Limits Continuity And Differentiability Question 91
Question: If $ {{(x-a)}^{2}}+{{(y-b)}^{2}}=c^{2} $ , for some c > 0, then $ \frac{{{[ 1+{{( \frac{dy}{dx} )}^{2}} ]}^{\frac{3}{2}}}}{\frac{d^{2}y}{dx^{2}}} $ is
Options:
A) Is a constant dependent on a
B) Is a constant dependent on b
C) Is a constant independent of a and b
D) 0
Show Answer
Answer:
Correct Answer: C
Solution:
Given relation is $ {{(x-a)}^{2}}+{{(y-b)}^{2}}=c^{2},c>0 $
Let $ x-a=c\cos \theta $ and $ y-b=c\sin \theta $ . Therefore,
$ \frac{dx}{d\theta }=-c\sin \theta $ and $ \frac{dy}{d\theta }=c\cos \theta $
$ \therefore \frac{dy}{dx}=-\cot \theta $
Differentiating both sides with respect to $ \theta $ , we get
$ \frac{d}{d\theta }( \frac{dy}{dx} )=\frac{d}{d\theta }(-cot\theta ) $
or $ \frac{d}{dx}( \frac{dy}{dx} )\frac{dx}{d\theta }=\cos ec^{2}\theta $
or $ \frac{d^{2}y}{dx^{2}}(-csin\theta )=\cos ec^{2}\theta $
or $ \frac{d^{2}y}{dx^{2}}=\frac{\cos ec^{2}\theta }{c} $
$ \therefore \frac{{{[ 1+{{( \frac{dy}{dx} )}^{2}} ]}^{\frac{3}{2}}}}{\frac{d^{2}y}{dx^{2}}}=\frac{c{{[ 1+{{\cot }^{2}}\theta ]}^{\frac{3}{2}}}}{-\cos ec^{3}\theta }=\frac{c{{(cosec^{2}\theta )}^{\frac{3}{2}}}}{-\cos ec^{3}\theta } $
$ =-c. $
Which is constant and is independent of a and b.
BETA
