Problem 9 - Calculus for Biology and Medicine

Question

Suppose that the volume \(V(t)\) of a cell at time \(t\) changes according to $$\frac{d V}{d t}=1+\cos t \quad \text { with } V(0)=5$$ Find \(V(t)\).

Step-by-Step Solution

Answer

The volume function is \( V(t) = t + \sin t + 5 \).

Step 1: Identify the Type of Differential Equation

The given differential equation \( \frac{d V}{d t} = 1 + \cos t \) is a first-order ordinary differential equation (ODE). This is a straightforward example of a separable differential equation, meaning it can be integrated directly with respect to \(t\).

Step 2: Integrate Both Sides

To solve \( \frac{d V}{d t} = 1 + \cos t \), integrate both sides with respect to \( t \):\[V(t) = \int (1 + \cos t) \, dt.\]

Step 3: Perform the Integration

Calculate the integral: \[\int (1 + \cos t) \, dt = \int 1 \, dt + \int \cos t \, dt.\]Integrating, we find:\[V(t) = t + \sin t + C,\]where \( C \) is the integration constant.

Step 4: Apply Initial Condition

Use the initial condition \( V(0) = 5 \) to find \( C \):\[V(0) = 0 + \sin(0) + C = 5.\]Simplifying, we have \( C = 5 \).

Step 5: Write the Particular Solution

Substitute \( C = 5 \) back into the equation for \( V(t) \):\[V(t) = t + \sin t + 5.\]

Key Concepts

Initial Condition

In the world of differential equations, an initial condition is a piece of information that allows us to find a specific solution to the equation. Think of it as a starting point. In the exercise, the initial condition is given as \(V(0)=5\). This means when time \(t = 0\), the volume \(V\) has a value of 5.
Without this information, we would end up with a general solution filled with unknown constants. But with an initial condition, we find a particular solution that matches the exact scenario described.
In mathematical terms, the initial condition helps us solve for the integration constant \(C\). Once we know \(C\), we can express the solution more specifically to reflect reality as described by the problem.

Integration

Integration is a fundamental concept in calculus. It's the process of finding the integral of a function, which in essence allows us to find the area under a curve or convey an accumulated quantity.
In this context, integration is used to solve the differential equation \(\frac{dV}{dt} = 1 + \cos t\). By integrating both sides, we express the change in volume \(V\) over time \(t\) as a function.
Let's break it down:

• The left side \(\int dV\) represents an accumulation leading to \(V(t)\).
• The right side involves breaking down the integral \(\int (1 + \cos t) \, dt\) into individual parts that we can solve: \(\int 1 \, dt\) and \(\int \cos t \, dt\).
• Calculating these gives us \(V(t) = t + \sin t + C\), where \(C\) is the constant of integration, representing the unknown starting value.

Integration is a powerful tool in differential equations, serving as the key to unlock unknown quantities from rates of change. Once integration is performed, you can adjust the result using initial conditions.

First-Order Ordinary Differential Equation

A first-order ordinary differential equation (ODE) involves the first derivatives, which indicate the rate of change. These are the simplest type of ODEs encountered in differential equations.
For our exercise, the equation \(\frac{dV}{dt} = 1 + \cos t\) is a first-order ODE. Here, \(\frac{dV}{dt}\) tells us how the volume changes with small increments of time. We want to transition from this rate of change to an explicit form expressing \(V\) as a function of \(t\).
Key traits of a first-order ODE:

• They contain only first derivatives (no higher-order derivatives).
• They often describe processes evolving over time (like growth, decay, or oscillation).
• Separation of variables and direct integration are common methods used to solve them.

Once you identify a differential equation as first-order, solving it typically involves recognizing the structure and applying methods like integration, as demonstrated in our given exercise.