点 $P$ 在曲线 $y = \frac{kx^2}{\sqrt{2-x}}$ 上，其中 $k$ 是常数。$P$ 的 $x$ 坐标为 $-2$。

1. Find the gradient of the tangent at $P$.

Let $m_t$ be the gradient of the tangent and $m_n$ be the gradient of the normal.
$m_t \times m_n = -1$
$m_t \times \frac{1}{7} = -1 \implies m_t = -7$

2. Differentiate the equation of the curve with respect to $x$.

$y = \frac{kx^2}{\sqrt{2-x}}$
Using the quotient rule:
$\frac{dy}{dx} = \frac{\sqrt{2-x} \cdot \frac{d}{dx}(kx^2) - kx^2 \cdot \frac{d}{dx}(\sqrt{2-x})}{(\sqrt{2-x})^2}$
$\frac{dy}{dx} = \frac{\sqrt{2-x}(2kx) - kx^2\left[\frac{1}{2}(2-x)^{-\frac{1}{2}}(-1)\right]}{2-x}$
$\frac{dy}{dx} = \frac{2kx\sqrt{2-x} + \frac{kx^2}{2\sqrt{2-x}}}{2-x}$

3. Substitute $x = -2$ to find the gradient of the tangent in terms of $k$.

When $x = -2$:
$\frac{dy}{dx} = \frac{2k(-2)\sqrt{2-(-2)} + \frac{k(-2)^2}{2\sqrt{2-(-2)}}}{2-(-2)}$
$\frac{dy}{dx} = \frac{-4k\sqrt{4} + \frac{4k}{2\sqrt{4}}}{4}$
$\frac{dy}{dx} = \frac{-4k(2) + \frac{4k}{2(2)}}{4}$
$\frac{dy}{dx} = \frac{-8k + k}{4} = -\frac{7k}{4}$

4. Equate the gradient to $-7$ and solve for $k$.

$-\frac{7k}{4} = -7$
$-7k = -28$
$k = 4$ (shown)

1. Express $(OP)^2$ in terms of $x$.

Let $L = (OP)^2$.
Since $O$ is the origin $(0,0)$ and $P$ is $(x, y)$,
$L = x^2 + y^2$
From part (i), $k = 4$, so $y = \frac{4x^2}{\sqrt{2-x}}$.
$L = x^2 + \left(\frac{4x^2}{\sqrt{2-x}}\right)^2 = x^2 + \frac{16x^4}{2-x}$

2. Differentiate $(OP)^2$ with respect to $x$.

$\frac{dL}{dx} = 2x + \frac{(2-x)\frac{d}{dx}(16x^4) - 16x^4\frac{d}{dx}(2-x)}{(2-x)^2}$
$\frac{dL}{dx} = 2x + \frac{(2-x)(64x^3) - 16x^4(-1)}{(2-x)^2}$
$\frac{dL}{dx} = 2x + \frac{128x^3 - 64x^4 + 16x^4}{(2-x)^2}$
$\frac{dL}{dx} = 2x + \frac{128x^3 - 48x^4}{(2-x)^2}$

3. Evaluate $\frac{dL}{dx}$ at $x = 0.5$.

Substitute $x = 0.5$:
$\frac{dL}{dx} = 2(0.5) + \frac{128(0.5)^3 - 48(0.5)^4}{(2-0.5)^2}$
$\frac{dL}{dx} = 1 + \frac{128(0.125) - 48(0.0625)}{1.5^2}$
$\frac{dL}{dx} = 1 + \frac{16 - 3}{2.25} = 1 + \frac{13}{2.25} = 1 + \frac{52}{9} = \frac{61}{9}$

4. Apply the chain rule to find the rate of change of $(OP)^2$ with respect to time.

Given $\frac{dx}{dt} = 0.9$,
$\frac{dL}{dt} = \frac{dL}{dx} \times \frac{dx}{dt}$
$\frac{dL}{dt} = \frac{61}{9} \times 0.9 = 6.1$