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JEE Main 2025 · Mathematics · Differentiation and Application of Derivatives · medium

If x=2(t-sin t) and y=2(1-cos t), then (d^2y)/(dx^2 at t=(π)/(2) equals

Single correct option

If x=2(t−sin⁡t)x=2(t-\sin t) and y=2(1−cos⁡t)y=2(1-\cos t), then d2ydx2\dfrac{d^2y}{dx^2} at t=π2t=\dfrac{\pi}{2} equals
  1. A. 12\dfrac12
  2. B. −1-1
  3. C. −14-\dfrac14
  4. D. −12-\dfrac12
Show answer and solution

Answer: (D) −12-\dfrac12

Differentiate the parametric equations: dxdt=2(1−cos⁡t)\dfrac{dx}{dt}=2(1-\cos t) and dydt=2sin⁡t\dfrac{dy}{dt}=2\sin t, so dydx=sin⁡t1−cos⁡t=cot⁡t2\dfrac{dy}{dx}=\dfrac{\sin t}{1-\cos t}=\cot\dfrac t2. Next, d2ydx2=ddt(cot⁡t2)/dxdt=−12csc⁡2t22(1−cos⁡t)\dfrac{d^2y}{dx^2}=\dfrac{d}{dt}\left(\cot\dfrac t2\right)\Big/\dfrac{dx}{dt}=\dfrac{-\tfrac12\csc^2\tfrac t2}{2(1-\cos t)}. At t=π2t=\dfrac\pi2: csc⁡2π4=2\csc^2\dfrac\pi4=2 and 1−cos⁡π2=11-\cos\dfrac\pi2=1, so the value is −12⋅22⋅1=−12\dfrac{-\tfrac12\cdot2}{2\cdot1}=-\dfrac12. Forgetting to divide by dxdt\dfrac{dx}{dt} a second time gives the wrong value −1-1. Evaluating numerically near t=π2t=\dfrac\pi2 also gives approximately −0.5-0.5.
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