CUET UG 2025 — Mathematics Calculus

Given $x = t^2$x=t2 and $y = t^3$y=t3.

To find $\frac{dy}{dx}$dxdy​ for parametric equations:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$dxdy​=dx/dtdy/dt​

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Finding $\frac{dy}{dt}$dtdy​:

$$y = t^3$$y=t3

$$\frac{dy}{dt} = 3t^2$$dtdy​=3t2

Finding $\frac{dx}{dt}$dtdx​:

$$x = t^2$$x=t2

$$\frac{dx}{dt} = 2t$$dtdx​=2t

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Therefore:

$$\frac{dy}{dx} = \frac{3t^2}{2t}$$dxdy​=2t3t2​

$$\frac{dy}{dx} = \frac{3t}{2}$$dxdy​=23t​

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To find $\frac{d^2y}{dx^2}$dx2d2y​:

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

Finding $\frac{d}{dt}\left(\frac{3t}{2}\right)$dtd​(23t​):

$$\frac{d}{dt}\left(\frac{3t}{2}\right) = \frac{3}{2}$$dtd​(23t​)=23​

Finding $\frac{dt}{dx}$dxdt​:

$$\frac{dt}{dx} = \frac{1}{dx/dt}$$dxdt​=dx/dt1​

$$\frac{dt}{dx} = \frac{1}{2t}$$dxdt​=2t1​

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Therefore:

$$\frac{d^2y}{dx^2} = \frac{3}{2} \times \frac{1}{2t}$$dx2d2y​=23​×2t1​

$$\frac{d^2y}{dx^2} = \frac{3}{4t}$$dx2d2y​=4t3​