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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If $\text{x}=\text{a}\sin\text{t}\ \text{and}\ \text{y}=\text{a}(\cos\text{t}+\log\tan\frac{\text{t}}{2}),$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
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Answer

$\text{x}=\text{a}\sin\text{t}\ \text{and}\ \text{y}=\text{a}(\cos\text{t}+\log\tan\frac{\text{t}}{2}),$
$\frac{\text{dx}}{\text{dt}}=\text{a}\cos\text{t}$
$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{a}\sin\text{t}$
$\frac{\text{dy}}{\text{dt}}=-\text{a}\sin\text{t}+\text{a}\frac{1}{\tan\frac{\text{t}}{2}}\times\sec^2\frac{\text{t}}{2}\times\frac{1}{2}$
$=-\text{a}\sin\text{t}+\text{a}\frac{1}{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}$
$=-\text{a}\sin\text{t}+\text{a}\ \text{cosec}\ \text{t}$
$\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\text{a}\cos\text{t}-\text{a cosec t}\cot\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\frac{\text{dx}}{\text{dt}}\frac{\text{d}^2\text{y}}{\text{dt}^2}-\frac{\text{dy}}{\text{dt}}\frac{\text{d}^2\text{x}}{\text{dt}^2}}{\Big(\frac{\text{dx}}{\text{dt}}\Big)^3}$
$=\frac{\text{a}\cos\text{t}(-\text{a}\cos\text{t}-\text{a}\ \text{cosec t}\cot\text{t})-(-\text{a}\sin\text{t}+\text{a}\text{cosec t})(-\text{a}\sin\text{t})}{(\text{a}\cos\text{t})^3}$
$=\frac{-\text{a}^2\cos^2\text{t}-\text{a}^2\cot^2\text{t}-\text{a}^2\sin^2\text{t}+\text{a}^2}{\text{a}^3\cos^3\text{t}}$
$=\frac{-\text{a}^2\cos^2\text{t}-\text{a}^2\sin^2\text{t}-\text{a}^2\cot^2\text{t}+\text{a}^2}{\text{a}^3\cos^3\text{t}}$
$=\frac{-\text{a}^2(\cos^2\text{t}+\sin^2\text{t})-\text{a}^2\cot^2\text{t}+\text{a}^2}{\text{a}^3\cos^3\text{t}}$
$=-\frac{1}{\text{a}\sin^2\text{t}\cos\text{t}}$

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