Problem 60 - Calculus Early Transcendentals

Question

For the following functions \(f\), find the antiderivative \(F\) that satisfies the given condition. $$f(t)=\sec ^{2} t ; F(\pi / 4)=1$$

Step-by-Step Solution

Answer

Question: Find the antiderivative \(F(t)\) of the function \(f(t) = \sec^2 t\) with the condition \(F(\pi / 4) = 1\). Answer: The antiderivative \(F(t)\) of the function \(f(t) = \sec^2 t\) that satisfies the given condition is \(F(t) = \tan t\).

Step 1: Find the general antiderivative of \(f(t)\)

To find the antiderivative of the given function \(f(t)=\sec^2 t\), we will use the integral: $$\int \sec^2 t \, dt$$ Recall that the antiderivative of \(\sec^2 t\) is the tangent function, i.e., \(\int \sec^2 t \, dt = \tan t + C\). Therefore, we have the following general antiderivative: $$F(t) = \tan t + C$$

Step 2: Determine the specific antiderivative satisfying the given condition

We know that \(F(t) = \tan t + C\), and we are given the condition \(F(\pi / 4)=1\). Plug in this information and solve for \(C\): $$1 = \tan(\pi / 4) + C$$ Since \(\tan(\pi / 4) = 1\), we have: $$1 = 1 + C$$ Therefore, \(C = 0\).

Step 3: Write the final solution

Now that we have found the constant \(C\), we can write the specific antiderivative which satisfies the condition: $$F(t) = \tan t + 0$$ So, the antiderivative \(F(t)\) is: $$F(t) = \tan t$$

Key Concepts

Antiderivative

The concept of an antiderivative involves finding a function whose derivative is the given function. In simpler terms, if you have a function and take its derivative, you end up with another function. Finding the antiderivative means reversing this process. This is closely related to integration.
To illustrate, consider our function \(f(t) = \sec^{2} t\). We are tasked with finding \(F(t)\) such that its derivative gives \(\sec^{2} t\). The key step here is recognizing that the derivative of \(\tan t\) is \(\sec^{2} t\). Hence, \(F(t) = \tan t + C\) is an antiderivative, where \(C\) is a constant.

Once you find a general antiderivative, you often need to find a particular solution using initial conditions, like in our problem \(F(\pi/4) = 1\). This step helps in determining the specific constant value in the solution.

Integration

Integration is a foundational concept in calculus that refers to the process of finding an integral, or the antiderivative, of a function. Integrating a function can be seen as accumulating quantities; imagine summing infinitely tiny quantities to find the whole.
When we integrate \(\sec^{2} t\) with respect to \(t\), the result is \(\tan t + C\). The \(C\) here represents an arbitrary constant reflecting that integration can result in more than one function differing only by a constant.

This constant arises because the derivative of any constant is zero, so many functions share the same derivative. When additional conditions like \(F(\pi/4) = 1\) are given, they allow us to solve for \(C\) to find a particular solution.

Trigonometric functions

Trigonometric functions are an essential part of calculus, especially in solving problems involving integrals and derivatives. Functions like sine, cosine, and tangent have distinct derivatives and integrals.
In the context of the given problem, \(f(t) = \sec^2 t\), it is crucial to recognize the relationship between \(\sec^2 t\) and \(\tan t\). This connection stems from the derivative of \(\tan t\), which is \(\sec^2 t\), making it straightforward to conclude the antiderivative of \(\sec^2 t\) is \(\tan t + C\).

Remembering these fundamental relationships between derivatives and integrals of trigonometric functions is key in solving such calculus problems efficiently.