Question

For the following functions \(f\), find the anti-derivative \(F\) that satisfies the given condition. $$f(v)=\sec v \tan v ; F(0)=2,-\pi / 2 < v < \pi / 2$$

Short answer

Question: Find the anti-derivative \(F(v)\) of the function \(f(v) = \sec v \tan v\), given that \(F(0) = 2\) and \(-\pi/2 < v < \pi/2\). Solution: The anti-derivative \(F(v)\) of the function \(f(v) = \sec v \tan v\) is given by \(F(v) = \sec v + 1\).

Step-by-step solution

Integrate the function

To find the anti-derivative \(F(v)\), we need to integrate the function \(f(v)\). Let's compute the integral of \(f(v) = \sec v \tan v\) with respect to \(v\): $$ F(v) = \int \sec v \tan v dv $$ Here, we have a classic integration problem of the product of secant and tangent functions.

Recognize the common integral

Recall that the derivative of \(\sec v\) is \(\sec v \tan v\). Thus, the integral of \(\sec v \tan v\) gives us back the secant function (up to a constant): $$ \int \sec v \tan v dv = \sec v + C $$

Apply the given condition

We're given the condition that \(F(0) = 2\). Now we need to find the constant \(C\): $$ 2 = \sec(0) + C $$ Since \(\sec(0) = 1\), we have: $$ 2 = 1 + C $$ Solving for \(C\), we get \(C = 1\).

Write the final anti-derivative

Now we have everything we need to write the final anti-derivative: $$ F(v) = \sec v + 1 $$ This is the anti-derivative of \(f(v) = \sec v \tan v\) that satisfies the given condition \(F(0) = 2\) and is defined within the range \(-\pi/2 < v < \pi/2\).

Key concepts

Integration

Integration is the process of finding the anti-derivative, also known as an integral, of a function. In essence, it is the reverse of differentiation. While differentiation deals with finding the rate of change, integration is about finding the accumulated quantity. In this exercise, we are tasked with finding the anti-derivative of the trigonometric function \(f(v) = \sec v \tan v\).

The integral here is set up as \(F(v) = \int \sec v \tan v \, dv\), which means we are trying to reverse the process of differentiation that originally gave us \(f(v)\). The antiderivative of a function represents an infinite number of functions since integrating introduces an unknown constant \(C\). This constant is determined using initial or boundary conditions provided in the exercise.

Remember, recognizing integrals and understanding their relationship with derivative functions (like knowing that the derivative of \( \sec v \) is \( \sec v \tan v \)) helps solve these problems effectively.

Trigonometric functions

Trigonometric functions like sine, cosine, and tangent play a crucial role in calculus, often showing up in differentiation and integration problems. In this problem, we are dealing with \(\sec v\) and \(\tan v\). These functions are reciprocals of cosine and sine, respectively, and are used frequently in calculus for their unique properties.

Knowing the derivatives and integrals of these trigonometric functions is key. For instance, we utilized the fact that the derivative of \(\sec v\) is \(\sec v \tan v\) to find the antiderivative in the exercise. This form is well-known and commonly encountered in calculus, which makes it a recognizable pattern.

Understanding these relationships helps in not only solving problems but also in understanding the behaviors of waveforms and angles, which is crucial in physics and engineering applications.

Initial condition

The initial condition in calculus problems is a specific requirement that helps determine the constant of integration \(C\). When we integrate a function to find its anti-derivative, the result is not a single unique function. Instead, it represents a family of functions due to the presence of an arbitrary constant \(C\). The initial condition is a given point that the solution must pass through, allowing us to solve for \(C\).

In our exercise, we were given \(F(0)=2\) as the initial condition. It means that when \(v=0\), the value of the anti-derivative is 2. By substituting the value of \(v\) and \(F(v)\) into the anti-derivative expression, we were able to solve for \(C\). Specifically, knowing \(\sec(0) = 1\), we adjusted the equation to find \(C = 1\).

Initial conditions are crucial because they tailor the general solution, shaped by the integral, to fit a particular problem.