Applications Of Derivatives Question 214
Question: If $ x=\sin t $ and $ y=\sin pt $ , then the value of $ ( 1-x^{2} )\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}+p^{2}y $ is equal to
[Pb. CET 2002]
Options:
A) 0
B) 1
C) -1
D) $ \sqrt{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
Given that $ x=\sin t $ , $ y=\sin pt $
$ \frac{dx}{dt}=\cos t $ , $ \frac{dy}{dt}=p\cos pt $
$ \frac{dy}{dx}=\frac{p\cos pt}{\cos t}=\frac{p\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}} $
Again differentiate w.r.t. x,
$ \frac{d^{2}y}{dx^{2}}=\frac{p\sqrt{1-x^{2}}.\frac{1}{2\sqrt{1-y^{2}}}.(-2y)\frac{dy}{dx}-p\sqrt{1-y^{2}}.\frac{1}{2\sqrt{1-x^{2}}}(-2x)}{(1-x^{2})} $
$ (1-x^{2})\frac{d^{2}y}{dx^{2}}=-py\frac{\sqrt{1-x^{2}}}{\sqrt{1-y^{2}}}\frac{dy}{dx}+px\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}} $
$ (1-x^{2})\frac{d^{2}y}{dx^{2}}=-p^{2}y+x\frac{dy}{dx} $
$ (1-x^{2})\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}+p^{2}y=0 $ -
BETA
