Question

Write the angle between the curves y = e^−x and y = e^x at their point of intersections.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ y=e^{-x}...\left(1\right) \\ y=e^x...\left(2\right) \\ \mathrm{On}\mathrm{s}\text{ubstituting the value of}\mathit{\text{y}}\text{in (1), we get} \\ {e}^{-x}=e^x \\ \Rightarrow x=0 \\ \mathrm{and} \\ y=1[\text{From (2)]} \\ \mathrm{On}\mathrm{d}\text{ifferentiating (1) w.r.t.}\mathit{\text{x}}\text{,we get} \\ \frac{dy}{dx}=\mathit{-}{e}^{\mathit{-}x} \\ \Rightarrow m_{\mathit{1}}={\left(\frac{dy}{dx}\right)}_{\left(0,1\right)}=-1 \\ \mathrm{On}\mathrm{d}\text{ifferentiating (2) w.r.t}\mathit{\text{.}}\mathit{\text{x,}}\text{we get} \\ \frac{dy}{dx}\mathit{=}{e}^x \\ \Rightarrow m_2={\left(\frac{dy}{dx}\right)}_{\left(0,1\right)}=1 \\ \because m_{\mathit{1}}\times m_{\mathit{2}}=-1 \\ \mathrm{Since}\mathrm{the}\mathrm{multiplication}\mathrm{of}\mathrm{the}\mathrm{slopes}\mathrm{is}-1\mathrm{so}\mathrm{the}\mathrm{slopes}\mathrm{are}\mathrm{perpendicular}\mathrm{to}\mathrm{each}\mathrm{other}. \\ \therefore \mathrm{Required}\mathrm{angle}=\frac{\mathrm{\pi }}{2} \end{gathered}$$