在一个化学反应中，化合物 X 由两种化合物 Y 和 Z 反应生成。 在反应开始后 t 秒时，存在的 X、Y 和 Z 的质量（以克为单位）分别为 $x$、$15 - x$ 和 $25 - x$。在任何时刻，X 的生成速率与此时存在的 Y 和 Z 的质量乘积成正比。已知当 $t = 0$ 时，$x = 0$ 且 $\frac{dx}{dt} = 3$。

1. Translate the physical relationship into a differential equation.

Let $k$ be the constant of proportionality.
The rate of formation of X is represented by $\frac{dx}{dt}$.
Since this rate is proportional to the product of the masses of Y and Z present,
$\frac{dx}{dt} = k(15 - x)(25 - x)$

2. Substitute the initial conditions to find the constant of proportionality.

Given that $x = 0$ and $\frac{dx}{dt} = 3$ when $t = 0$, substitute these values into the equation:
$3 = k(15 - 0)(25 - 0)$
$3 = 375k$
$k = \frac{3}{375} = \frac{1}{125}$

1. State the differential equation from part (i) and separate the variables.

From part (i), the differential equation is $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{125}(15-x)(25-x)$.

Separating the variables:
$\int \frac{1}{(15-x)(25-x)} \, \mathrm{d}x = \int \frac{1}{125} \, \mathrm{d}t$

2. Express the rational function in partial fractions.

Let $\frac{1}{(15-x)(25-x)} = \frac{A}{15-x} + \frac{B}{25-x}$
$1 = A(25-x) + B(15-x)$

Substitute $x = 15$:
$1 = 10A \implies A = \frac{1}{10}$
Substitute $x = 25$:
$1 = -10B \implies B = -\frac{1}{10}$

Therefore, the integral becomes:
$\frac{1}{10} \int \left( \frac{1}{15-x} - \frac{1}{25-x} \right) \, \mathrm{d}x = \int \frac{1}{125} \, \mathrm{d}t$

3. Integrate both sides to form an equation involving logarithms.

$\frac{1}{10} \left( -\ln(15-x) + \ln(25-x) \right) = \frac{1}{125}t + C$
$\ln\left(\frac{25-x}{15-x}\right) = \frac{2}{25}t + 10C$

(Note: Modulus signs are omitted because $x$ starts at $0$ and cannot exceed $15$, ensuring $15-x > 0$ and $25-x > 0$.)

4. Substitute the initial conditions to find the constant of integration.

Given $x = 0$ when $t = 0$:
$\ln\left(\frac{25-0}{15-0}\right) = \frac{2}{25}(0) + 10C$
$10C = \ln\left(\frac{25}{15}\right) = \ln\left(\frac{5}{3}\right)$

Substituting this back:
$\ln\left(\frac{25-x}{15-x}\right) = \frac{2}{25}t + \ln\left(\frac{5}{3}\right)$

5. Rearrange the equation to express x as the subject.

$\frac{25-x}{15-x} = \mathrm{e}^{0.08t + \ln(5/3)}$
$\frac{25-x}{15-x} = \frac{5}{3}\mathrm{e}^{0.08t}$
$3(25-x) = 5(15-x)\mathrm{e}^{0.08t}$
$75 - 3x = 75\mathrm{e}^{0.08t} - 5x\mathrm{e}^{0.08t}$
$x(5\mathrm{e}^{0.08t} - 3) = 75\mathrm{e}^{0.08t} - 75$
$x = \frac{75(\mathrm{e}^{0.08t} - 1)}{5\mathrm{e}^{0.08t} - 3}$

1. Extract the key features of the curve from the solution

From part (ii), $x = 0$ when $t = 0$, so the curve passes through the origin.

Given $\dfrac{dx}{dt} = 3$ at $t = 0$, the curve starts with gradient $3$ (rising).

Since $x$ increases towards the limiting mass of $Y$, as $t \to \infty$, $\dfrac{25-x}{15-x} = \dfrac{5}{3}e^{2t/25} \to \infty$, which forces $x \to 15$. Hence $x = 15$ is a horizontal asymptote.

