If x = e^t sin t and y = e^t cos t,where t is | vedclass

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(A) Given $x = e^t \sin t$ and $y = e^t \cos t$. At point $(1, 1)$,we have $1 = e^t \sin t$ and $1 = e^t \cos t$. Dividing these,we get $	an t = 1$,w

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$-\frac{1}{2}$

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(A) Given $x = e^t \sin t$ and $y = e^t \cos t$. At point $(1, 1)$,we have $1 = e^t \sin t$ and $1 = e^t \cos t$. Dividing these,we get $	an t = 1$,which implies $t = \frac{\pi}{4}$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = e^t \sin t + e^t \cos t = e^t(\sin t + \cos t)$$\frac{dy}{dt} = e^t \cos t - e^t \sin t = e^t(\cos t - \sin t)$Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\cos t - \sin t}{\cos t + \sin t}$.Now,$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{\cos t - \sin t}{\cos t + \sin t}ight) \cdot \frac{dt}{dx}$.Using the quotient rule for $\frac{d}{dt}\left(\frac{\cos t - \sin t}{\cos t + \sin t}ight)$:$= \frac{(\cos t + \sin t)(-\sin t - \cos t) - (\cos t - \sin t)(-\sin t + \cos t)}{(\cos t + \sin t)^2} = \frac{-(\cos t + \sin t)^2 - (\cos t - \sin t)^2}{(\cos t + \sin t)^2} = \frac{-2}{(\cos t + \sin t)^2}$.Thus,$\frac{d^2y}{dx^2} = \frac{-2}{(\cos t + \sin t)^2} \cdot \frac{1}{e^t(\sin t + \cos t)} = \frac{-2}{e^t(\sin t + \cos t)^3}$.At $t = \frac{\pi}{4}$,$e^t \sin t = 1$ and $e^t \cos t = 1$,so $e^t = \sqrt{2}$ and $\sin t = \cos t = \frac{1}{\sqrt{2}}$.Substituting these values: $\frac{d^2y}{dx^2} = \frac{-2}{\sqrt{2}(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})^3} = \frac{-2}{\sqrt{2}(\sqrt{2})^3} = \frac{-2}{\sqrt{2} \cdot 2\sqrt{2}} = \frac{-2}{4} = -\frac{1}{2}$.

If $x = e^t \sin t$ and $y = e^t \cos t$,where $t$ is a parameter,then $\frac{d^2y}{dx^2}$ at $(1, 1)$ is equal to

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