大卫操控一架无人机，其飞行轨迹为曲线 $C$，由方程 $y = 2^x$ 建模，其中 $x 60; 0$ 且 $y > 0$。他站在原点 $O$ 处，并在点 $P(a,b)$ 处看到了无人机。假设大卫的视线是曲线 $C$ 在点 $P$ 处且通过原点的切线，求 $P$ 的精确坐标。

1. Express lengths $SQ$ and $QR$ in terms of $t$.

Given that David starts at $O$ and moves at $5\text{ m/s}$, the distance he travels in $t$ seconds is $OS = 5t$.
Since the total distance $OQ = 500\text{ m}$, the remaining distance $SQ$ is:
$SQ = 500 - 5t$

The drone ascends vertically from $Q$ at $20\text{ m/s}$, so its height after $t$ seconds is:
$QR = 20t$

2. Use trigonometry in $\triangle SQR$ to express $\tan\theta$.

In the right-angled triangle $SQR$, the angle at $Q$ is $90^\circ$. For angle $SRQ = \theta$, the opposite side is $SQ$ and the adjacent side is $QR$.
$\tan \theta = \frac{SQ}{QR}$
$\tan \theta = \frac{500 - 5t}{20t}$
Dividing the numerator and the denominator by 5:
$\tan \theta = \frac{100 - t}{4t} \quad \text{(Shown)}$

1. Differentiate both sides of the equation from (b)(i) with respect to $t$ using implicit differentiation and the quotient rule.

$\frac{d}{dt}(\tan\theta) = \frac{d}{dt}\left(\frac{100 - t}{4t}\right)$
$\sec^2\theta \frac{d\theta}{dt} = \frac{4t(-1) - (100 - t)(4)}{(4t)^2}$
$\sec^2\theta \frac{d\theta}{dt} = \frac{-4t - 400 + 4t}{16t^2} = -\frac{25}{t^2}$

2. Express $\sec^2\theta$ in terms of $t$ using the identity $\sec^2\theta = 1 + \tan^2\theta$.

$\sec^2\theta = 1 + \left(\frac{100 - t}{4t}\right)^2$
$\sec^2\theta = 1 + \frac{10000 - 200t + t^2}{16t^2}$
$\sec^2\theta = \frac{16t^2 + 10000 - 200t + t^2}{16t^2}$
$\sec^2\theta = \frac{17t^2 - 200t + 10000}{16t^2}$

3. Substitute the expression for $\sec^2\theta$ into the differentiated equation and simplify to isolate $\frac{d\theta}{dt}$.

$\left(\frac{17t^2 - 200t + 10000}{16t^2}\right) \frac{d\theta}{dt} = -\frac{25}{t^2}$
$\frac{d\theta}{dt} = -\frac{25}{t^2} \times \frac{16t^2}{17t^2 - 200t + 10000}$
$\frac{d\theta}{dt} = -\frac{400}{17t^2 - 200t + 10000}$