Limit Continuity And Differentiability Ques 10
- If $x^2+y^2=1$, then
(2000, 1M)
(a) $y y^{\prime \prime}-2\left(y^{\prime}\right)^2+1=0$
(b) $y y^{\prime \prime}+\left(y^{\prime}\right)^2+1=0$
(c) $y y^{\prime \prime}+\left(y^{\prime}\right)^2-1=0$
(d) $y y^{\prime \prime}+2\left(y^{\prime}\right)^2+1=0$
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Answer:
Correct Answer: 10.(b)
Solution: (b) Given, $x^2+y^2=1$
On differentiating w.r.t. $x$, we get
$ \begin{aligned} 2 x+2 y y^{\prime} & =0 \\ x+y y^{\prime} & =0 . \end{aligned} $
Again, differentiating w.r.t. $x$, we get
$ \begin{aligned} 1+y^{\prime} y^{\prime}+y y^{\prime \prime} & =0 \\ 1+\left(y^{\prime}\right)^2+y y^{\prime \prime} & =0 \end{aligned} $
BETA
