A curve $C$ has parametric equations $x = \cos(e^t), \quad y = \sin(e^t), \quad \text{where } t \le \ln\frac{\pi}{2}.$

1. Find the Cartesian equation of the curve.

Given $x = \cos(e^t)$ and $y = \sin(e^t)$, we can square both and add them:
$x^2 + y^2 = \cos^2(e^t) + \sin^2(e^t)$
$x^2 + y^2 = 1$
This shows the curve lies on a circle centered at the origin with radius 1.

2. Determine the domain and range based on the parameter limits.

The parameter is given as $t \le \ln\left(\frac{\pi}{2}\right)$.
Since the exponential function is always positive, we have:
$0 < e^t \le \frac{\pi}{2}$
Let the angle be $\theta = e^t$. For $0 < \theta \le \frac{\pi}{2}$:
$0 \le \cos \theta < 1 \implies 0 \le x < 1$
$0 < \sin \theta \le 1 \implies 0 < y \le 1$

3. Identify the axial intercepts and endpoints for the sketch.

At the upper bound $e^t = \frac{\pi}{2}$:
$x = \cos\left(\frac{\pi}{2}\right) = 0, \quad y = \sin\left(\frac{\pi}{2}\right) = 1$
So, $(0, 1)$ is the $y$-intercept and is included on the curve.
As $t \to -\infty$, $e^t \to 0^+$:
$x \to \cos(0) = 1, \quad y \to \sin(0) = 0$
Since $e^t$ never reaches $0$, the point $(1, 0)$ is strictly not on the curve. There is no $x$-intercept.

Sketch details: Draw a quarter-circle arc in the first quadrant from the open circle at $(1,0)$ to the solid dot at $(0,1)$, labeling the axes and the $y$-intercept $1$.

1. Differentiate the parametric equations with respect to $t$.

$x = \cos(e^t) \implies \frac{dx}{dt} = -e^t \sin(e^t)$
$y = \sin(e^t) \implies \frac{dy}{dt} = e^t \cos(e^t)$

2. Find the gradient of the tangent to the curve, $\frac{dy}{dx}$.

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{e^t \cos(e^t)}{-e^t \sin(e^t)} = -\cot(e^t)$

3. Find the gradient of the normal to the curve.

$m_{\text{normal}} = -\frac{1}{\frac{dy}{dx}} = \tan(e^t) = \frac{\sin(e^t)}{\cos(e^t)}$

4. Form the equation of the normal and show it passes through $(0,0)$.

At a general point $(\cos(e^t), \sin(e^t))$, the equation of the normal is:
$y - \sin(e^t) = \frac{\sin(e^t)}{\cos(e^t)}(x - \cos(e^t))$
To account for all valid $t$ (including where $\cos(e^t) = 0$), multiply by $\cos(e^t)$:
$y \cos(e^t) - \sin(e^t)\cos(e^t) = x \sin(e^t) - \sin(e^t)\cos(e^t)$
$y \cos(e^t) = x \sin(e^t)$
Substitute the origin $(x,y) = (0,0)$ into the equation:
$0 \cdot \cos(e^t) = 0 \cdot \sin(e^t) \implies 0 = 0$
Since this holds true for any parameter $t$, any normal to the curve $C$ passes through the origin.

1. Deduce the gradient of the tangent from the equation of the normal.

The equation of the normal at $P$ is $y = x$, which has a gradient of $m_n = 1$.
Since the tangent is perpendicular to the normal ($m_n \times m_t = -1$), the gradient of the tangent at $P$ is $m_t = -1$.

2. Find the parameter value at point $P$.

Since $P$ lies on the normal $y = x$, we equate the parametric equations:
$\sin(e^t) = \cos(e^t)$
$\tan(e^t) = 1$
Given the domain $t \le \ln\frac{\pi}{2}$, we have $0 < e^t \le \frac{\pi}{2}$.
Therefore, $e^t = \frac{\pi}{4}$.

3. Determine the Cartesian coordinates of $P$.

Substitute $e^t = \frac{\pi}{4}$ back into the parametric equations:
$x = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
$y = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
So, $P$ is the point $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$.

4. Form the equation of the tangent at $P$.

Using the point-slope form with $m = -1$ and point $P\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$:
$y - \frac{\sqrt{2}}{2} = -1\left(x - \frac{\sqrt{2}}{2}\right)$
$y = -x + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$
$y = -x + \sqrt{2}$