Problem 31 - Essential Calculus Early Transcendentals

Question

Find \(f\) $$f^{\prime \prime}(\theta)=\sin \theta+\cos \theta, \quad f(0)=3, \quad f^{\prime}(0)=4$$

Step-by-Step Solution

Answer

The function is \(f(\theta) = -\sin \theta - \cos \theta + 5\theta + 4\).

Step 1: Integrate the Second Derivative Once

Start with the second derivative, \(f''(\theta) = \sin \theta + \cos \theta\). Integrate this function with respect to \(\theta\) to find the first derivative, \(f'(\theta)\).\[f'(\theta) = \int (\sin \theta + \cos \theta) \, d\theta = -\cos \theta + \sin \theta + C_1\]where \(C_1\) is the constant of integration.

Step 2: Use Initial Condition on First Derivative

Use the given initial condition \(f'(0) = 4\) to solve for the constant \(C_1\).Substitute \(\theta = 0\) into the found expression:\[-\cos(0) + \sin(0) + C_1 = 4 \Rightarrow -1 + 0 + C_1 = 4\]Solving for \(C_1\), we get:\[C_1 = 5\]

Step 3: Integrate the First Derivative

Now, integrate \(f'(\theta)\) to find \(f(\theta)\).\[f(\theta) = \int (-\cos \theta + \sin \theta + 5) \, d\theta = -\sin \theta - \cos \theta + 5\theta + C_2\]where \(C_2\) is another constant of integration.

Step 4: Use Initial Condition on Original Function

Use the given initial condition \(f(0) = 3\) to find \(C_2\).Substitute \(\theta = 0\) into the expression for \(f(\theta)\):\[-\sin(0) - \cos(0) + 5(0) + C_2 = 3 \Rightarrow 0 - 1 + 0 + C_2 = 3\]Solving for \(C_2\), we get:\[C_2 = 4\]

Step 5: Write the Final Function

By substituting back the constants \(C_1\) and \(C_2\) into \(f(\theta)\) and \(f'(\theta)\), respectively, we obtain the final solution:Function \(f(\theta)\):\[f(\theta) = -\sin \theta - \cos \theta + 5\theta + 4\]

Key Concepts

Integration

Integration is a fundamental concept in calculus which is used to determine the antiderivative or the original function from its derivative. This is the opposite process of differentiation. In the given problem, we start off with the second derivative of a function, denoted as \( f''(\theta) = \sin \theta + \cos \theta \). To find the original function \( f(\theta) \), integration must be performed twice. First, integrating \( f''(\theta) \) with respect to \( \theta \) gives the first derivative \( f'(\theta) \), which includes a constant of integration, often denoted as \( C_1 \):

• \( f'(\theta) = \int (\sin \theta + \cos \theta) \, d\theta = -\cos \theta + \sin \theta + C_1 \)

Integrating the first derivative yields the original function \( f(\theta) \), adding another constant \( C_2 \):

• \( f(\theta) = \int (-\cos \theta + \sin \theta + 5) \, d\theta = -\sin \theta - \cos \theta + 5\theta + C_2 \)

Every integration step results in an additional constant. These constants are determined through initial conditions. Thus, integration not only helps in finding antiderivatives but bind closely with initial conditions to define functions precisely.

Initial Conditions

Initial conditions are specific values given for functions and their derivatives at certain points, which are crucial in solving specific types of differential equations such as initial value problems. They play a key role in determining the unknown constants introduced by the integration process. In the problem at hand, we have the initial conditions \( f(0) = 3 \) and \( f'(0) = 4 \).

• For the first derivative \( f'(\theta) = -\cos \theta + \sin \theta + C_1 \), using \( f'(0) = 4 \), we determine \( C_1 \) by substituting \( \theta = 0 \):
  • \( -1 + 0 + C_1 = 4 \)
  Solving gives: \( C_1 = 5 \).
• For the original function \( f(\theta) = -\sin \theta - \cos \theta + 5\theta + C_2 \), using \( f(0) = 3 \), we find \( C_2 \) by substituting \( \theta = 0 \):
  • \( 0 - 1 + 0 + C_2 = 3 \)
  Solution: \( C_2 = 4 \).

These values define the exact form of the function and its derivatives. Initial conditions thus ensure that the solution satisfies the given problem, leading us from a general family of functions to a specific problem-solving one.

Trigonometric Functions

Trigonometric functions, including sine and cosine, are essential components in calculus, influencing how we differentiate and integrate expressions involving them. In the given exercise, both \( \sin \theta \) and \( \cos \theta \) appear in the second derivative of the function. Understanding the derivatives and antiderivatives of these functions is key:

• The derivative of \( \sin \theta \) is \( \cos \theta \), while its antiderivative is \( -\cos \theta \).
• The derivative of \( \cos \theta \) is \(-\sin \theta \), with an antiderivative of \( \sin \theta \).

When integrating \( f''(\theta) = \sin \theta + \cos \theta \), these properties allow us to find the first derivative directly:

• \( f'(\theta) = \int (\sin \theta + \cos \theta) \, d\theta = -\cos \theta + \sin \theta + C_1 \)

An understanding of these basic trigonometric identities and their interactions through calculus operations is vital. They appear frequently in physics and engineering to describe periodic phenomena, making them a core part of any calculus curriculum.