CUET UG 2025 — Mathematics Calculus

Given $x = t^{1/2}$x=t1/2 and $y = t^{3/2}$y=t3/2, find $\frac{dy}{dx}$dxdy​.

When both $x$x and $y$y depend on $t$t:

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

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For $x = t^{1/2}$x=t1/2:

$$\frac{dx}{dt} = \frac{1}{2} \cdot t^{1/2 - 1}$$dtdx​=21​⋅t1/2−1

$$\frac{dx}{dt} = \frac{1}{2}t^{-1/2}$$dtdx​=21​t−1/2

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For $y = t^{3/2}$y=t3/2:

$$\frac{dy}{dt} = \frac{3}{2} \cdot t^{3/2 - 1}$$dtdy​=23​⋅t3/2−1

$$\frac{dy}{dt} = \frac{3}{2}t^{1/2}$$dtdy​=23​t1/2

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

$$\frac{dy}{dx} = \frac{\frac{3}{2}t^{1/2}}{\frac{1}{2}t^{-1/2}}$$dxdy​=21​t−1/223​t1/2​

$$\frac{dy}{dx} = \frac{3}{2}t^{1/2} \times \frac{2}{t^{-1/2}}$$dxdy​=23​t1/2×t−1/22​

$$\frac{dy}{dx} = \frac{3}{2} \times 2 \times t^{1/2} \times t^{1/2}$$dxdy​=23​×2×t1/2×t1/2

$$\frac{dy}{dx} = 3 \times t^{1/2 + 1/2}$$dxdy​=3×t1/2+1/2

$$\frac{dy}{dx} = 3 \times t^{1}$$dxdy​=3×t1

$$\frac{dy}{dx} = 3t$$dxdy​=3t

Therefore, $\frac{dy}{dx} = 3t$dxdy​=3t.