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Magazine Chapter 4: Problem 101
Solve the initial value problems in Exercises. $$\frac{d v}{d t}=\frac{1}{2} \sec t \tan t, \quad v(0)=1$$
Short Answer
Expert verified
The solution is \(v(t) = \frac{1}{2} \sec t + \frac{1}{2}\).
Step by step solution
01
Understand the Differential Equation
The given differential equation is \(\frac{dv}{dt} = \frac{1}{2} \sec t \tan t\). This is a first-order differential equation involving \(v\) and \(t\). We are also given an initial condition \(v(0) = 1\). Our goal is to find the function \(v(t)\) that satisfies both the differential equation and the initial condition.
02
Integrate the Right Side
We start by integrating the right-hand side of the equation with respect to \(t\). Since \(\frac{dv}{dt} = \frac{1}{2} \sec t \tan t\), we can integrate: \[ v = \int \frac{1}{2} \sec t \tan t \ dt. \]
03
Use an Antiderivative of \(\sec t \tan t\)
We know that the derivative of \(\sec t\) is \(\sec t \tan t\). So, the integral \(\int \sec t \tan t \ dt\) results in \(\sec t\). Therefore, the antiderivative for \(\frac{1}{2} \sec t \tan t\) is \(\frac{1}{2} \sec t + C\), where \(C\) is the constant of integration.
04
Apply Initial Condition
After integrating, we have: \[ v(t) = \frac{1}{2} \sec t + C. \] Now, we apply the initial condition \(v(0) = 1\) to find \(C\). Since \(\sec(0) = 1\), substituting \(t = 0\) into the equation gives: \[ 1 = \frac{1}{2} \cdot 1 + C \Rightarrow C = \frac{1}{2}. \]
05
Write the Solution
Substituting \(C = \frac{1}{2}\) back into the integrated function, we find: \[ v(t) = \frac{1}{2} \sec t + \frac{1}{2}. \] This function satisfies both the differential equation and the initial condition provided.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Value Problem
In the world of differential equations, an **Initial Value Problem** refers to a scenario where we not only have a differential equation to solve, but also a specific starting condition. This condition helps us find the exact solution that fits our problem.When tackling an initial value problem, we typically:
- Identify the differential equation given. In this case, we had \( \frac{dv}{dt} = \frac{1}{2} \sec t \tan t \).
- Note the initial condition, like \( v(0) = 1 \) in our example. This tells us the value of our unknown function at a specific point, crucial for finding the constant \( C \) later.
- After solving the differential equation, apply this initial condition to determine any constants introduced during the integration process.
Integration Techniques
When you see a differential equation like \( \frac{dv}{dt} = \frac{1}{2} \sec t \tan t \), you must use **Integration Techniques** to solve it. The main goal of integration here is to eliminate the derivative and find the function \( v(t) \).To perform the integration, you follow these steps:
- Set up the integral for the right-hand side of the equation. For our problem, this means solving \( \int \frac{1}{2} \sec t \tan t \, dt \).
- Utilize known integral formulas and substitution methods. Recognize the derivative patterns. Here, since the derivative of \( \sec t \) is \( \sec t \tan t \), we know the integral \( \int \sec t \tan t \, dt = \sec t \).
- Don't forget the constant of integration \( C \), which appears because indefinite integration accounts for all possible antiderivatives.
Antiderivatives
**Antiderivatives** are at the heart of solving differential equations. They allow us to reverse the process of differentiation, restoring the original function that, when differentiated, yields the integrand.Finding an antiderivative involves:
- Recognizing the form of the integrand. In our exercise, the given integrand \( \frac{dv}{dt} = \frac{1}{2} \sec t \tan t \) is linked to the derivative of \( \sec t \).
- Using known formulas or rules to identify the corresponding function. Here, since \( \frac{d}{dt}(\sec t) = \sec t \tan t \), we knew that an antiderivative would be \( \frac{1}{2} \sec t \) plus a constant \( C \).
- Applying initial conditions to solve for any constants introduced by the integration process, ensuring our antiderivative precisely fits the problem context.
