Question
$$x = t\cos \,t,\,y = t + \sin \,t$$ then $$\frac{{{d^2}x}}{{d{y^2}}}$$ at $$t = \frac{\pi }{2}$$ is equal to :
A.
$$\frac{{\pi + 4}}{2}$$
B.
$$ - \frac{{\pi + 4}}{2}$$
C.
$$-2$$
D.
none of these
Answer :
$$ - \frac{{\pi + 4}}{2}$$
Solution :
$$\eqalign{
& \frac{{dx}}{{dy}} = \frac{{\frac{{dx}}{{dt}}}}{{\frac{{dy}}{{dt}}}} = \frac{{\cos \,t - t\sin \,t}}{{1 + \cos \,t}} \cr
& \therefore \frac{{{d^2}x}}{{d{y^2}}} = \frac{{\frac{d}{{dt}}\left( {\frac{{dx}}{{dy}}} \right)}}{{\frac{{dy}}{{dt}}}} \cr
& = \frac{{\frac{{\left( { - 2\sin \,t - t\cos \,t} \right)\left( {1 + \cos \,t} \right) - \left( {\cos \,t - t\sin \,t} \right)\left( { - \sin \,t} \right)}}{{{{\left( {1 + \cos \,t} \right)}^2}}}}}{{1 + \cos \,t}} \cr
& {\text{Now, put }}t = \frac{\pi }{2} \cr} $$