Differentiation Question 377
Question: A curve is given by the equations $ x=a\cos \theta +\frac{1}{2}b\cos 2\theta , $
$ y=a\sin \theta +\frac{1}{2}b\sin 2\theta $ , then the points for which $ \frac{d^{2}y}{dx^{2}}=0, $ is given by
[Kurukshetra CEE 2002]
Options:
A) $ \sin \theta =\frac{2a^{2}+b^{2}}{5ab} $
B) $ \tan \theta =\frac{3a^{2}+2b^{2}}{4ab} $
C) $ \cos \theta =\frac{-( a^{2}+2b^{2} )}{3ab} $
D) $ \cos \theta =\frac{( a^{2}-2b^{2} )}{3ab} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ x=a\cos \theta +\frac{1}{2}b\cos 2\theta $ , $ y=a\sin \theta +\frac{1}{2}b\sin 2\theta $
$ \frac{dy}{d\theta }=a\cos \theta +b\cos 2\theta $ , $ \frac{dx}{d\theta }=-a\sin \theta -b\sin 2\theta $
\
$ \therefore $ $ \frac{d}{dx}( \frac{dy}{dx} )=\frac{d}{d\theta }.( \frac{dy}{dx} ).\frac{d\theta }{dx} $
= $ [ \frac{(a\sin \theta +b\sin 2\theta )(a\sin \theta +2b\sin 2\theta )}{{{(a\sin \theta +b\sin 2\theta )}^{2}}} . $
$ . +\frac{(a\cos \theta +b\cos 2\theta )(a\cos \theta +2b\cos 2\theta )}{{{(a\sin \theta +b\sin 2\theta )}^{2}}} ].\frac{d\theta }{dx} $ but $ \frac{d^{2}y}{dx^{2}}=0 $
Therefore $ a^{2}+2b^{2}+3ab[\sin 2\theta \sin \theta +\cos 2\theta .\cos \theta ]=0 $
Therefore $ a^{2}+2b^{2}=-3ab\cos (2\theta -\theta ) $
\ $ \cos \theta =-( \frac{a^{2}+2b^{2}}{3ab} ) $ .
BETA
