Differentiation Question 415
Question: If $ x=\frac{2t}{1+t^{2}},y=\frac{1-t^{2}}{1+t^{2}}, $ then $ \frac{dy}{dx} $ equals
[RPET 1999]
Options:
A) $ \frac{2t}{t^{2}+1} $
B) $ \frac{2t}{t^{2}-1} $
C) $ \frac{2t}{1-t^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ x=\frac{2t}{1+t^{2}},y=\frac{1-t^{2}}{1+t^{2}} $
Put $ t=\tan \theta $
$ x=\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }=\sin 2\theta $ , $ y=\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }=\cos 2\theta $
$ \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{-2\sin 2\theta }{2\cos 2\theta } $
= $ -\tan 2\theta $ = $ \frac{-2\tan \theta }{1-{{\tan }^{2}}\theta } $ = $ \frac{-2t}{1-t^{2}} $ = $ \frac{2t}{t^{2}-1} $ .
BETA
