Applications Of Derivatives Question 198
Question: If $ x=\sec \theta -\cos \theta $ and $ y={{\sec }^{n}}\theta -{{\cos }^{n}}\theta $ , then
[IIT 1989]
Options:
A) $ (x^{2}+4){{( \frac{dy}{dx} )}^{2}}=n^{2}(y^{2}+4) $
B) $ (x^{2}+4){{( \frac{dy}{dx} )}^{2}}=x^{2}(y^{2}+4) $
C) $ (x^{2}+4){{( \frac{dy}{dx} )}^{2}}=(y^{2}+4) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta } $
$ =\frac{n{{\sec }^{n}}\theta \tan \theta +n{{\cos }^{n-1}}\theta \sin \theta }{\sec \theta \tan \theta +\sin \theta } $
$ =\frac{n({{\sec }^{n}}\theta +{{\cos }^{n}}\theta )}{\sec \theta +\cos \theta } $ (Dividing $ N^{r} $ and $ D^{r} $ by $ \tan \theta $ )
therefore $ {{( \frac{dy}{dx} )}^{2}}=\frac{n^{2}{{({{\sec }^{n}}\theta +{{\cos }^{n}}\theta )}^{2}}}{{{(\sec \theta +\cos \theta )}^{2}}} $
$ =\frac{n^{2}[{{({{\sec }^{n}}\theta -{{\cos }^{n}}\theta )}^{2}}+4{{\sec }^{n}}\theta {{\cos }^{n}}\theta ]}{{{(\sec \theta -\cos \theta )}^{2}}+4\sec \theta .\cos \theta }=\frac{n^{2}(y^{2}+4)}{x^{2}+4} $
therefore $ (x^{2}+4){{( \frac{dy}{dx} )}^{2}}=n^{2}(y^{2}+4) $ .
BETA
