Problem 23 - Calculus Single Variable

Question

Compute \(F(c)\) from the given information. $$ F^{\prime}(x)=\cos (x), F(\pi / 2)=-1, c=\pi / 6 $$

Step-by-Step Solution

Answer

\( F(c) = -\frac{3}{2} \) at \( c = \pi/6 \).

Step 1: Understand the Problem

We need to calculate the value of the function \( F(c) \) where \( c = \pi/6 \), given that the derivative \( F'(x) = \, \cos(x) \) and \( F(\pi/2) = -1 \).

Step 2: Antiderivative of the Derivative

Since \( F'(x) = \cos(x) \), the antiderivative or integral of \( \cos(x) \) is \( F(x) = \sin(x) + C \), where \( C \) is an arbitrary constant of integration.

Step 3: Use the Initial Condition

We know that \( F(\pi/2) = -1 \). Substitute \( \pi/2 \) into the antiderivative: \( F(\pi/2) = \sin(\pi/2) + C = -1 \). Since \( \sin(\pi/2) = 1 \), it follows that:\[ 1 + C = -1 \]Solve for \( C \) to get \( C = -2 \).

Step 4: Write the Specific Antiderivative

Now that we know \( C = -2 \), the specific function is \( F(x) = \sin(x) - 2 \).

Step 5: Evaluate \( F \) at \( c = \pi/6 \)

Substitute \( c = \pi/6 \) into \( F(x) = \sin(x) - 2 \) to find \( F(\pi/6) \):\[ F(\pi/6) = \sin(\pi/6) - 2 \]We know \( \sin(\pi/6) = 1/2 \), so:\[ F(\pi/6) = \frac{1}{2} - 2 = -\frac{3}{2} \]

Key Concepts

Antiderivative

Understanding the concept of an antiderivative is crucial in calculus. When you have a function like \(F'(x)\), which represents the derivative of \(F(x)\), finding the antiderivative involves determining a function whose derivative would return you to \(F'(x)\). In the exercise, \(F'(x) = \, \cos(x)\), and the goal was to find \(F(x)\).
The antiderivative of \(\cos(x)\) is \(\sin(x)\). Therefore, \(F(x)\) initially becomes \(\sin(x) + C\), where \(C\) is a constant. This constant arises because when you derive \(F(x)\), any constant disappears. Thus, it's important to include this " + C" when integrating, as it's a placeholder for all potential vertical shifts of the graph of \(F(x)\). This provides the general form of the antiderivative.
Overall, taking the antiderivative essentially reverses differentiation. It's a crucial tool in finding the original function from its rate of change.

Initial Condition

The initial condition helps determine the exact form of a function when integrating. While taking an antiderivative gives us a family of functions with a constant \(C\), the initial condition specifies which member of this family we have. Think of it like solving a puzzle with a missing piece.In this exercise, the problem provides the condition \(F(\pi/2) = -1\). By substituting \(\pi/2\) into the generalized antiderivative \(F(x) = \sin(x) + C\) and setting it to \(-1\), we solve:

• \(\sin(\pi/2) + C = -1\)
• Since \(\sin(\pi/2) = 1\), the equation becomes \(1 + C = -1\)

This calculation reveals that \(C = -2\), specifying the exact antiderivative as \(F(x) = \sin(x) - 2\). Utilizing initial conditions is essential for pinpointing the correct function in real-world problems.

Trigonometric Functions

Trigonometric functions like \(\sin(x)\) and \(\cos(x)\) play a significant role in calculus due to their periodic nature and fundamental connections to angles and cycles. Each trigonometric function has a unique derivative and antiderivative.
In this task, you encountered \(\cos(x)\), which is the derivative of \(\sin(x)\). Recognizing these relationships is key:

• Derivative of \(\sin(x)\) is \(\cos(x)\)
• Antiderivative of \(\cos(x)\) is \(\sin(x) + C\)

Understanding these basic trigonometric identities and their derivatives and antiderivatives allows you to effortlessly transition between functions and their rates of change, which is vital in solving problems like these.

Definite Integration

Although not explicitly required in this problem, definite integration is a related concept often accompanying antiderivatives. While finding an antiderivative gives a general solution \(F(x)\), definite integration deals with the specific area under a curve between two points. This connects closely with the idea of calculating an exact difference or integral over a specified interval.For definite integration, if you know \(F(x)\) is the antiderivative of \(f(x)\), then the definite integral from \(a\) to \(b\) is calculated as:

• \(\int_a^b f(x) \, dx = F(b) - F(a)\)

Understanding definite integration is essential for applications requiring exact values over intervals, such as finding total distance traveled or total accumulated change, making it a powerful tool within calculus.