The second derivative of a $\sin ^3 t$ w.r.t $a \cos ^3 t$ at $t=\frac{\pi}{4}$ is

The second derivative of a $\sin ^3 t$ w.r.t $a \cos ^3 t$ at $t=\frac{\pi}{4}$ is
  1. $\frac{-4 \sqrt{2}}{3 a}$
  2. $\frac{4 \sqrt{2}}{9 a}$
  3. $\frac{4 \sqrt{2}}{3 a}$
  4. $\frac{1}{12 a}$

Solution

$\begin{aligned} & \frac{\mathrm{d}^2\left(a \sin ^3 t\right)}{\alpha\left(a \cos ^3 t\right)^2}=\frac{d\left(\frac{\mathrm{d}\left(a \sin ^3 t\right)}{\mathrm{d}\left(a \cos ^3 t\right)}\right)}{\mathrm{d}\left(a \cos ^3 t\right)}=\frac{\mathrm{d}\left\{\frac{\mathrm{d}\left(a \sin ^3 t\right) / \mathrm{d} t}{\mathrm{~d}\left(a \cos ^3 t / \mathrm{d} t\right)}\right\}}{\mathrm{d}\left(a \cos ^3 t\right)} \\ & =\frac{\mathrm{d}\left\{\frac{3 a \sin ^2 t \cdot \cos t}{-a \cdot 3 \cdot \cos ^2 t \cdot \sin t}\right\}}{\mathrm{d}\left(a \cos ^3 t\right)}=\frac{\mathrm{d}\{-\tan t\}}{\mathrm{d}\left(a \cos ^3 t\right)} \\ & =\frac{\frac{\mathrm{d}(-\tan t)}{\mathrm{d} t}}{\frac{\mathrm{d}\left(a \cos ^3 t\right)}{\mathrm{d} t}}=\frac{-\sec ^2 t}{-3 a \cos ^2 t \sin t}=\frac{1}{3 a \cos ^4 t \cdot \sin t} \\ & \text { at, } t=\frac{\pi}{4} ; \frac{1}{3 a \cos ^4 \frac{\pi}{4} \cdot \sin \frac{\pi}{4}}=\frac{1}{3 a\left(\frac{1}{\sqrt{2}}\right)^4 \frac{1}{\sqrt{2}}}=\frac{4 \sqrt{2}}{3 a}\end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 1)

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