Applications-Of-Derivatives Question 457
Question: The length of normal to the curve $ x=a,(\theta +\sin \theta ), $ $ y=a(1-\cos \theta ) $ at the point $ \theta =\pi /2 $ is
[RPET 1999]
Options:
A) $ 2a $
B) $ a/2 $
C) $ \sqrt{2},a $
D) $ a/\sqrt{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Length of normal = $ y\sqrt{1+{{( \frac{dy}{dx} )}^{2}}} $ Now, $ \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{a\sin \theta }{a(1+\cos \theta )}=\frac{\sin \theta }{1+\cos \theta }=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{{\cos }^{2}}\frac{\theta }{2}} $ \ $ {{( \frac{dy}{dx} )}{( \theta =\frac{\pi }{2} )}}=,{{[ \tan \frac{\theta }{2} ]}{( \theta =\frac{\pi }{2} )}}=1{{[y]}_{( \theta =\frac{\pi }{2} )}}=a,( 1-\cos \frac{\pi }{2} )=a $ \ Length of normal = $ a\sqrt{1+{{(1)}^{2}}}=\sqrt{2}a $ .
BETA
