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Second derivative d2y/dx2 of a curve given in parametric form

This type of question, asked 8 times: 15 May 2023 S1, 15 May 2023 S2, 14 May 2024 S2, 21 Apr 2025 S1, 19 Apr 2026 S1, 20 Apr 2026 S2, 12 May 2026 S1, 12 May 2026 S2. All 8 questions below, oldest first.

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  1. Question 1

    15 May 2023 S1

    133
    If x=2cos⁡θ−cos⁡2θx=2\cos\theta-\cos 2\theta and y=2sin⁡θ−sin⁡2θy=2\sin\theta-\sin 2\theta, then d2ydx2\frac{d^{2}y}{dx^{2}} is equal to

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  2. Question 2

    15 May 2023 S2

    101
    If x=sin⁡θx = \sin\theta, y=sin⁡3θy = \sin^3\theta, then d2ydx2\frac{d^2y}{dx^2} at θ=π2\theta = \frac{\pi}{2} is

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  3. Question 3

    14 May 2024 S2

    121
    If x=esin⁡−1tx = e^{\sin^{-1}t} and y=ecos⁡−1ty = e^{\cos^{-1}t}, then d2ydx2\frac{d^2y}{dx^2} is

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  4. Question 4

    21 Apr 2025 S1

    134
    If x=log⁡t,t>0x= \log t,t> 0 and y=1ty=\frac{1}{t} then d2ydx2=\frac{d^{2}y}{dx^{2}}=

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  5. Question 5

    19 Apr 2026 S1

    72
    Let x=at2−1x = at^2 - 1, where a>0a > 0 and y=t3+1y = t^3 + 1. If at t=1t = 1, d2ydx2=316\dfrac{d^2y}{dx^2} = \dfrac{3}{16}, then the value of aa is...

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  6. Question 6

    20 Apr 2026 S2

    74
    If x=asin⁡3tx = a\sin^3 t and y=acos⁡3ty = a\cos^3 t, then the value of d2ydx2\dfrac{d^2y}{dx^2} at t=π3t = \dfrac{\pi}{3} is equal to

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  7. Question 7

    12 May 2026 S1

    72
    If x=asin⁡t−bcos⁡t,y=acos⁡t+bsin⁡tx = a\sin t - b\cos t, y = a\cos t + b\sin t and it is given that d2ydx2=0\dfrac{d^2y}{dx^2} = 0, then

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  8. Question 8

    12 May 2026 S2

    71
    The second derivative of asin⁡3ta\sin^3 t with respect to acos⁡3ta\cos^3 t at t=π4t = \dfrac{\pi}{4} is...

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