Q107 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Mathematics

If $x=a(t+\sin t)$ and $y=a(1-\cos t)$, then $\frac{d^2 y}{d x^2}$

  • A
    $\frac{1}{4 a \sin ^4\left(\frac{t}{2}\right)}$
  • Correct answer
    B
    $\frac{1}{4 a \cos ^4\left(\frac{t}{2}\right)}$
  • C
    $4 a \operatorname{cosec}^4\left(\frac{t}{2}\right)$
  • D
    $4 a \sec ^4\left(\frac{t}{2}\right)$

Solution

$\begin{gathered}x=a(t+\sin t) \text { and } y=a(1-\cos t) \\ \frac{d x}{d t}=a(1+\cos t), \frac{d y}{d t}=a \sin t \\ \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a \sin t}{a(1+\cos t)}=\frac{2 \sin t / 2 \cos t / 2}{2 \cos ^2 t / 2} \\ \frac{d y}{d x}=\tan \frac{t}{2} \\ \frac{d^2 y}{d x^2}=\sec ^2 t / 2 \cdot \frac{1}{2} \cdot \frac{d t}{d x} \\ =\frac{\sec ^2 \frac{t}{2}}{2 a(1+\cos t)}=\frac{1}{4 a \cos ^4(t / 2)} .\end{gathered}$

Hence, the correct answer is option: 'B'.

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