Q96 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics

If $x=3 \sqrt{2} \cos ^3 \theta$ and $y=4 \tan ^2 \theta$ then $\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{4}}=$

  • A
    $\frac{32 \sqrt{2}}{9}$
  • B
    $\frac {16}{9}$
  • C
    $-\frac {16}{9}$
  • Correct answer
    D
    $\frac {32}{9}$

Solution

$x=3 \sqrt{2} \cos ^3 \theta$ $\begin{aligned} & \frac{d x^{\prime}}{d \theta}=3 \sqrt{2} \cdot 3 \cos ^2 \theta(-\sin \theta) \\ & y=4 \tan ^2 \theta\end{aligned}$ $\begin{aligned} & \frac{d y}{d \theta}=8 \tan \theta\left(\sec ^2 \theta\right) \\ & \begin{aligned} \frac{d y}{d x}= & \frac{d y / d \theta}{d x / d \theta}=\frac{8 \tan \theta \sec ^2 \theta}{-9 \sqrt{2} \sin \theta \cos ^2 \theta} \\ \left.\frac{d y}{d x}\right|_{\theta=\frac{\pi}{4}} & =\frac{8 \tan \frac{\pi}{4} \sec ^2 \frac{\pi}{4}}{-9 \sqrt{2} \cdot \sin \frac{\pi}{4} \cos ^2 \frac{\pi}{4}} \\ & =\frac{8 \times 1 \times 2}{-9 \sqrt{2} \cdot \frac{1}{\sqrt{2}} \times \frac{1}{2}}=\frac{-32}{9}\end{aligned}\end{aligned}$

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

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