Glucose is a type of sugar in our blood and our body uses it for energy. Having low sugar in the blood for long periods of time can cause health problems if it is not treated. At 10:00 am, a patient is given glucose via an intravenous drip at a constant rate of $r$ units of glucose per hour. It is known that the rate at which the glucose in the blood stream is absorbed by the body, is proportional to the amount of glucose present in the blood stream. At the start of the treatment, the patient has $x_0$ units of glucose in his blood stream. It is given that $x$ denotes the amount of glucose present in the blood stream of the patient at time $t$ hours after he starts the treatment.

1. Formulate the differential equation based on the given rates.

The rate at which glucose is added is $r$.
The rate at which it is absorbed is proportional to $x$, so the rate out is $kx$, where $k$ is a positive constant.
$\frac{dx}{dt} = \text{Rate in} - \text{Rate out}$
$\frac{dx}{dt} = r - kx$

2. Separate variables and integrate both sides.

$\int \frac{1}{r - kx} \, dx = \int 1 \, dt$
$-\frac{1}{k} \ln |r - kx| = t + C$

3. Express the general solution without logarithms.

$\ln |r - kx| = -kt - kC$
$r - kx = \pm e^{-kC} e^{-kt}$
Let $A = \pm e^{-kC}$.
$r - kx = A e^{-kt}$

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

At $t = 0, x = x_0$:
$r - kx_0 = A e^{0} \implies A = r - kx_0$
Substituting $A$ back into the equation:
$r - kx = (r - kx_0) e^{-kt}$
$kx = r - (r - kx_0)e^{-kt}$
$x = \frac{r}{k} - \left(\frac{r}{k} - x_0\right)e^{-kt}$
$x = \left(x_0 - \frac{r}{k}\right)e^{-kt} + \frac{r}{k} \quad \text{(shown)}$

1. Determine the behaviour of $x$ as $t \to \infty$.

As $t \to \infty$, since $k$ is a positive constant, $e^{-kt} \to 0$.

Therefore, $x \to \left( x_0 - \frac{r}{k} \right)(0) + \frac{r}{k} = \frac{r}{k}$.

1. Determine the intercepts and asymptotes of the graph.

At $t = 0$, $x = x_0$. Thus, the vertical intercept is $(0, x_0)$.
From part (b), as $t \to \infty$, the theoretical limiting value is $x = \frac{r}{k}$. Thus, there is a horizontal asymptote at $x = \frac{r}{k}$.

2. Deduce the shape of the curve based on the context and sketch the graph.

Since the patient is receiving treatment for low blood sugar, the initial amount $x_0$ is less than the target stable level $\frac{r}{k}$. This means $\left( x_0 - \frac{r}{k} \right)$ is a negative constant.
As $t$ increases, $e^{-kt}$ decreases, so the expression $\left( x_0 - \frac{r}{k} \right)e^{-kt}$ approaches $0$ from below. Therefore, $x$ increases over time, starting from $x_0$ and asymptotically approaching $\frac{r}{k}$.

Key features for the sketch:

1. Axes correctly labelled $t$ and $x$.
2. A horizontal dotted line for the asymptote at $x = \frac{r}{k}$.
3. A concave-down curve starting at $(0, x_0)$ and increasing towards $x = \frac{r}{k}$.