A wafer fabrication company uses the floating-zone method to purify polysilicon ingots, each having a uniform cross-sectional area and a length of 200 cm. The method involves placing a polysilicon ingot with impurity concentration $C_0$ atoms/$\text{cm}^3$ on top of a single seed crystal. The polysilicon ingot is then heated externally by an RF coil, which locally melts the ingot. The impurities prefer to stay in the molten state than in the solid state and thus as the coil and the melt zone move upwards, a single crystal, that is purer, solidifies on top of the seed crystal. A schematic illustration of the method is shown below in Fig. 1 and Fig. 2. For a 'floating' melt zone of length $L$ cm, the concentration of impurities in the melt zone, $C$ atoms/$\text{cm}^3$, and the distance moved by the RF coil, $x$ cm, are related by the differential equation
$\frac{\text{d}C}{\text{d}x} = \frac{1}{L}(C_0 - kC),$
where $k$ is a constant such that $0 < k < 1$. The length of the "floating" melt zone, $L$ cm, adopted by the company is 2 cm and $0 \le x \le 198$. It is also given that when $x = 0$, $C = C_0$.

1. Substitute the given value of $L$ into the differential equation and separate the variables.

Given $L = 2$, the differential equation becomes:
$\frac{\text{d}C}{\text{d}x} = \frac{1}{2}(C_0 - kC)$
Separating the variables $C$ and $x$:
$\int \frac{1}{C_0 - kC} \, \text{d}C = \int \frac{1}{2} \, \text{d}x$

2. Integrate both sides to find the general solution.

$-\frac{1}{k} \ln |C_0 - kC| = \frac{x}{2} + A$

3. Apply the initial condition to find the constant of integration, $A$.

Given that when $x = 0$, $C = C_0$:
$-\frac{1}{k} \ln |C_0 - kC_0| = \frac{0}{2} + A$
Since $0 < k < 1$ and $C_0 > 0$, we have $C_0 - kC_0 = C_0(1 - k) > 0$. Therefore:
$A = -\frac{1}{k} \ln [C_0(1 - k)]$
Substituting $A$ back into the equation:
$-\frac{1}{k} \ln |C_0 - kC| = \frac{x}{2} - \frac{1}{k} \ln [C_0(1 - k)]$

4. Rearrange the equation to make $C$ the subject.

Multiply the entire equation by $-k$:
$\ln |C_0 - kC| = -\frac{kx}{2} + \ln [C_0(1 - k)]$
$\ln |C_0 - kC| - \ln [C_0(1 - k)] = -\frac{kx}{2}$
$\ln \left| \frac{C_0 - kC}{C_0(1 - k)} \right| = -\frac{kx}{2}$
Since $C_0 - kC$ is positive initially and continuous, we can drop the absolute value:
$\frac{C_0 - kC}{C_0(1 - k)} = e^{-\frac{kx}{2}}$
$C_0 - kC = C_0(1 - k)e^{-\frac{kx}{2}}$
$kC = C_0 - C_0(1 - k)e^{-\frac{kx}{2}}$
$C = \frac{C_0}{k} \left[ 1 - (1 - k)e^{-\frac{kx}{2}} \right]$

1. Identify the initial point and horizontal asymptote from the solution for $C$.

From part (a), the solution is $C = \frac{C_0}{k} \left( 1 - (1-k) e^{-\frac{kx}{2}} \right)$.
When $x = 0$, $C = C_0$.
As $x \to \infty$, $e^{-\frac{kx}{2}} \to 0$ (since $k > 0$).
Thus, $C \to \frac{C_0}{k}$. The horizontal asymptote is $C = \frac{C_0}{k}$.

2. Determine the shape of the curve based on the first derivative.

From the given differential equation, $\frac{\text{d}C}{\text{d}x} = \frac{1}{2}(C_0 - kC)$.
Since $0 < k < 1$ and $C$ starts at $C_0$, the initial gradient is $\frac{1}{2}C_0(1-k) > 0$.
As $C$ increases towards $\frac{C_0}{k}$, the term $(C_0 - kC)$ remains positive but decreases towards $0$.
Therefore, the gradient is positive but strictly decreasing, meaning the curve is increasing and concave downwards.

3. Sketch the graph incorporating the valid domain.

Draw the horizontal $x$-axis and vertical $C$-axis.
Plot the $C$-intercept at $(0, C_0)$.
Draw a horizontal dotted line at $C = \frac{C_0}{k}$ (note that $\frac{C_0}{k} > C_0$ since $0 < k < 1$).
Draw a smooth, increasing, concave down curve from the intercept, approaching the asymptote.
Terminate the graph at $x = 198$, matching the domain given in the context.

