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  4. x= cos θ, y= sin 5 θ ⇒ (1-x2) (d2 y/d x2)-x (d y/d x) is equal to

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Q. $x=\cos \theta, y=\sin 5 \theta \Rightarrow \left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}$ is equal to

EAMCETEAMCET 2007

Solution:

Given, $x =\cos \theta,\, y=\sin 5 \theta$
$\frac{d x}{d \theta} =-\sin \theta, \frac{d y}{d \theta}=5 \cos 5 \theta$
$\therefore \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=-\frac{5 \cos 5 \theta}{\sin \theta}$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \cdot \frac{d \theta}{d x}$
$=\frac{d}{d \theta}\left(\frac{-5 \cos 5 \theta}{\sin \theta}\right) \cdot \frac{1}{-\sin \theta}$
$=\left(\frac{\sin \theta \sin 5 \theta \cdot 25+5 \cos 5 \theta \cos \theta}{\sin ^{2} \theta}\right) \cdot \frac{1}{-\sin \theta}$
$=-\frac{25 \sin 5 \theta}{\sin ^{2} \theta}-\frac{5 \cos 5 \theta \cos \theta}{\sin ^{3} \theta}$
Now,
$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}$
$=\left(1-\cos ^{2} \theta\right)\left(\frac{-25 \sin 5 \theta}{\sin ^{2} \theta}-\frac{5 \cos 5 \theta \cos \theta}{\sin ^{3} \theta}\right)$
$-\cos \theta\left(\frac{-5 \cos 5 \theta}{\sin \theta}\right)$
$=\sin ^{2} \theta\left(\frac{-25 \sin 5 \theta}{\sin ^{2} \theta}-\frac{5 \cos \theta \cos 5 \theta}{\sin ^{3} \theta}\right)$
$+\frac{5 \cos \theta \cos 5 \theta}{\sin \theta}$
$=-25 \sin 5 \theta-\frac{5 \cos \theta \cos 5 \theta}{\sin \theta}+\frac{5 \cos \theta \cos 5 \theta}{\sin \theta}$
$=-25 \sin 5 \theta$
$=-25\, y$