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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find0 $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\text{a}\Big(\cos\text{t}+\log\tan\frac{\text{t}}{2}\Big)\text{y}=\text{a}\sin\text{t}$
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Answer

The given equations are $\text{x}=\text{a}\Big(\cos\text{t}+\log\tan\frac{\text{t}}{2}\Big)\text{ and y}=\text{a}\sin\text{t}$
Then, $\frac{\text{dx}}{\text{dt}}= \text{a}.\Big[\frac{\text{d}}{\text{dt}}(\cos\text{t})+\frac{\text{d}}{\text{dt}}\Big(\log\tan\frac{\text{t}}{2}\Big)\Big]$
$=\text{a}\Bigg[-\sin\text{t}+\frac{1}{\tan\frac{\text{t}}{2}}.\frac{\text{d}}{\text{dt}}\Big(\tan\frac{\text{t}}{2}\Big)\Bigg]$
$=\text{a}\Big[-\sin\text{t}+\cot\frac{\text{t}}{2}.\sec^2\frac{\text{t}}{2}.\frac{\text{d}}{\text{dt}}\Big(\frac{\text{t}}{2}\Big)\Big]$
$=\text{a}\Bigg[-\sin\text{t}+\frac{\cos\frac{\text{t}}{2}}{\sin\frac{\text{t}}{2}}\times\frac{1}{\cos^2\frac{\text{t}}{2}}\times\frac{1}{2}\Bigg]$
$=\text{a}\Bigg[-\sin\text{t}+\frac{1}{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}\Bigg]$
$=\text{a}\Big(-\sin\text{t}+\frac{1}{\sin\text{t}}\Big)$
$=\text{a}\Big(\frac{-\sin^2\text{t}+1}{\sin\text{t}}\Big)$
$=\text{a}\frac{\cos^2\text{t}}{\sin\text{t}}$
$\frac{\text{dy}}{\text{dt}}=\text{a}\frac{\text{d}}{\text{dt}}(\sin\text{t})=\text{a}\cos\text{t}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{\text{a}\cos\text{t}}{\Big(\text{a}\frac{\cos^2\text{t}}{\sin\text{t}}\Big)}=\frac{\sin\text{t}}{\cos\text{t}}=\tan\text{t}$

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