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Question Number 130918 by bramlexs22 last updated on 30/Jan/21

$$\{\begin{array}{l}y=\sin \theta -\cos {}^3\theta \\ x=\cos \theta -\sin {}^3\theta \end{array}$$$$\frac{d^{2}y}{{dx}^2}=?$$

Answered by benjo_mathlover last updated on 30/Jan/21

$$\{\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{d}\theta }=\cos \theta +3\cos {}^2\theta \sin \theta \\ \frac{\mathrm{dx}}{\mathrm{d}\theta }=-\sin \theta -3\sin {}^2\theta \cos \theta \end{array}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{d}\theta }.\frac{\mathrm{d}\theta }{\mathrm{dx}}=\frac{\cos \theta +3\cos {}^2\theta \sin \theta }{-\sin \theta -3\sin {}^2\theta \cos \theta }$$$$=\frac{\cos \theta (1+3\cos \theta \sin \theta )}{-\sin \theta (1+3\cos \theta \sin \theta )}=-\cot \theta$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{\mathrm{d}}{\mathrm{d}\theta }(-\cot \theta ).\frac{\mathrm{d}\theta }{\mathrm{dx}}={\csc }^2\theta .\left(\frac{1}{-\sin \theta -3\sin {}^2\theta \cos \theta }\right)$$$$=-\frac{1}{\sin {}^3\theta (1+\frac{3}{2}\sin 2\theta )}$$$$=-\frac{2}{\sin {}^3\theta (2+3\sin 2\theta )}$$

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