l Find d^2 y d x^2 : x=a ( t+ t 2 ), y=a t

Question

$\begin{array}{l}\text { Find } \frac{d^2 y}{d x^2}: \\ x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t\end{array}$

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Answer

Differentiating with respect to $t$,
$\begin{aligned}
\frac{d x}{d t} & =a\left(-\sin t+\frac{1}{\tan \frac{t}{2}} \cdot \sec ^2 \frac{t}{2} \cdot \frac{1}{2}\right) \\
& =a\left(-\sin t+\frac{1}{\frac{\sin \frac{t}{2}}{\cos \frac{t}{2}}} \cdot \frac{1}{\cos ^2 \frac{t}{2}} \cdot \frac{1}{2}\right) \\
& =a\left(-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right) \\
& =a\left(-\sin t+\frac{1}{\sin t}\right) \\
& =a\left(\frac{-\sin ^2 t+1}{\sin t}\right) \\
& =\frac{a \cos ^2 t}{\sin t}
\end{aligned}$
$\begin{aligned} \frac{d y}{d t} & =\frac{d}{d t}(a \sin t) \\ & =a \cos t \\ \frac{d y}{d x} & =\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ & =\frac{a \cos t}{\frac{a \cos ^2 t}{\sin t}} \\ & =\frac{\sin t}{\cos t}\end{aligned}$
$\begin{array}{l}\therefore \quad \frac{d y}{d x}=\tan t \\ \therefore \quad \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}(\tan t)\end{array}$
$\begin{array}{l}=\sec ^2 t \frac{d t}{d x} \\ =\frac{1}{\cos ^2 t} \cdot \frac{1}{\frac{d x}{d t}} \\ =\frac{1}{\cos ^2 t} \cdot \frac{1}{\frac{a \cos ^2 t}{\sin t}} \\ =\frac{1}{a} \sec ^3 t \cdot \tan t\end{array}$

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