2. Sketch the graph of x against t

Draw axes ($t$ horizontal, $x$ vertical). Sketch a curve that:

• starts at the origin $(0,0)$ with positive gradient $3$;
• rises throughout (always increasing);
• bends over (concave down) and flattens out, approaching the dashed horizontal line $x = 15$ from below without touching it.

1. Derive the relationship between $x$ and $t$ from the earlier parts of the question.

From part (i), the differential equation is $\frac{dx}{dt} = k(15-x)(25-x)$.
Given $x=0, \frac{dx}{dt}=3$ when $t=0$:
$3 = k(15)(25) \implies k = \frac{1}{125}$
Solving the differential equation for part (ii) by separating variables:
$\int \frac{1}{(15-x)(25-x)} \, dx = \int \frac{1}{125} \, dt$
Using partial fractions:
$\frac{1}{10} \int \left( \frac{1}{15-x} - \frac{1}{25-x} \right) \, dx = \frac{t}{125} + C$
$\ln(25-x) - \ln(15-x) = \frac{2t}{25} + C'$
Substituting $t=0, x=0$ gives $C' = \ln\left(\frac{25}{15}\right) = \ln\left(\frac{5}{3}\right)$.
Thus, the relationship is:
$\frac{2t}{25} = \ln\left(\frac{25-x}{15-x}\right) - \ln\left(\frac{5}{3}\right) = \ln\left(\frac{3(25-x)}{5(15-x)}\right)$

2. Determine the value of $x$ when the mass of X equals the mass of Y.

The mass of X is $x$ and the mass of Y is $15 - x$.
Equating the two masses:
$x = 15 - x$
$2x = 15 \implies x = 7.5$

3. Calculate the corresponding time $t$.

Substitute $x = 7.5$ into the derived equation:
$\frac{2t}{25} = \ln\left( \frac{3(25 - 7.5)}{5(15 - 7.5)} \right)$
$\frac{2t}{25} = \ln\left( \frac{3(17.5)}{5(7.5)} \right) = \ln\left( \frac{52.5}{37.5} \right)$
$\frac{2t}{25} = \ln(1.4)$
$t = 12.5 \ln(1.4) \approx 4.21 \text{ s}$

1. Evaluate the limit of $x$ as $t$ approaches infinity using the expression from part (ii).

Using the expression for $x$ derived in part (ii), $x = \frac{75(e^{\frac{2t}{25}} - 1)}{5e^{\frac{2t}{25}} - 3}$.

Dividing the numerator and the denominator by $e^{\frac{2t}{25}}$:
$x = \frac{75(1 - e^{-\frac{2t}{25}})}{5 - 3e^{-\frac{2t}{25}}}$

As $t \to \infty$, $e^{-\frac{2t}{25}} \to 0$.
$\lim_{t \to \infty} x = \frac{75(1 - 0)}{5 - 0} = 15$

2. Interpret the mathematical limit in the context of the problem.

The mass of compound X approaches 15 grams (as the limiting compound Y is completely used up).

1. Identify the parameters of the geometric progression representing the distances.

Let the distance run in the first session be $a = 2000\text{ m}$.
Since the distance decreases by $10\%$ in each subsequent session, the common ratio $r = 1 - 0.10 = 0.9$.

2. Calculate the sum to infinity of the geometric progression.

$S_\infty = \frac{a}{1 - r}$
$S_\infty = \frac{2000}{1 - 0.9} = \frac{2000}{0.1} = 20000\text{ m}$

3. Convert units to kilometres and deduce that the total accumulated distance never exceeds this limit.

$20000\text{ m} = 20\text{ km}$
For any finite number of sessions $n$, the accumulated distance is $S_n = 20000(1 - 0.9^n)$.
Since $0.9^n > 0$ for all integers $n > 0$, $S_n < 20000\text{ m} = 20\text{ km}$.
Therefore, Wendie will never accumulate a total distance exceeding $20\text{ km}$ (Shown).