Determine the intervals of monotonicity of the curve

Determine the intervals of monotonicity of the curve given parametrically by $x = te^{t}$, $y = te^{-t}$.

Step 1: The two derivatives in $t$

\[ \dot{x} = e^{t}+te^{t} = e^{t}\bigl(1+t\bigr), \tag{1} \]

\[ \dot{y} = e^{-t}-te^{-t} = e^{-t}\bigl(1-t\bigr). \tag{2} \]

Both exponentials are positive, so each sign is decided by its linear factor.

Step 2: Monotonicity of $x(t)$

From (1), $\dot{x}\lt 0$ for $t\lt -1$ and $\dot{x}\gt 0$ for $t\gt -1$. Thus $x$ decreases, then increases, with a minimum

\[ x(-1) = -e^{-1} = -0.3678794. \]

Step 3: Monotonicity of $y(t)$

From (2), $\dot{y}\gt 0$ for $t\lt 1$ and $\dot{y}\lt 0$ for $t\gt 1$. Thus $y$ increases, then decreases, with a maximum

\[ y(1) = e^{-1} = 0.3678794. \]

Step 4: The slope $\dfrac{dy}{dx}$

\[ \frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} = \frac{e^{-t}\bigl(1-t\bigr)}{e^{t}\bigl(1+t\bigr)} = \frac{1-t}{e^{2t}\bigl(1+t\bigr)}, \qquad t\neq-1. \tag{3} \]

Since $e^{2t}\gt 0$, the sign is that of $\dfrac{1-t}{1+t}$: \[t\lt -1:\ \frac{(+)}{(-)} = -\ (\searrow); \qquad -1\lt t\lt 1:\ \frac{(+)}{(+)} = +\ (\nearrow); \qquad t\gt 1:\ \frac{(-)}{(+)} = -\ (\searrow).\]

Step 5: The two special points

At $t = -1$: $\dot{x} = 0$ while $\dot{y} = 2e\neq0$, so the tangent is vertical, at the point $\bigl(-e^{-1},\,-e\bigr)$. At $t = 1$: $\dot{y} = 0$ while $\dot{x} = 2e\neq0$, so the tangent is horizontal, at $\bigl(e,\,e^{-1}\bigr)$.

Step 6: Verification

\[t = -2:\ \bigl(-0.270671,\,-14.7781\bigr); \quad t = -1:\ \bigl(-0.367879,\,-2.71828\bigr); \quad t = 0:\ \bigl(0,0\bigr) \;\;\checkmark\]

\[t = 1:\ \bigl(2.71828,\,0.367879\bigr); \quad t = 2:\ \bigl(14.7781,\,0.270671\bigr) \;\;\checkmark\]

\[\text{Symmetry: } x(-t) = -y(t),\ \text{so the curve is symmetric about the line } y = -x \;\;\checkmark\]