Question

Write the angle made by the tangent to the curve x = e^t cos t, y = e^t sin t at $t=\frac{\mathrm{\pi }}{4}$ with the x-axis.

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ x=e^t\cos t\mathrm{and}y=e^t\sin t \\ \frac{dx}{dt}=e^{t}c\mathrm{os}t-e^t\sin t\mathrm{and}\frac{dy}{dt}=e^t\sin t+e^t\cos t \\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{e^t\sin t+e^t\cos t}{e^{t}c\mathrm{os}t-e^t\sin t}=\frac{\sin t+\cos t}{\cos t-\sin t} \\ \mathrm{Now}, \\ \text{Slope of the tangent =}{\left(\frac{dy}{dx}\right)}_{t=\frac{\mathrm{\pi }}{4}}\text{=}\frac{\sin \frac{\mathrm{\pi }}{4}+\cos \frac{\mathrm{\pi }}{4}}{\cos \frac{\mathrm{\pi }}{4}-\sin \frac{\mathrm{\pi }}{4}}\text{=}\frac{{\displaystyle \frac{1}{\sqrt{2}}}+\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}}\text{=}\frac{{\displaystyle \frac{2}{\sqrt{2}}}}{0}\text{=∞} \\ \text{Let}\mathit{\text{θ}}\text{be the angle made by the tangent with the}\mathit{\text{x}}\text{-axis.} \\ \therefore \text{tan}\mathit{\text{θ}}\text{=∞} \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{2} \end{gathered}$$