Applications Of Derivatives Question 140
Question: The length of the normal at point -t- of the curve $ x=a(t+\sin t), $ $ y=a(1-\cos t) $ is
[RPET 2001]
Options:
A) $ a\sin t $
B) $ 2a{{\sin }^{3}}(t/2)\sec (t/2) $
C) $ 2a\sin (t/2)\tan ,(t/2) $
D) $ 2a\sin (t/2) $
Show Answer
Answer:
Correct Answer: C
Solution:
$ x=a(t+\sin t) $ , $ y=a(1-\cos t) $ \ $ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a(\sin t)}{a(1+\cos t)} $ = $ \tan \frac{t}{2} $
Length of the normal = $ y\sqrt{1+{{( \frac{dy}{dx} )}^{2}}} $ = $ a(1-\cos t),\sqrt{1+{{\tan }^{2}}(t/2)} $ = $ a(1-\cos t),\sec (t/2) $ = $ 2a{{\sin }^{2}}(t/2),\sec ,(t/2) $ = $ 2a\sin (t/2),\tan (t/2) $ .
BETA
