Question
Write the coordinates of the point on the curve $\text{x}=\text{e}^\text{t}\cos\text{t},\text{y}=\text{e}^\text{t}$ where the tangent line makes an angle $\frac{\pi}{4}$ with x-axis.

Answer

Here,
$\text{x}=\text{e}^\text{t}\cos\text{t}\text{ and }\text{y}=\text{e}^{\text{t}}\sin\text{t}$
$\frac{\text{dx}}{\text{dt}}=\text{e}^{\text{t}}\cos\text{t}-\text{e}^{\text{t}}\sin\text{t}\text{ and }\frac{\text{dy}}{\text{dt}}=\text{e}^{\text{t}}\sin\text{t}+\text{e}^{\text{t}}\cos\text{t}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{e}^\text{t}\sin\text{t}+\text{e}^\text{t}\cos\text{t}}{\text{e}^\text{t}\sin\text{t}+\text{e}^\text{t}\cos\text{t}}=\frac{\sin\text{t}+\cos\text{t}}{\cos\text{t}-\sin\text{t}}$
Now,
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{t}=\frac{\pi}{4}}=\frac{\sin\frac{\pi}{4}+\cos\frac{\pi}{4}}{\cos\frac{\pi}{4}-\sin\frac{\pi}{4}}=\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}}=\frac{\frac{2}{\sqrt{2}}}{0}=\infty$
Let $\theta$ be the angle made by the tangent with the x-axis.
$\therefore\tan\theta=\infty$
$\Rightarrow\theta=\frac{\pi}{2}$

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