In an experiment to test the design theory of launching a new space telescope into space, scientists monitor the movement of a charged particle in an experimental system with the help of a computer analytical programme that presents its path in the form of a curve on an x-y plane. The x- and y-coordinates of the particle, t minutes from launch are described by the parametric equations: $x = t - \frac{1}{t} + 2; \quad y = t + \frac{1}{t} \quad\text{for } t \ge 0.5.$

1. Differentiate the x-coordinate with respect to t

$x = t - t^{-1} + 2$

$\frac{dx}{dt} = 1 - (-1)t^{-2} = 1 + \frac{1}{t^2} = \frac{t^2 + 1}{t^2}$

2. Differentiate the y-coordinate with respect to t

$y = t + t^{-1}$

$\frac{dy}{dt} = 1 + (-1)t^{-2} = 1 - \frac{1}{t^2} = \frac{t^2 - 1}{t^2}$

3. Calculate the gradient of the curve using the chain rule

$\frac{dy}{dx} = \frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$

$\frac{dy}{dx} = \frac{\left(\frac{t^2 - 1}{t^2}\right)}{\left(\frac{t^2 + 1}{t^2}\right)} = \frac{t^2 - 1}{t^2 + 1}$ (Shown)

1. Equate the gradient to zero to find the parameter $t$ at the stationary point.

$\frac{t^2 - 1}{t^2 + 1} = 0$
$t^2 - 1 = 0$
Since $t \ge 0.5$, we reject $t = -1$ and take $t = 1$.

2. Substitute $t = 1$ into the parametric equations to find the coordinates of the stationary point.

$x = 1 - \frac{1}{1} + 2 = 2$
$y = 1 + \frac{1}{1} = 2$
The coordinates of the stationary point are $(2, 2)$.

3. Find the second derivative $\frac{d^2y}{dx^2}$ to determine the nature of the stationary point.

Express $\frac{dy}{dx}$ in a simpler form for differentiation:
$\frac{dy}{dx} = \frac{t^2 - 1}{t^2 + 1} = 1 - \frac{2}{t^2 + 1}$
Differentiate with respect to $t$:
$\frac{d}{dt}\left(\frac{dy}{dx}\right) = -2(-1)(t^2+1)^{-2}(2t) = \frac{4t}{(t^2+1)^2}$
Differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = 1 + \frac{1}{t^2} = \frac{t^2+1}{t^2}$
Using the chain rule for the second parametric derivative:
$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{4t}{(t^2+1)^2} \div \frac{t^2+1}{t^2} = \frac{4t^3}{(t^2+1)^3}$

4. Evaluate $\frac{d^2y}{dx^2}$ at $t = 1$ to conclude the nature of the point.

Substitute $t = 1$ into the second derivative:
$\frac{d^2y}{dx^2} = \frac{4(1)^3}{(1^2+1)^3} = \frac{4}{8} = 0.5$
Since $\frac{d^2y}{dx^2} > 0$, the stationary point is a minimum point.

1. Determine the behaviour of the gradient for large values of $t$.

As $t \to \infty$, we evaluate the limit of the gradient:
$\lim_{t \to \infty} \frac{dy}{dx} = \lim_{t \to \infty} \frac{t^2 - 1}{t^2 + 1} = \lim_{t \to \infty} \frac{1 - \frac{1}{t^2}}{1 + \frac{1}{t^2}}$
Since $\frac{1}{t^2} \to 0$ as $t \to \infty$,
$\lim_{t \to \infty} \frac{dy}{dx} = \frac{1 - 0}{1 + 0} = 1$
The gradient approaches $1$.

2. Find coordinates of the starting point and use the stationary point to anchor the sketch.

The path is defined for $t \ge 0.5$. At the starting point $t = 0.5$:
$x = 0.5 - \frac{1}{0.5} + 2 = 0.5$
$y = 0.5 + \frac{1}{0.5} = 2.5$
So the curve starts at the coordinate $(0.5, 2.5)$.

From part (ii), the stationary point occurs at $t = 1$:
$x = 1 - \frac{1}{1} + 2 = 2$
$y = 1 + \frac{1}{1} = 2$
This is the minimum point $(2, 2)$.

3. Describe and construct the sketch of the path.

Plot the starting point $(0.5, 2.5)$ and the minimum point $(2, 2)$ on Cartesian axes.
Draw a smooth curve that decreases from the start to the minimum, then increases continuously for $t > 1$.
As $x$ and $y$ increase, the curve should visually straighten out, reflecting that its gradient tends towards $1$ (it asymptotically approaches the line $y = x - 2$).

1. Express the rate of change of $y$ and $x$ with respect to $t$ and evaluate $\frac{1}{t^2}$ using the given condition.

$y = t + t^{-1} \implies \frac{dy}{dt} = 1 - \frac{1}{t^2}$
$x = t - t^{-1} + 2 \implies \frac{dx}{dt} = 1 + \frac{1}{t^2}$
Given $\frac{dy}{dt} = 0.5$:
$1 - \frac{1}{t^2} = 0.5 \implies \frac{1}{t^2} = 0.5$

2. Calculate the corresponding rate of change of the $x$-coordinate.

Substitute $\frac{1}{t^2} = 0.5$ into the expression for $\frac{dx}{dt}$:
$\frac{dx}{dt} = 1 + 0.5 = 1.5$
(Note: $t^2 = 2 \implies t = \sqrt{2}$, which satisfies the condition $t \ge 0.5$)