Problem 42 - Calculus Single Variable

Question

In each of Exercises \(41-48\), use the given information to find \(F(c)\). $$ F^{\prime}(x)=4 \sin (x), \quad F(\pi / 3)=3, \quad c=\pi $$

Step-by-Step Solution

Answer

\( F(\pi) = 9 \)

Step 1: Understand the Problem

The problem requires us to find the value of the function \( F(c) \) given its derivative \( F'(x) = 4 \sin(x) \), a specific value \( F(\pi / 3) = 3 \), and that \( c = \pi \).

Step 2: Find the General Solution

Since \( F'(x) = 4 \sin(x) \), we can find \( F(x) \) by integrating the derivative. \[ F(x) = \int 4 \sin(x) \; dx = -4 \cos(x) + C \] where \( C \) is the constant of integration.

Step 3: Use Initial Condition

Utilize the given condition \( F(\pi / 3) = 3 \) to find \( C \).Substitute \( x = \pi/3 \) into the general solution:\[ 3 = -4 \cos(\pi/3) + C \]Knowing \( \cos(\pi/3) = 1/2 \), the equation becomes:\[ 3 = -4 \times \frac{1}{2} + C \]\[ 3 = -2 + C \]\[ C = 5 \]

Step 4: Substitute Back to Find F(c)

Now substitute \( C = 5 \) into the general solution:\[ F(x) = -4 \cos(x) + 5 \]Finally, substitute \( c = \pi \) to find \( F(c) \):\[ F(\pi) = -4 \cos(\pi) + 5 \]Given that \( \cos(\pi) = -1 \), this yields:\[ F(\pi) = -4(-1) + 5 \]\[ F(\pi) = 4 + 5 \]\[ F(\pi) = 9 \]

Step 5: Conclusion

Therefore, the value of \( F(c) \) when \( c = \pi \) is 9.

Key Concepts

Derivative

In calculus, a derivative represents how a function changes as its input changes. It's like the speed of a car, showing how fast a position changes. For a given function \( f(x) \), the derivative \( f'(x) \) indicates the rate of change with respect to \( x \).

• The derivative is a fundamental tool in calculus used to analyze the behavior of functions.
• It helps in finding slopes of curves, which is crucial in understanding geometrical features of graphs.
• In the problem above, we have \( F'(x) = 4 \sin(x) \) which reveals that the rate of change of \( F(x) \) is dependent on \( \sin(x) \).

Understanding derivatives is key to moving forward into finding integrals, which are essentially the reverse operations.

Integration

Integration is the reverse process of differentiation. If the derivative of a function represents a rate of change, then integration accumulates those changes to find the original function.

• In mathematical terms, if \( F'(x) \) is the derivative of \( F(x) \), then integrating \( F'(x) \) will give \( F(x) \), up to a constant.
• In our exercise, we integrated \( 4 \sin(x) \) to get \( F(x) = -4 \cos(x) + C \).

By understanding integration, you're able to find the function that a given derivative comes from, unraveling the history of the function’s behavior.

Trigonometric Functions

Trigonometric functions like \( \sin(x) \) and \( \cos(x) \) describe relationships in right-angled triangles and are used to model periodic phenomena.

• These functions cycle through values, hence are often used in calculus problems involving periodic elements like waves.
• In this problem, we used \( \cos(x) \) in the integration process. It's a common technique since sine and cosine have known derivatives and integrals.
• Familiarity with basic trigonometric identities, such as \( \cos(\pi/3) = 1/2 \) and \( \cos(\pi) = -1 \), was pivotal in solving for \( C \) and subsequently finding \( F(\pi) \).

Knowing these trigonometric values offhand can greatly simplify calculus problems, as it did here.

Constant of Integration

The constant of integration, represented as \( C \), arises when integrating a function. This constant accounts for all the possible vertical shifts of the antiderivative function.

• The integration process naturally introduces this constant since differentiating a constant gives zero.
• Without specific initial conditions, the constant cannot be uniquely determined, leading to a family of solutions.
• For our problem, the condition \( F(\pi/3) = 3 \) allowed us to solve for \( C \), resulting in \( C = 5 \).

Understanding the role of this constant is crucial, as it provides the means to tailor the general antiderivative to specific scenarios, as demonstrated by incorporating the initial condition provided.