$$\frac{d}{dt}\left(-t^{-2}\right) = \frac{2}{t^3}$$dtd​(−t−2)=t32​

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Since $\frac{dx}{dt} = 4$dtdx​=4:

$$\frac{dt}{dx} = \frac{1}{4}$$dxdt​=41​

Therefore:

$$\frac{d^2y}{dx^2} = \frac{2}{t^3} \times \frac{1}{4}$$dx2d2y​=t32​×41​

$$\frac{d^2y}{dx^2} = \frac{2}{4t^3}$$dx2d2y​=4t32​

$$\frac{d^2y}{dx^2} = \frac{1}{2t^3}$$dx2d2y​=2t31​