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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives5 Marks
Question
If $\text{x}=\text{a}(\cos\text{t}+\log\tan\frac{\text{t}}{2})\ \text{and}\ \text{y}=\text{a}(\sin\text{t}),$evaluate $\frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{3}.$
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Answer

We have,
$\text{x}=\text{a}(\cos\text{t}+\log\tan\frac{\text{t}}{2})\ \text{and}\ \text{y}=\text{a}(\sin\text{t}),$
On differentiating with respect to t, we get
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\text{a}\Big(\cos\text{t}+\log\tan\frac{\text{t}}{2}\Big)\Big]=\text{a}\Bigg(-\sin\text{t}+\frac{1}{\tan\frac{\text{t}}{2}}\times\sec^2\frac{\text{t}}{2}\times\frac{1}{2}\Bigg)$
$=\text{a}\Bigg(-\sin\text{t}+\frac{1}{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}\Bigg)=\text{a}\Big(-\sin\text{t}+\frac{1}{\sin\text{t}}\Big)$
$=\text{a}\Big(\frac{-\sin^2\text{t}+1}{\sin\text{t}}\Big)=\text{a}\Big(\frac{\cos^2\text{t}}{\sin\text{t}}\Big)$
And
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\sin\text{t})=\text{a}\cos\text{t}$
Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dX}}{\text{dt}}}=\frac{\text{a}\cos\text{t}}{\text{a}\frac{\cos^2\text{t}}{\sin\text{t}}}=\tan\text{t}$
Therefore
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}(\tan(\text{t}))$
$=\frac{\text{d}}{\text{dt}}(\tan(\text{t}))\times\frac{\text{dt}}{\text{dx}}=\sec^2\text{t}\times\frac{\sin\text{t}}{\text{a}\cos^2\text{t}}$
$=\Big(\frac{\sin\text{t}}{\text{a}\cos^4\text{t}}\Big)$
$\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)_{\text{t}=\frac{\pi}{3}}=\Bigg(\frac{\sin\Big(\frac{\pi}{3}\Big)}{\text{a}\cos^4\Big(\frac{\pi}{3}\Big)}\Bigg)=\frac{\frac{\sqrt{3}}{2}}{\text{a}\Big(\frac{1}{16}\Big)}=\frac{8\sqrt{3}}{\text{a}}$
Hence, $\text{at}\ \text{t}=\frac{\pi}{3},\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{8\sqrt{3}}{\text{a}}$

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