Question:
Write the angle between the curves $y=e^{-x}$ and $y=e^{x}$ at their point of intersection.
Solution:
Given that $y=e^{-x} \ldots(1)$ and $y=e^{x}$. .......(2)
Substituting the value of $y$ in (1),
$e^{-x}=e^{x}$
$\Rightarrow x=0$
And $y=1($ from 2$)$
On differentiating (1) w.r.t. $x$, we get
$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{-x}$
$\Rightarrow \mathrm{m}_{2}=\frac{\mathrm{dy}}{\mathrm{dx}}=1$
$\because m_{1} \times m_{2}=-1$
Since the multiplication of both the slopes is $-1$ so the slopes are perpendicular to each other.
$\therefore$ Required angle $=90^{\circ}$