Problem 61 - Calculus Early Transcendentals

Question

For the following functions \(f\), find the antiderivative \(F\) that satisfies the given condition. $$f(v)=\sec v \tan v ; F(0)=2$$

Step-by-Step Solution

Answer

Question: Find the antiderivative F(v) of the function f(v) = sec v tan v with the initial condition F(0) = 2. Answer: F(v) = sec v + 1

Step 1: Integrate the function

To find the antiderivative, we'll integrate the given function with respect to \(v\): $$\int \sec v \tan v \, dv$$ This is a straightforward integral since the derivative of \(\sec v\) is \(\sec v \tan v\). So the integral becomes: $$F(v) = \int \sec v \tan v \, dv = \sec v + C$$

Step 2: Use the initial condition to find the constant of integration

We know that \(F(0) = 2\). We can plug in this value and solve for the constant \(C\): $$2 = F(0) = \sec(0) + C$$ Since \(\sec(0) = 1\), we get: $$2 = 1 + C$$ $$C = 1$$

Step 3: Write the antiderivative with the constant of integration

Now that we have found the constant of integration \(C = 1\), we can write the antiderivative \(F(v)\) as: $$F(v) = \sec v + 1$$

Key Concepts

Integral Calculus

Integral calculus is a fundamental part of mathematics that deals with finding antiderivatives, also known as integrals. Essentially, it’s about the reverse process of differentiation. Where differentiation gives us the rate of change of a function, integration provides the accumulation of quantities.

When you come across a problem involving the integration of a function, as in \(f(v) = \sec v \tan v\), you're tasked with finding a new function, let's call it \(F(v)\), whose derivative is equal to the given function \(f(v)\). This process is called finding the antiderivative or indefinite integral.

As you work through integration problems, you'll notice patterns, like how the integral of \(\sec v \tan v\) is \(\sec v\), which directly relates to the fact that the derivative of \(\sec v\) yields \(\sec v \tan v\). Recognizing these patterns is key in solving these problems efficiently.

Initial Condition

The initial condition is the piece of information that allows us to find a specific solution from a family of antiderivatives. When you find an antiderivative, you're not getting one fixed function, but rather a set of functions that differ by a constant. This is because the derivative of a constant is zero, and thus, it doesn't appear in the original function.

Using the Initial Condition

When you have an initial condition, such as \(F(0) = 2\), it provides a specific point the function must pass through. This means that we have an additional piece of the puzzle to find the unique function that not only has the correct derivative but also fits the given point. Plugging this initial condition into our general antiderivative allows us to solve for the constant of integration, yielding one unique solution in the context of the problem.

Constant of Integration

The constant of integration, typically denoted as \(C\), represents an unknown constant that we add to the antiderivative to account for all possible antiderivatives of a given function. When integrating a function, unlike differentiation, we must consider that any function \(F(v)\) could have been shifted vertically by a constant amount.

Role of the Constant of Integration

In the step by step solution given, after integrating \(\sec v \tan v\), we get \(\sec v + C\). Here \(C\) represents all the possible vertical translations of \(\sec v\). To find the specific value of \(C\), we looked at the initial condition \(F(0) = 2\) and used it to determine that \(C = 1\). Finally, with that value, we were able to write the unique antiderivative that satisfies both the integral and the initial condition: \(F(v) = \sec v + 1\). Understanding and correctly applying the constant of integration is crucial for accurately solving problems in integral calculus.