一个小金属球从一个高圆柱形容器内的液体表面被释放。金属球在液体中垂直下落，并测得在时间 $t$ 秒内小球下落的距离为 $x$ cm。小球在时间 $t$ 秒时的速度为 $v\text{ cm s}^{-1}$。释放小球的方式使得当 $t = 0$ 时，$x = 0$ 且 $v = 0$。

1. Express the second derivative of $x$ in terms of $v$.

Given $v = \frac{dx}{dt}$, differentiating both sides with respect to $t$ yields:
$\frac{dv}{dt} = \frac{d^2x}{dt^2}$

2. Substitute into the original differential equation and rearrange to obtain the required form.

Substitute $\frac{d^2x}{dt^2} = \frac{dv}{dt}$ and $\frac{dx}{dt} = v$ into the given differential equation:
$\frac{dv}{dt} + \frac{1}{2}v - 10 = 0$
$\frac{dv}{dt} = 10 - \frac{1}{2}v \quad \text{(shown)}$

1. Separate variables and integrate the differential equation to find a relation for $v$

$\frac{dv}{dt} = \frac{20 - v}{2}$
$\int \frac{1}{20 - v} \, dv = \int \frac{1}{2} \, dt$
$-\ln|20 - v| = 0.5t + C$

2. Apply the initial condition to find the constant and express $v$ explicitly in terms of $t$

Given $v = 0$ when $t = 0$:
$-\ln|20 - 0| = 0.5(0) + C \implies C = -\ln 20$
$-\ln|20 - v| = 0.5t - \ln 20$
$\ln|20 - v| = \ln 20 - 0.5t$
$|20 - v| = e^{\ln 20 - 0.5t} = 20e^{-0.5t}$
Since $v = 0$ initially and $\frac{dv}{dt} > 0$, velocity is increasing but bounded below 20, so $20 - v > 0$. We drop the modulus:
$20 - v = 20e^{-0.5t}$
$v = 20 - 20e^{-0.5t}$

3. Substitute $v = \frac{dx}{dt}$ and integrate to find an expression for $x$

$\frac{dx}{dt} = 20 - 20e^{-0.5t}$
$x = \int (20 - 20e^{-0.5t}) \, dt$
$x = 20t - \frac{20}{-0.5}e^{-0.5t} + D$
$x = 20t + 40e^{-0.5t} + D$

4. Apply the initial condition to find the constant and express $x$ exactly in terms of $t$

Given $x = 0$ when $t = 0$:
$0 = 20(0) + 40e^{0} + D \implies D = -40$
$x = 20t + 40e^{-0.5t} - 40$

1. Separate the variables to set up the integrals.

$\frac{dv}{dt} = 10 - k^{2} v^{2}$
$\int \frac{1}{10 - k^{2} v^{2}} \,dv = \int 1 \,dt$

2. Integrate both sides using the standard formula for $\int \frac{1}{a^2 - x^2} \,dx$.

Factor out $k^2$ in the denominator to match the standard form:
$\frac{1}{k^2} \int \frac{1}{\left(\frac{\sqrt{10}}{k}\right)^2 - v^2} \,dv = \int 1 \,dt$
$\frac{1}{k^2} \left[ \frac{1}{2\left(\frac{\sqrt{10}}{k}\right)} \ln \left| \frac{\frac{\sqrt{10}}{k} + v}{\frac{\sqrt{10}}{k} - v} \right| \right] = t + C$
$\frac{1}{2\sqrt{10}k} \ln \left| \frac{\sqrt{10} + kv}{\sqrt{10} - kv} \right| = t + C$

3. Use the initial conditions to find the constant of integration.

Given that $v = 0$ when $t = 0$:
$\frac{1}{2\sqrt{10}k} \ln \left| \frac{\sqrt{10} + 0}{\sqrt{10} - 0} \right| = 0 + C$
$\frac{1}{2\sqrt{10}k} \ln(1) = C \implies C = 0$

4. Make $v$ the subject of the formula.

$\ln \left| \frac{\sqrt{10} + kv}{\sqrt{10} - kv} \right| = 2\sqrt{10}kt$
Since $v$ starts at $0$ and the liquid exerts resistance, the term inside the modulus is positive:
$\frac{\sqrt{10} + kv}{\sqrt{10} - kv} = e^{2\sqrt{10}kt}$
$\sqrt{10} + kv = \sqrt{10}e^{2\sqrt{10}kt} - kve^{2\sqrt{10}kt}$
$kv + kve^{2\sqrt{10}kt} = \sqrt{10}e^{2\sqrt{10}kt} - \sqrt{10}$
$kv(1 + e^{2\sqrt{10}kt}) = \sqrt{10}(e^{2\sqrt{10}kt} - 1)$
$v = \frac{\sqrt{10}}{k} \left( \frac{e^{2\sqrt{10}kt} - 1}{e^{2\sqrt{10}kt} + 1} \right)$

1. Concise derivation of $v$ in terms of $t$ (result needed from part (b)(i)).

From part (b)(i), separating variables for $\frac{dv}{dt} = 10 - k^2 v^2$:
$\int \frac{1}{\left(\frac{\sqrt{10}}{k}\right)^2 - v^2} \,dv = \int k^2 \,dt$
$\frac{k}{2\sqrt{10}} \ln \left| \frac{\frac{\sqrt{10}}{k} + v}{\frac{\sqrt{10}}{k} - v} \right| = k^2 t + C$
Using $v=0$ when $t=0$, we find $C = 0$. Rearranging yields:
$\frac{\sqrt{10} + kv}{\sqrt{10} - kv} = e^{2\sqrt{10}kt}$
$v = \frac{\sqrt{10}}{k} \left( \frac{e^{2\sqrt{10}kt} - 1}{e^{2\sqrt{10}kt} + 1} \right) = \frac{\sqrt{10}}{k} \left( \frac{1 - e^{-2\sqrt{10}kt}}{1 + e^{-2\sqrt{10}kt}} \right)$

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

The terminal speed is attained after a long time, so we evaluate the limit as $t \to \infty$.
Since $k$ is a given positive constant ($k > 0$), the exponent $-2\sqrt{10}kt \to -\infty$ as $t \to \infty$.
Thus, $e^{-2\sqrt{10}kt} \to 0$.
Substituting this limit into the expression for $v$:
$v \to \frac{\sqrt{10}}{k} \left( \frac{1 - 0}{1 + 0} \right) = \frac{\sqrt{10}}{k}$
(Note: An alternative acceptable justification is to state that as terminal speed is reached, $v$ approaches a constant and acceleration $\frac{dv}{dt} \to 0$. Solving $0 = 10 - k^2v^2$ directly yields $v = \frac{\sqrt{10}}{k}$.)