一名“红狮”（Red Lion）跳伞运动员从直升机上跳下。他离开直升机时的速度为零，他的速度 $v\text{ ms}^{-1}$ 满足微分方程 $\frac{\text{d}v}{\text{d}t} = 2\text{e}^{-0.1t}.$

1. Integrate the differential equation to find the general solution for $v$.

$\frac{\text{d}v}{\text{d}t} = 2\text{e}^{-0.1t}$
$v = \int 2\text{e}^{-0.1t} \,\text{d}t$
$v = \frac{2}{-0.1}\text{e}^{-0.1t} + C$
$v = -20\text{e}^{-0.1t} + C$

2. Apply the initial condition to find the particular solution.

The skydiver leaves with zero speed, so when $t = 0$, $v = 0$.
$0 = -20\text{e}^{0} + C$
$0 = -20 + C \implies C = 20$
Therefore,
$v = 20 - 20\text{e}^{-0.1t}$

3. Substitute $v = 10$ to find the exact time taken to reach $10\text{ ms}^{-1}$.

$10 = 20 - 20\text{e}^{-0.1t}$
$20\text{e}^{-0.1t} = 10$
$\text{e}^{-0.1t} = 0.5$
Taking the natural logarithm on both sides:
$-0.1t = \ln(0.5)$
$t = -10\ln\left(\frac{1}{2}\right)$
$t = 10\ln 2$

1. Derive the expression for $v$ in terms of $t$ (required from part i).

$v = \int 2\text{e}^{-0.1t} \,\text{d}t = -20\text{e}^{-0.1t} + C$
Given $v = 0$ when $t = 0$, $0 = -20(1) + C \implies C = 20$.
Thus, $v = 20 - 20\text{e}^{-0.1t}$.

2. Evaluate the limit of $v$ as $t \to \infty$ to determine the eventual speed.

As $t \to \infty$, $\text{e}^{-0.1t} \to 0$.
Therefore, $v \to 20 - 20(0) = 20$.

1. Separate the variables to set up the integral.

$\int \frac{1}{(w-3)(w+2)} \, \text{d}w = \int -\frac{1}{2} \, \text{d}t$

2. Express the integrand using partial fractions.

$\frac{1}{(w-3)(w+2)} = \frac{A}{w-3} + \frac{B}{w+2}$
$1 = A(w+2) + B(w-3)$
Substituting $w = 3 \implies A = \frac{1}{5}$
Substituting $w = -2 \implies B = -\frac{1}{5}$
$\int \left( \frac{1}{5(w-3)} - \frac{1}{5(w+2)} \right) \, \text{d}w = \int -\frac{1}{2} \, \text{d}t$

3. Integrate both sides to obtain an equation relating $w$ and $t$.

$\frac{1}{5} \ln|w-3| - \frac{1}{5} \ln|w+2| = -\frac{1}{2}t + C_1$
$\ln \left| \frac{w-3}{w+2} \right| = -\frac{5}{2}t + C$

4. Use the initial condition to find the constant of integration.

Given $w = 18$ when $t = 0$:
$\ln \left| \frac{18-3}{18+2} \right| = -\frac{5}{2}(0) + C$
$C = \ln \left( \frac{15}{20} \right) = \ln \frac{3}{4}$
$\ln \left( \frac{w-3}{w+2} \right) = -2.5t + \ln \frac{3}{4}$

5. Solve for $w$ to express it explicitly in terms of $t$.

$\frac{w-3}{w+2} = \text{e}^{-2.5t + \ln(0.75)}$
$\frac{w-3}{w+2} = \frac{3}{4} \text{e}^{-2.5t}$
$4(w-3) = 3(w+2)\text{e}^{-2.5t}$
$4w - 12 = 3w\text{e}^{-2.5t} + 6\text{e}^{-2.5t}$
$w(4 - 3\text{e}^{-2.5t}) = 12 + 6\text{e}^{-2.5t}$
$w = \frac{12 + 6\text{e}^{-2.5t}}{4 - 3\text{e}^{-2.5t}}$

1. Concisely find $w$ in terms of $t$ (result of part iii) to determine the shape of the graph.

Separating the variables from $-2\frac{\text{d}w}{\text{d}t} = (w - 3)(w + 2)$:
$\int \left( \frac{1}{w-3} - \frac{1}{w+2} \right) \text{d}w = \int -2.5 \,\text{d}t$
$\ln\left|\frac{w-3}{w+2}\right| = -2.5t + C$
Using $t = 0, w = 18$:
$C = \ln\left(\frac{15}{20}\right) = \ln\left(\frac{3}{4}\right)$
$\frac{w-3}{w+2} = \frac{3}{4}\text{e}^{-2.5t}$
Rearranging for $w$:
$w = \frac{12 + 6\text{e}^{-2.5t}}{4 - 3\text{e}^{-2.5t}}$

2. Determine the initial value and the horizontal asymptote to sketch the graph.

From the initial condition, the $w$-intercept is $(0, 18)$.
To find the long-term behaviour, take the limit as $t \to \infty$:
As $t \to \infty$, $\text{e}^{-2.5t} \to 0$.
$w \to \frac{12 + 0}{4 - 0} = 3$
Thus, $w = 3$ is a horizontal asymptote. The graph is a smooth curve starting at $(0, 18)$ and continually decreasing, tending towards $w=3$ from above.

3. State the speed that the skydiver will not fall below.

From the sketch and asymptotic behaviour, the velocity $w$ approaches $3\text{ ms}^{-1}$ but never goes below it.
Therefore, the skydiver will not fall below $3\text{ ms}^{-1}$.