2022年 H2 数学 CT 未考的最大最小值与变化率问题。 在一项测试将新型空间望远镜发射到太空的设计理论的实验中，科学家们借助计算机分析程序，在一个实验系统中监测一个带电粒子的运动，该程序以 $x$-$y$ 平面上的曲线形式展示其路径。粒子在发射后 $t$ 分钟的 $x$ 和 $y$ 坐标由以下参数方程描述：
$x = t - rac{1}{t} + 2 \ ; \ y = t + rac{1}{t} \quad \text{for } t \ge 0.5 \ .$

1. Differentiate $x$ and $y$ with respect to the parameter $t$.

$x = t - t^{-1} + 2 \implies \frac{\mathrm{d}x}{\mathrm{d}t} = 1 + t^{-2} = \frac{t^2 + 1}{t^2}$

$y = t + t^{-1} \implies \frac{\mathrm{d}y}{\mathrm{d}t} = 1 - t^{-2} = \frac{t^2 - 1}{t^2}$

2. Apply the chain rule for parametric equations to find $\frac{\mathrm{d}y}{\mathrm{d}x}$.

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{t^2 - 1}{t^2}}{\frac{t^2 + 1}{t^2}}$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{t^2 - 1}{t^2 + 1} \quad \text{(Shown)}$

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

$\frac{dy}{dx} = \frac{t^2 - 1}{t^2 + 1}$
For stationary points, $\frac{dy}{dx} = 0$:
$\frac{t^2 - 1}{t^2 + 1} = 0$
$t^2 - 1 = 0$
$t = 1 \quad \text{or} \quad t = -1$
Given $t \ge 0.5$, we accept only $t = 1$.

2. Calculate the $x$ and $y$ coordinates of the stationary point.

Substitute $t = 1$ into the parametric equations:
$x = 1 - \frac{1}{1} + 2 = 2$
$y = 1 + \frac{1}{1} = 2$
The stationary point is $(2, 2)$.

3. Find the second derivative $\frac{d^2y}{dx^2}$ to apply the second derivative test.

To find $\frac{d^2y}{dx^2}$, we use the formula $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$.
First, differentiate $x = t - t^{-1} + 2$ with respect to $t$:
$\frac{dx}{dt} = 1 + t^{-2} = \frac{t^2 + 1}{t^2}$
Next, differentiate $\frac{dy}{dx}$ with respect to $t$ using the quotient rule:
$\frac{d}{dt}\left(\frac{t^2 - 1}{t^2 + 1}\right) = \frac{(t^2 + 1)(2t) - (t^2 - 1)(2t)}{(t^2 + 1)^2}$
$= \frac{2t^3 + 2t - 2t^3 + 2t}{(t^2 + 1)^2} = \frac{4t}{(t^2 + 1)^2}$
Now, compute $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{ \frac{4t}{(t^2 + 1)^2} }{ \frac{t^2 + 1}{t^2} } = \frac{4t^3}{(t^2 + 1)^3}$

4. Evaluate the second derivative at $t = 1$ to conclude the nature of the stationary point.

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

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

From part (i), the gradient is $\frac{dy}{dx} = \frac{t^2 - 1}{t^2 + 1}$.
Divide numerator and denominator by $t^2$:
$\frac{dy}{dx} = \frac{1 - \frac{1}{t^2}}{1 + \frac{1}{t^2}}$

As $t \to \infty$, $\frac{1}{t^2} \to 0$, so $\lim_{t \to \infty} \frac{dy}{dx} = \frac{1 - 0}{1 + 0} = 1$.
Thus, for large values of $t$, the gradient of the path approaches 1.

2. Find the equation of the corresponding slant asymptote.

Since the gradient approaches 1, the curve approaches an oblique asymptote of the form $y = x + c$.
Consider the difference $y - x$:
$y - x = \left(t + \frac{1}{t}\right) - \left(t - \frac{1}{t} + 2\right) = \frac{2}{t} - 2$

As $t \to \infty$, $\frac{2}{t} \to 0$, so $y - x \to -2$.
The curve therefore approaches the line $y = x - 2$.

3. Determine the coordinates of the starting point and verify the minimum point for the sketch.

For the starting point at $t = 0.5$:
$x = 0.5 - \frac{1}{0.5} + 2 = 0.5 - 2 + 2 = 0.5$
$y = 0.5 + \frac{1}{0.5} = 0.5 + 2 = 2.5$
The curve starts at the point $(0.5, 2.5)$.

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

4. Describe and provide the sketch of the particle's path.

The required sketch is plotted on the $x$-$y$ plane:

1. Draw a pair of axes and mark the start point $(0.5, 2.5)$ with a solid dot.
2. Draw the minimum point at $(2, 2)$.
3. Draw a dashed line for the oblique asymptote $y = x - 2$.
4. Connect the start point down to the minimum point, then draw a smooth curve extending upwards and to the right, growing steadily closer to the asymptote $y = x - 2$ from above (since $\frac{2}{t} > 0$).