CUET UG 2025 — Mathematics Calculus

Given $x = at^2$x=at2 and $y = 2at$y=2at, where both are expressed in terms of parameter $t$t.

For parametric equations, the first derivative is:

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

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Finding $\frac{dy}{dt}$dtdy​:

$$y = 2at$$y=2at

$$\frac{dy}{dt} = 2a$$dtdy​=2a

Finding $\frac{dx}{dt}$dtdx​:

$$x = at^2$$x=at2

$$\frac{dx}{dt} = 2at$$dtdx​=2at

Therefore:

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

$$\frac{dy}{dx} = \frac{2a}{2at}$$dxdy​=2at2a​

$$\frac{dy}{dx} = \frac{1}{t}$$dxdy​=t1​

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For the second derivative in parametric form:

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

Finding $\frac{d}{dt}\left(\frac{dy}{dx}\right)$dtd​(dxdy​):

$$\frac{d}{dt}\left(\frac{1}{t}\right) = \frac{d}{dt}(t^{-1})$$dtd​(t1​)=dtd​(t−1)

$$= -t^{-2}$$=−t−2

$$= -\frac{1}{t^2}$$=−t21​

Finding $\frac{dt}{dx}$dxdt​:

$$\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}}$$dxdt​=dtdx​1​

$$\frac{dt}{dx} = \frac{1}{2at}$$dxdt​=2at1​

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Therefore:

$$\frac{d^2y}{dx^2} = -\frac{1}{t^2} \times \frac{1}{2at}$$dx2d2y​=−t21​×2at1​

$$\frac{d^2y}{dx^2} = -\frac{1}{2at^3}$$dx2d2y​=−2at31​