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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{x}=\text{a}(\cos\theta+\theta\sin\theta),\text{y}=\text{a}(\sin\theta-\theta\cos\theta)$ prove that $\frac{\text{d}^2\text{x}}{\text{d}\theta^2}=\text{a}(\cos\theta-\theta\sin\theta),\frac{\text{d}^2}{\text{d}\theta^2}$ $=\text{a}(\sin\theta-\theta\cos\theta)\ \text{and}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\sec^3\theta}{\text{a}\theta}$
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Answer

It is given that, $\text{x}=\text{a}(\cos\text{t}+\text{t}\sin\text{t}),\text{y}=\text{a}(\sin\text{t}-\text{t}\cos\text{t})$
$\therefore\frac{\text{dx}}{\text{dt}}=\text{a}.\frac{\text{d}}{\text{dt}}(\cos\text{t}+\text{t}\sin\text{t})$
$=\text{a}\Big[-\sin\text{t}+\sin\text{t}.\frac{\text{d}}{\text{dt}.}(\text{t})+\frac{\text{d}}{\text{dt}}(\sin\text{t})\Big]$
$=\text{a}[-\sin\text{t}+\sin\text{t}+t\cos\text{t}]=\text{at}\cos\text{t}$
$\frac{\text{dy}}{\text{dt}}=\text{a}.\frac{\text{d}}{\text{dt}}(\sin\text{t}-\text{t}\cos\text{t})$
$=\text{a}\Big[\cos\text{t}-\Big\{\cos\text{t}\frac{\text{d}}{\text{dt}}(\text{t})+\text{t}.\frac{\text{d}}{\text{dt}}(\cos\text{t})\Big\}\Big]$
$=\text{a}[\cos\text{t}-\{\cos\text{t}-\text{t}\sin\text{t}\}]=\text{at}\sin\text{t}$
then, $\therefore\frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{\text{at}\sin\text{t}}{\text{at}\cos\text{t}}=\tan\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}(\tan\text{t})=\sec^2\text{t}.\frac{\text{dt}}{\text{dx}}$
$=\sec^2\text{t}.\frac{1}{\text{at}\cos\text{t}}\ \Big[\frac{\text{dx}}{\text{dt}}=\text{at}\cos\text{t}\Rightarrow\frac{\text{dt}}{\text{dx}}=\frac{1}{\text{at}\cos\text{t}}\Big]$
$=\frac{\sec^3\text{t}}{\text{at}},0<\text{t}<\frac{\pi}{2}$

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