If x=3 - 3t,y=3 3t- 3t find dy dx at t= 3 - Vidyadip

Question

If $\text{x}=3\sin\text{t}-\sin3\text{t},\text{y}=3\cos3\text{t}-\cos3\text{t}$ find $\frac{\text{dy}}{\text{dx}}\text{ at t}=\frac{\pi}{3}$

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Answer

$\text{x}=3\sin\text{t}-\sin3\text{t and } \text{y}=3\cos3\text{t}-\cos3\text{t}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=3\cos\text{t}-3\cos3\text{t}\text{ and} \\ \frac{\text{dy}}{\text{dt}}=-3\sin\text{t}-3\sin3\text{t}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-3\sin\text{t}+3\sin3\text{t}}{3\cos\text{t}-3\cos3\text{t}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{t}=\frac{\pi}{3}}=\frac{-3\sin\frac{\pi}{3}+3\sin\pi}{3\cos\frac{\pi}{3}-3\cos\pi}$
$=\frac{3\times\frac{\sqrt{3}}{2}+0}{3\times\frac{1}{2}+3}$
$=\frac{\frac{-3\sqrt{3}}{2}}{\frac{9}{2}}$
$=-\frac{1}{\sqrt{3}}$

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