MHT CET · Maths · Differentiation
The second derivative of a \(\sin ^3 t\) w.r.t \(a \cos ^3 t\) at \(t=\frac{\pi}{4}\) is
- A \(\frac{-4 \sqrt{2}}{3 a}\)
- B \(\frac{-1}{12 a}\)
- C \(\frac{4 \sqrt{2}}{3 a}\)
- D \(\frac{1}{12 a}\)
Answer & Solution
Correct Answer
(C) \(\frac{4 \sqrt{2}}{3 a}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{\mathrm{d}^2\left(a \sin ^3 t\right)}{\alpha\left(a \cos ^3 t\right)^2}=\frac{d\left(\frac{\mathrm{d}\left(a \sin ^3 t\right)}{\mathrm{d}\left(a \cos ^3 t\right)}\right)}{\mathrm{d}\left(a \cos ^3 t\right)}=\frac{\mathrm{d}\left\{\frac{\mathrm{d}\left(a \sin ^3 t\right) / \mathrm{d} t}{\mathrm{~d}\left(a \cos ^3 t / \mathrm{d} t\right)}\right\}}{\mathrm{d}\left(a \cos ^3 t\right)} \\ & =\frac{\mathrm{d}\left\{\frac{3 a \sin ^2 t \cdot \cos t}{-a \cdot 3 \cdot \cos ^2 t \cdot \sin t}\right\}}{\mathrm{d}\left(a \cos ^3 t\right)}=\frac{\mathrm{d}\{-\tan t\}}{\mathrm{d}\left(a \cos ^3 t\right)} \\ & =\frac{\frac{\mathrm{d}(-\tan t)}{\mathrm{d} t}}{\frac{\mathrm{d}\left(a \cos ^3 t\right)}{\mathrm{d} t}}=\frac{-\sec ^2 t}{-3 a \cos ^2 t \sin t}=\frac{1}{3 a \cos ^4 t \cdot \sin t} \\ & \text { at, } t=\frac{\pi}{4} ; \frac{1}{3 a \cos ^4 \frac{\pi}{4} \cdot \sin \frac{\pi}{4}}=\frac{1}{3 a\left(\frac{1}{\sqrt{2}}\right)^4 \frac{1}{\sqrt{2}}}=\frac{4 \sqrt{2}}{3 a}\end{aligned}\)
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