Problem 26 - Essential Calculus Early Transcendentals

Question

Find \(f\) $$f^{\prime}(t)=2 \cos t+\sec ^{2} t, \quad-\pi / 2

Step-by-Step Solution

Answer

The function is \(f(t) = 2 \sin t + \tan t + 4 - 2\sqrt{3}\).

Step 1: Integrate the Derivative

To find the function \(f(t)\), let's first integrate its derivative. We have \(f'(t) = 2 \cos t + \sec^2 t\). Integrating each term separately:\[\int 2 \cos t \, dt = 2 \sin t + C_1\]\[\int \sec^2 t \, dt = \tan t + C_2\]So, combining these results:\[ f(t) = 2 \sin t + \tan t + C \] where \(C = C_1 + C_2\) is the constant of integration.

Step 2: Use the Initial Condition to Find the Constant

We are given the condition \(f(\pi / 3) = 4\). Substitute \(t = \pi / 3\) into the integrated function:\[f(\pi / 3) = 2 \sin(\pi / 3) + \tan(\pi / 3) + C = 4\]Since \(\sin(\pi / 3) = \frac{\sqrt{3}}{2}\) and \(\tan(\pi / 3) = \sqrt{3}\), substitute these values:\[2 \times \frac{\sqrt{3}}{2} + \sqrt{3} + C = 4\]This simplifies to:\[\sqrt{3} + \sqrt{3} + C = 4\]\[2\sqrt{3} + C = 4\]\[C = 4 - 2\sqrt{3}\]

Step 3: Write the Final Function

Now, substitute the value of \(C\) back into the function \(f(t)\):\[f(t) = 2 \sin t + \tan t + (4 - 2\sqrt{3})\]This is the final function for \(f(t)\).

Key Concepts

Integration Techniques

Integrating functions is an essential part of calculus. In this problem, we started with the derivative of a function and needed to find the original function. This process is referred to as integration. When we integrate, we are essentially reversing the process of differentiation.

For the given exercise, the derivative provided was a combination of terms: \(f'(t) = 2 \cos t + \sec^2 t\). To integrate, we handled each term separately:

• The integral of \(2 \cos t\) is \(2 \sin t + C_1\), because differentiation of \(\sin t\) yields \(\cos t\).
• The integral of \(\sec^2 t\) is \(\tan t + C_2\), as differentiating \(\tan t\) results in \(\sec^2 t\).

After integrating, these terms were summed up to get the general form of the function \(f(t) = 2 \sin t + \tan t + C\), where \(C\) is the constant of integration determined by the initial value problem.

Initial Value Problems

An initial value problem is a type of differential equation along with a specific condition at a particular point. This condition helps find unique solutions to problems that otherwise would have infinitely many solutions due to the constant of integration.

In this exercise, we are given \(f(\pi/3) = 4\). This means that when \(t = \pi/3\), \(f(t)\) should equal 4.

To find the constant \(C\), substitute \(\pi/3\) into the function: \(f(\pi/3) = 2 \sin(\pi/3) + \tan(\pi/3) + C = 4\). Knowing that \(\sin(\pi/3) = \frac{\sqrt{3}}{2}\) and \(\tan(\pi/3) = \sqrt{3}\), we found \(C\) by simplifying the equation. Thus, the specific value was found to be \(C = 4 - 2\sqrt{3}\).

This methodology ensures we have a well-defined function that satisfies the given conditions.

Trigonometric Functions

Trigonometric functions often appear in integral calculus as they are fundamental in various applications. In this exercise, the integrals involved are \(\cos t\) and \(\sec^2 t\). Understanding how to integrate these standard trigonometric functions is crucial:

• For \(\cos t\), the integral \(\int \cos t \, dt\) gives \(\sin t + C\).
• For \(\sec^2 t\), the integral \(\int \sec^2 t \, dt\) yields \(\tan t + C\).

The knowledge of these trigonometric function properties simplifies many calculus problems. They are beneficial in representing periodic phenomena, solving differential equations, and in this case, flipping back from a derivative to its antiderivative. By practicing integrals of trigonometric functions, you develop a better intuition for solving similar calculus problems.