If theta is the angle made by the normal draw | vedclass

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(D) Given the parametric equations of the curve: $x=e^t \cos t$ and $y=e^t \sin t$. First,we find the derivatives with respect to $t$: $\frac{dx}{dt}

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$3 \pi / 4$

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(D) Given the parametric equations of the curve: $x=e^t \cos t$ and $y=e^t \sin t$. First,we find the derivatives with respect to $t$: $\frac{dx}{dt} = e^t(\cos t - \sin t)$ and $\frac{dy}{dt} = e^t(\sin t + \cos t)$. The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sin t + \cos t}{\cos t - \sin t}$. At the point $(1,0)$,we have $y=0$,which implies $e^t \sin t = 0$. Since $e^t 
eq 0$,we have $\sin t = 0$,so $t=0$. At $t=0$,the slope of the tangent is $\frac{dy}{dx} = \frac{\sin 0 + \cos 0}{\cos 0 - \sin 0} = \frac{0+1}{1-0} = 1$. The slope of the normal $m$ is given by $m = -\frac{1}{dy/dx} = -\frac{1}{1} = -1$. Since the slope of the normal is $	an 	heta = -1$,and $	heta$ is the angle with the $X$-axis,we have $	heta = \frac{3\pi}{4}$.

If $\theta$ is the angle made by the normal drawn to the curve $x=e^{t} \cos t, y=e^{t} \sin t$ at the point $(1,0)$,with the $X$-axis,then $\theta=$

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