1. Briefly solve the differential equation from part (a) to express $C$ in terms of $x$, as the result is required.

As required from part (a), we first find the expression for $C$ by separating variables and integrating:
$\int \frac{1}{C_0 - kC} \,\text{d}C = \int \frac{1}{2} \,\text{d}x$
$-\frac{1}{k} \ln |C_0 - kC| = \frac{x}{2} + A$
Using the initial condition $C = C_0$ when $x = 0$:
$A = -\frac{1}{k} \ln(C_0 - kC_0) = -\frac{1}{k} \ln(C_0(1-k))$
$\ln\left(\frac{C_0 - kC}{C_0(1-k)}\right) = -\frac{kx}{2} \implies C = \frac{C_0}{k}\left(1 - (1-k)e^{-\frac{kx}{2}}\right)$

2. Substitute $C = 2C_0$ and $k = 0.3$ to find the distance $x$ moved.

$2C_0 = \frac{C_0}{0.3}\left(1 - (1 - 0.3)e^{-\frac{0.3x}{2}}\right)$
$0.6 = 1 - 0.7e^{-0.15x}$
$0.7e^{-0.15x} = 0.4 \implies e^{-0.15x} = \frac{4}{7}$
$x = -\frac{1}{0.15}\ln\left(\frac{4}{7}\right) = \frac{20}{3}\ln 1.75 \approx 3.7308 \text{ cm}$

3. Convert the speed to consistent units and calculate the time taken $t$.

The distance $x$ is in cm, so the upward speed must be in cm/h:
$v = 8 \text{ mm/h} = 0.8 \text{ cm/h}$
$t = \frac{x}{v} = \frac{3.7308}{0.8} \approx 4.6635 \text{ hours}$
So the time taken is approximately $4.66$ hours.

4. Set up the relationship for the rate of change of $C$ with respect to time using the chain rule.

$\frac{\text{d}C}{\text{d}t} = \frac{\text{d}C}{\text{d}x} \times \frac{\text{d}x}{\text{d}t}$
We know $\frac{\text{d}x}{\text{d}t} = 0.8 \text{ cm/h}$ and the differential equation gives $\frac{\text{d}C}{\text{d}x} = \frac{1}{2}(C_0 - kC)$.

5. Evaluate the rate of change of concentration at the instant when $C = 2C_0$.

At the instant when $C = 2C_0$ and $k = 0.3$:
$\frac{\text{d}C}{\text{d}x} = \frac{1}{2}(C_0 - 0.3(2C_0)) = \frac{1}{2}(C_0 - 0.6C_0) = 0.2C_0$
$\frac{\text{d}C}{\text{d}t} = (0.2C_0) \times 0.8 = 0.16C_0$

1. Concisely solve the differential equation from part (a) to find the expression for $C$.

Separating variables for the given differential equation:
$\int \frac{1}{C_0 - kC} \, \mathrm{d}C = \int \frac{1}{L} \, \mathrm{d}x$
$-\frac{1}{k} \ln(C_0 - kC) = \frac{x}{L} + A$
Using the initial condition $C = C_0$ when $x = 0$:
$A = -\frac{1}{k} \ln(C_0(1-k))$
$\ln \left( \frac{C_0 - kC}{C_0(1-k)} \right) = -\frac{kx}{L} \implies C_0 - kC = C_0(1-k)\mathrm{e}^{-\frac{kx}{L}}$
$C = \frac{C_0}{k} \left[ 1 - (1-k)\mathrm{e}^{-\frac{kx}{L}} \right]$

2. Analyze the effect of a shorter melt zone length $L$ on the impurity concentration $C$.

From the expression for $C$, for a shorter length $L$, the value of $\frac{kx}{L}$ is larger, making $\mathrm{e}^{-\frac{kx}{L}}$ smaller.
Since $1-k > 0$, the term $1 - (1-k)\mathrm{e}^{-\frac{kx}{L}}$ becomes larger.
Therefore, a shorter length $L$ results in a larger $C$ for any given $x > 0$.
(Alternatively, directly from $\frac{\mathrm{d}C}{\mathrm{d}x} = \frac{1}{L}(C_0 - kC)$, a smaller $L$ yields a larger positive $\frac{\mathrm{d}C}{\mathrm{d}x}$, meaning $C$ increases more rapidly.)

3. Relate the higher concentration $C$ to the purity of the solidifying crystal to draw a conclusion.

A larger $C$ means a higher concentration of impurities in the melt zone.
Because the newly formed single crystal draws its material from this melt zone, a higher impurity concentration in the melt leads to a higher concentration of impurities solidifying into the crystal.
As this makes the single crystal less pure, a shorter length is not preferable.