$If x = a{\cos }^3\theta and y = a{\sin }^3\theta , then find the value of \frac{d^{2}y}{d{x}^2} at \theta = \frac{\mathrm{\pi }}{6}.$

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$$\begin{gathered} We have: x = a{\cos }^3\theta \\ Differentiating with respect to \theta , \\ \frac{dx}{d\theta } = 3a{\cos }^2\theta \left(-\sin \theta \right) = -3a{\cos }^2\theta \sin \theta \\ Also, y = a{\sin }^3\theta \\ Differentiating with respect to \theta , \\ \frac{dy}{d\theta } = 3a{\sin }^2\theta \cos \theta \\ \therefore \frac{dy}{dx} = \frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}} = \frac{3a{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta } = -\frac{\sin \theta }{\cos \theta } = -\tan \theta \\ Differentiating \frac{dy}{dx} with respect to x, \\ \frac{d^{2}y}{d{x}^2} = -se{c}^2\theta \frac{d\theta }{dx} \\ = \frac{-1}{{\cos }^2\theta } \times \frac{-1}{3a{\cos }^2\theta sin\theta } \\ = \frac{1}{3a{\cos }^4\theta \sin \theta } \\ Thus, at \theta = \frac{\mathrm{\pi }}{6}, \\ \frac{d^{2}y}{d{x}^2} = \frac{1}{3a{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^4\left({\displaystyle \frac{1}{2}}\right)} = \boxed{\frac{\mathbf{32}}{\mathbf{27}\mathbf{a}}} \end{gathered}$$

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