If $\mathrm{x}=\mathrm{e}^{\mathrm{t}}(\sin \mathrm{t}-\cos \mathrm{t})$ and $\mathrm{y}=\mathrm{e}^{\mathrm{t}}(\sin \mathrm{t}+\cos \mathrm{t})$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{t}=\frac{\pi}{3}$ is

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Aicharya · Accepted Answer

$\begin{aligned} & x=e^t(\sin t-\cos t) 	ext { and } y=e^t(\sin t+\cos t) \ & 	herefore \quad \frac{d x}{d t}=e^t(\sin t-\cos t)+e^t(\cos t+\sin t)=2 e^t \sin t \ & \frac{d y}{d t}=e^t(\sin t+\cos t)+e^t(\cos t-\sin t)=2 e^t \sin t \ & 	herefore \quad \frac{d y}{d x}=\frac{2 e^t \cos t}{2 e^t \sin t} \Rightarrow\left(\frac{d y}{d x}ight)_{t=\frac{\pi}{3}}=\cot \left(\frac{\pi}{3}ight)=\frac{1}{\sqrt{3}} \ & \end{aligned}$

If $\mathrm{x}=\mathrm{e}^{\mathrm{t}}(\sin \mathrm{t}-\cos \mathrm{t})$ and…

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