If x=e^{t}(sin t-cos t) and y=e^{t}(sin t+cos | vedclass

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(B) Given $x=e^t(\sin t-\cos t)$ and $y=e^t(\sin t+\cos t)$.Applying the product rule for differentiation with respect to $t$:$\frac{dx}{dt} = e^t(\si

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$\frac{1}{\sqrt{3}}$

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(B) Given $x=e^t(\sin t-\cos t)$ and $y=e^t(\sin t+\cos t)$.Applying the product rule for differentiation with respect to $t$:$\frac{dx}{dt} = e^t(\sin t-\cos t) + e^t(\cos t+\sin t) = e^t(\sin t - \cos t + \cos t + \sin t) = 2e^t \sin t$.$\frac{dy}{dt} = e^t(\sin t+\cos t) + e^t(\cos t-\sin t) = e^t(\sin t + \cos t + \cos t - \sin t) = 2e^t \cos t$.Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2e^t \cos t}{2e^t \sin t} = \cot t$.At $t = \frac{\pi}{3}$,$\frac{dy}{dx} = \cot(\frac{\pi}{3}) = \frac{1}{\sqrt{3}}$.

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

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