Problem 10 - Calculus for Biology and Medicine

Question

Suppose that the amount of phosphorus in a lake at time \(t\). denoted by \(P(t)\), follows the equation $$ \frac{d P}{d t}=3 t+1 \quad \text { with } P(0)=0 $$

Step-by-Step Solution

Answer

The phosphorus amount is \(P(t) = \frac{3}{2}t^2 + t\).

Step 1: Understanding the Problem

We are given a differential equation \(\frac{dP}{dt} = 3t + 1\), which represents the rate of change of phosphorus \(P(t)\) in a lake over time. We also have the initial condition \(P(0) = 0\). Our task is to find the expression for \(P(t)\).

Step 2: Integrate the Differential Equation

To find \(P(t)\), we need to integrate the differential equation. Integrating \(\frac{dP}{dt} = 3t + 1\) with respect to \(t\), we have: \(P(t) = \int (3t + 1) \, dt\).

Step 3: Solve the Integral

Now, solve the integral: \(\int (3t + 1) \, dt\) results in \(\frac{3}{2}t^2 + t + C\), where \(C\) is the constant of integration.

Step 4: Use Initial Condition to Find Constant

We use the initial condition \(P(0) = 0\) to find the constant \(C\). Substitute \(t = 0\) and \(P = 0\) into the equation \(P(t) = \frac{3}{2}t^2 + t + C\). Solving \(0 = \frac{3}{2}(0)^2 + 0 + C\), we find \(C = 0\).

Step 5: Write the Final Expression for P(t)

Substitute \(C = 0\) back into the expression for \(P(t)\), giving \(P(t) = \frac{3}{2}t^2 + t\). This is the final expression for the amount of phosphorus at time \(t\).

Key Concepts

Integration

In differential equations, integration is a crucial technique for finding an unknown function when we know its rate of change. When we have a differential equation like \(\frac{dP}{dt} = 3t + 1\), this tells us how the function \(P(t)\) changes over time. To find \(P(t)\), we need to reverse the process of differentiation, which is done through integration.

When we integrate the expression \(3t + 1\) with respect to \(t\), we calculate the antiderivative, which gives us

• \(\int 3t \, dt = \frac{3}{2} t^2\)
• \(\int 1 \, dt = t\)

This yields \(\frac{3}{2}t^2 + t + C\), where \(C\) is the integration constant.

Integration turns the rate of change back into the original function, plus any adjustments needed to fit specific circumstances, expressed by \(C\). Calculating this precisely is why integration is essential in solving differential equations.

Initial Conditions

Initial conditions are specified values that allow us to find the unique solution to a differential equation. Here, we know that at time \(t = 0\), the phosphorus amount is \(P(0) = 0\). This condition is essential because the process of integration introduces an arbitrary constant \(C\).

Without knowing \(C\), we would have many possible solutions for \(P(t)\). Initial conditions allow us to pinpoint the correct \(C\). By substituting \(t = 0\) into the expression \(P(t) = \frac{3}{2}t^2 + t + C\), and knowing \(P(0) = 0\), we simplify to:

• \(0 = \frac{3}{2}(0)^2 + 0 + C\)

This calculation shows \(C = 0\).

Thanks to the initial condition, we've determined the exact form of our function, \(P(t) = \frac{3}{2}t^2 + t\). Initial conditions ensure our solution accurately represents the real-world scenario described by the problem.

Rate of Change

The term "rate of change" refers to how a quantity, like phosphorus in a lake, changes over time. In differential equations, the rate of change is represented by derivatives. Here, \(\frac{dP}{dt} = 3t + 1\) is such a rate of change equation.

This tells us that the phosphorus increases in a manner proportional to time with an additional constant increment. Derivatives give us an immediate snapshot of how quickly or slowly the quantity is changing at any moment in time.

Understanding the rate of change is essential for predicting behavior over time. When integrated, it not only provides the total change over a period but also helps differentiate dynamic systems.

• The component \(3t\) means the phosphorus increase speeds up over time insignificance to time's square.
• The added \(+1\) describes a constant steady increase aside from time.

Recognizing the nuances of rate of change in the equation helps to model and forecast the system's future status, making it a key element of differential equations.