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Mathematics Question on Continuity and differentiability

If x=a(cos⁡θ+θsin⁡θ)x= a\left(\cos \theta +\theta \sin \theta\right) and y=a(sin⁡θ−θcos⁡θ)y = a \left(\sin \theta- \theta \cos \theta \right) where 0<θ<π20 < \theta < \frac{\pi}{2} , then d2ydx2 \frac{d^{2}y}{dx^{2}} at θ=π4 \theta = \frac{\pi}{4} is equal to

A

82aπ\frac{8\sqrt{2}}{a\pi}

B

42aπ\frac{4\sqrt{2}}{a\pi}

C

4aπ2\frac{4}{a\pi \sqrt{2}}

D

none of thesenone\, of\, these

Answer

82aπ\frac{8\sqrt{2}}{a\pi}

Explanation

Solution

We have, x−a(cos⁡θ+θsin⁡θ)x-a\left(\cos \theta +\theta \sin \theta\right)
y=a(sin⁡θ−cos⁡θ),0<θ<π2y = a\left( \sin \theta - \cos \theta \right), 0 < \theta< \frac{\pi}{2}
Differentiating w.r.t. 0, we get
dxdθ=a(−sin⁡θ+cos⁡θ+θsin⁡θ)=aθcos⁡θ\frac{dx}{d\theta} = a\left( - \sin \theta + \cos \theta +\theta \sin \theta \right) = a \theta \cos \theta
dydθ=a(cos⁡θ−cos⁡θ+θsin⁡θ)=aθsin⁡θ\frac{dy}{d\theta } = a\left(\cos \theta - \cos \theta +\theta \sin \theta \right) = a \theta \sin \theta
dydx=dydθdxdθ=aθsin⁡θaθcos⁡θ=tan⁡θ\frac{dy}{dx}= \frac{dy d\theta}{dx d\theta} = \frac{a \theta \sin \theta}{a \theta \cos \theta} = \tan \theta
Now, d2ydx2=ddx(tan⁡θ)\frac{d^{2}y}{dx^{2}} = \frac{d}{dx} \left( \tan \theta\right)
=sec⁡2θ.dθdx=sec⁡2θ1aθcos⁡θ=sec⁡3θaθ=\sec^{2} \theta . \frac{d\theta}{dx} = \sec ^{2} \theta \frac{1}{a\theta \cos\theta } = \frac{\sec^{3} \theta}{a\theta}
d2ydx2θ=π4=(sec⁡π4)3a(π4)=4aπ(2)3=82aπ\frac{d^{2} y}{dx^{2} } _{\theta =\pi 4} = \frac{\left(\sec \frac{\pi}{4}\right)^{3}}{a\left(\frac{\pi}{4}\right)} = \frac{4}{a\pi} \left(\sqrt{2}\right)^{3} = \frac{8\sqrt{2}}{a\pi}