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  • If x = 3sint - sin3t, y = 3cost - cos 3t, find dy / dx at t= pi/3

If x = 3sint - sin3t, y = 3cost - cos 3t, find \frac{\mathrm{dy}}{\mathrm{dx}} \text { at } \mathrm{t}=\frac{\pi}{3}

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x = 3\sin t - \sin 3t, y = 3\cos t - \cos 3t

Differentiate w.r.t t in both equation

x = 3sint - sin3t
 

 \begin{aligned} &\frac{d x}{d t}=\frac{d}{d t}(3 \sin t-\sin 3 t)\\ &\frac{d x}{d t}=\frac{d}{d t}(3 \sin t)-\frac{d}{d t}(\sin 3 t)\\ &\left.\frac{d x}{d t}=3 \cos t-3 \cos 3 t\right.\\ &\text { Now, for y }\\ &y=3 \operatorname{cost}-\cos 3 t \end{aligned}

\\ \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(3 \cos \mathrm{t}- \cos 3 \mathrm{t}) \\ \frac{\mathrm{dy}}{\mathrm{dt}}=3(-\sin \mathrm{t})- \frac{\mathrm{d}}{\mathrm{dt}}(\cos 3 \mathrm{t}) \\ \frac{\mathrm{dy}}{\mathrm{dt}}=3(-\sin \mathrm{t})-3(-\sin 3 \mathrm{t}) \\ \frac{\mathrm{dy}}{\mathrm{dt}}=-3 \sin \mathrm{t}+3 \sin 3 \mathrm{t} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}}

\\ \frac{d y}{d x}=\frac{-3 \sin t+3 \sin 3 t}{3 \cos t-3 \cos 3 t} \\ \text { At } t=\pi / 3 \\ \frac{d y}{d x}=\frac{-3 \sin \frac{\pi}{3}+3 \sin \frac{3 \pi}{3}}{3 \cos \frac{\pi}{3}-3 \cos \frac{3 \pi}{3}} \\ \frac{d y}{d x}=\frac{-\sin \frac{\pi}{3}+\sin \pi}{\cos \frac{\pi}{3}-\cos \pi} \\ \frac{d y}{d x}=\frac{-\frac{\sqrt{3}}{2}+0}{\frac{1}{2}-(-1)}

\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}+1} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\frac{\sqrt{3}}{2}}{\frac{3}{2}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\sqrt{3}}{2} \times \frac{2}{3} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{\sqrt{3}}

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