If x = sqrt{2} e^t(sin t - cos t) and y = sqr | vedclass

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(A) Given $x = \sqrt{2} e^t(\sin t - \cos t)$ and $y = \sqrt{2} e^t(\sin t + \cos t)$.First,find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:$\frac{dx}{dt} =

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$-e^{-\pi/4}$

Vedclass · Answer

(A) Given $x = \sqrt{2} e^t(\sin t - \cos t)$ and $y = \sqrt{2} e^t(\sin t + \cos t)$.First,find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:$\frac{dx}{dt} = \sqrt{2} [e^t(\sin t - \cos t) + e^t(\cos t + \sin t)] = \sqrt{2} e^t(2 \sin t) = 2\sqrt{2} e^t \sin t$.$\frac{dy}{dt} = \sqrt{2} [e^t(\sin t + \cos t) + e^t(\cos t - \sin t)] = \sqrt{2} e^t(2 \cos t) = 2\sqrt{2} e^t \cos t$.Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\sqrt{2} e^t \cos t}{2\sqrt{2} e^t \sin t} = \cot t$.Next,find $\frac{d^2 y}{dx^2} = \frac{d}{dx}(\cot t) = \frac{d}{dt}(\cot t) \cdot \frac{dt}{dx} = -\csc^2 t \cdot \frac{1}{dx/dt}$.Substitute $\frac{dx}{dt} = 2\sqrt{2} e^t \sin t$:$\frac{d^2 y}{dx^2} = -\csc^2 t \cdot \frac{1}{2\sqrt{2} e^t \sin t} = -\frac{1}{2\sqrt{2} e^t \sin^3 t}$.At $t = \frac{\pi}{4}$,$\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\sin^3(\frac{\pi}{4}) = (\frac{1}{\sqrt{2}})^3 = \frac{1}{2\sqrt{2}}$.$\left(\frac{d^2 y}{dx^2}\right)_{t = \pi/4} = -\frac{1}{2\sqrt{2} e^{\pi/4} \cdot (1 / 2\sqrt{2})} = -\frac{1}{e^{\pi/4}} = -e^{-\pi/4}$.

If $x = \sqrt{2} e^t(\sin t - \cos t)$ and $y = \sqrt{2} e^t(\sin t + \cos t)$,then $\left(\frac{d^2 y}{d x^2}\right)_{t = \pi/4} = $

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