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If $\theta$ is the angle made by the normal drawn to the curve $\mathrm{x}=\mathrm{e}^{\mathrm{t}} \cos \mathrm{t}, \mathrm{y}=\mathrm{e}^{\mathrm{t}} \sin \mathrm{t}$ at the point $(1,0)$, with the $\mathrm{X}$-axis, then $\theta=$
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The correct answer is:
$3 \pi / 4$
$\because x=e^t \cos t, \mathrm{y}=e^t \sin t$
$\frac{d x}{d t}=e^t(\cos t-\sin t)$ and $\frac{d y}{d t}=e^t(\sin t+\cos t)$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{e^t(\sin t+\cos t)}{e^t(\cos t-\sin t)}$
$\Rightarrow \frac{d y}{d t}=\frac{\sin t+\cos t}{\cos t-\sin t}$
At point $(1,0): x=1 \& y=0$
$\begin{aligned} & \because y=0 \Rightarrow e^t \sin t=0 \\ & \Rightarrow \sin t=0 \Rightarrow t=0\end{aligned}$
Now, slope of normal is
$\begin{aligned} & m=-\frac{1}{\frac{d y}{d x}}=-\frac{\cos t-\sin t}{\sin t+\cos t} \\ & m=\frac{\cos 0-\sin 0}{\sin 0+\cos 0}=\frac{-1+0}{0+1}=-1 \\ & \Rightarrow \tan \theta=-1 \Rightarrow \theta=\frac{3 \pi}{4}\end{aligned}$
$\frac{d x}{d t}=e^t(\cos t-\sin t)$ and $\frac{d y}{d t}=e^t(\sin t+\cos t)$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{e^t(\sin t+\cos t)}{e^t(\cos t-\sin t)}$
$\Rightarrow \frac{d y}{d t}=\frac{\sin t+\cos t}{\cos t-\sin t}$
At point $(1,0): x=1 \& y=0$
$\begin{aligned} & \because y=0 \Rightarrow e^t \sin t=0 \\ & \Rightarrow \sin t=0 \Rightarrow t=0\end{aligned}$
Now, slope of normal is
$\begin{aligned} & m=-\frac{1}{\frac{d y}{d x}}=-\frac{\cos t-\sin t}{\sin t+\cos t} \\ & m=\frac{\cos 0-\sin 0}{\sin 0+\cos 0}=\frac{-1+0}{0+1}=-1 \\ & \Rightarrow \tan \theta=-1 \Rightarrow \theta=\frac{3 \pi}{4}\end{aligned}$
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