91Ó°ÊÓ

In the context of our problem, the rate of conversion refers to how quickly the substance transforms from its original state to a converted state over time. Since it's given that this rate is proportional to the square root of the unconverted amount, it suggests that as less of the substance remains, the rate at which it disappears also decreases. Mathematically, this can be expressed with the equation: \[\frac{dx}{dt} = -k\sqrt{x}\]Here,

• \(x\) represents the amount of unconverted substance.
• \(k\) is the proportionality constant.
• The negative sign indicates that the quantity of the unconverted substance decreases over time.

This relationship implies that the conversion slows down as the substance gets converted. It's an important feature because it naturally limits how fast the substance can disappear, linked directly to the remaining amount.

Separation of variables is a technique used to solve differential equations, like the one given in the problem. It is particularly useful when you can rewrite the equation to isolate the functions of different variables on opposite sides.

The primary goal is to rearrange the equation: \[\frac{dx}{dt} = -k\sqrt{x}\]into a form where each side depends only on one variable. We achieve this by moving terms around: \[\frac{dx}{\sqrt{x}} = -k \, dt\]Now, you have a perfect form for integration: the left-hand side includes everything with \(x\), and the right-hand side includes everything with \(t\). By separating variables like this, we make it easier to perform the next step: integration.

Integration is the process of finding the integral of a function, which in this context, means finding the antiderivative. Once we separate variables, the next logical step is to integrate both sides of the equation.

For the equation \[\int \frac{dx}{\sqrt{x}} = \int -k \, dt\]the integration results are:

• The left side integrates to \(2\sqrt{x}\).
• The right side integrates to \(-kt + C\).

This gives us the integrated equation:\[2\sqrt{x} = -kt + C\]Integration helps us move from a derivative (rate of change) to a function representing the quantity itself. Thus, following integration, we get a clearer view of how \(x\) changes with time.

Initial conditions are crucial as they allow you to find the particular solution to a differential equation. In our case, we need to find the value of the integration constant \(C\). To do this, we use the known value of \(x\) at a known time, usually when time \(t = 0\).

For this problem, suppose the initial amount of substance is \(x_0\) at \(t = 0\). We substitute these into our integrated equation: \[2\sqrt{x_0} = C\]Thus, \(C = 2\sqrt{x_0}\).Substituting back, we get:\[2\sqrt{x} = -kt + 2\sqrt{x_0}\]Initial conditions personalize the general solution to fit the specific scenario. It ensures that our solution appropriately reflects the situation given in the problem, helping us predict exactly when the substance will disappear, which we found to be at \(t = \frac{2\sqrt{x_0}}{k}\).