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Magazine Chapter 4: Problem 42
In each of Exercises \(41-48\), use the given information to find \(F(c)\). $$ F^{\prime}(x)=4 \sin (x), \quad F(\pi / 3)=3, \quad c=\pi $$
Short Answer
Expert verified
\( F(\pi) = 9 \)
Step by step solution
01
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 \( c = \pi \).
02
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.
03
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:\[ 3 = -4 \times \frac{1}{2} + C \]\[ 3 = -2 + C \]\[ C = 5 \]
04
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:\[ F(\pi) = -4(-1) + 5 \]\[ F(\pi) = 4 + 5 \]\[ F(\pi) = 9 \]
05
Conclusion
Therefore, the value of \( F(c) \) when \( c = \pi \) is 9.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( \sin(x) \).
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 \( F(x) = -4 \cos(x) + C \).
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 \( F(\pi) \).
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 \( C = 5 \).
