Question
If $\text{x}=\text{a}(1-\cos^3\theta),\text{y}=\text{a}\sin^3\theta,$ Prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{32}{27\text{a}}\text{ at}\ \theta=\frac{\pi}{6}$

Answer

Here,
$\text{x}=\text{a}(1-\cos^3\theta),\text{y}=\text{a}\sin^3\theta$
Differentiating w.r.t. x, we get
$\frac{\text{dx}}{\text{d}\theta}=3\text{a}\cos2\theta\sin\theta\text{ and }\frac{\text{dy}}{\text{d}\theta}=3\text{a}\sin2\theta\cos\theta$]
$\frac{\text{dy}}{\text{dx}}=\frac{3\text{a}\sin^2\theta\cos\theta}{3\text{a}\cos^2\theta\sin\theta}=\tan\theta$
Differentiating w.r.t. x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\sec^2\theta\frac{\text{d}\theta}{\text{dx}}$
$=\frac{\sec^2\theta}{3\text{a}\cos^2\theta\sin\theta}$
$=\frac{\sec^4\theta}{3\text{a}\sin\theta}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}}\text{ at }\theta=\frac{\pi}{6}$
$\frac{\text{d}^2\text{y}}{\text{dx}}=\frac{\Big(\sec\frac{\text{x}}{6}\Big)^4}{3\text{a}\sin\frac{\text{x}}{6}}=\frac{32}{27\text{a}}$

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