Problem 57 - Calculus Single Variable

Question

For what a, \(0 \leq a \leq \pi,\) does the function \(f(x)=\sin (x)-\) \(\cos (x)\) have the greatest average over the interval \([a, a+\pi]\) of length \(\pi ?\)

Step-by-Step Solution

Answer

The value of \(a\) that maximizes the average is \(\frac{\pi}{4}\).

Step 1: State the Problem

We need to find the value of \(a\) within the interval \([0, \pi]\) for which the average value of the function \(f(x) = \sin(x) - \cos(x)\) over the interval \([a, a + \pi]\) is maximized. The function is composed of sine and cosine functions and we seek its average over intervals of length \(\pi\).

Step 2: Establish the Average Value Formula

The average value of a function \(f(x)\) over an interval \([a, b]\) is given by the formula \[ \text{Average} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx. \] For our function and interval, this becomes \[ \frac{1}{\pi} \int_{a}^{a+\pi} (\sin(x) - \cos(x)) \, dx. \]

Step 3: Compute the Integral

Calculate the integral \( \int_{a}^{a+\pi} (\sin(x) - \cos(x)) \, dx \). Compute separately: - \(\int \sin(x) \, dx = -\cos(x) + C\)- \(\int \cos(x) \, dx = \sin(x) + C\)Thus, the definite integral from \(a\) to \(a+\pi\) is \[-\cos(a+\pi) + \cos(a) - (\sin(a+\pi) - \sin(a)).\]

Step 4: Simplify Using Trigonometric Identities

Use trigonometric identities: \(\cos(a+\pi) = -\cos(a)\) and \(\sin(a+\pi) = -\sin(a)\). Substituting back, the definite integral becomes:\[- (-\cos(a)) + \cos(a) - (-\sin(a) + \sin(a)) = 2\cos(a) + 2\sin(a).\]

Step 5: Find the Average Value

Now, substitute back into the average formula:\[ \frac{1}{\pi}(2\cos(a) + 2\sin(a)) = \frac{2}{\pi} (\cos(a) + \sin(a)).\]This expression represents the average value of the function \(f(x)\) over the interval \([a, a+\pi]\).

Step 6: Maximize the Average Value

To maximize the average value, we maximize the expression \(\cos(a) + \sin(a)\). Recall the identity \(R \sin(a + \phi) = \cos(a) + \sin(a)\) where \(R = \sqrt{2}\).The maximum value of a sine function is 1, so \(\sqrt{2}\sin(a + \phi)\) achieves its maximum when \(\sin(a + \phi) = 1\). Thus, the maximum value is \(\sqrt{2}\), occurring when \(a + \phi = \frac{\pi}{2}\). Set \(R = \sqrt{2}\):\[ a + \frac{\pi}{4} = \frac{\pi}{2} \Rightarrow a = \frac{\pi}{4}. \]

Step 7: Conclusion

Therefore, the value of \(a\) that maximizes the average value of the function over the interval \([a, a+\pi]\) is \(a = \frac{\pi}{4}\).

Key Concepts

Trigonometric Identities

Trigonometric identities are vital tools in simplifying expressions involving sine and cosine functions. In our context, these identities make it easier to evaluate and manipulate the integral. Two crucial trigonometric identities used here include:

• \( \cos(a + \pi) = -\cos(a) \)
• \( \sin(a + \pi) = -\sin(a) \)

These identities can be derived from the symmetry and periodicity of the trigonometric functions. For example, the cosine of any angle increased by \( \pi \) (half a rotation) results in the opposite direction on the unit circle. This understanding gives us:

• If \( a = 0 \), then \( \cos(a + \pi) = -1 = -\cos(0) \).
• For \( a + \pi \), we simply reflect across the origin to get \( -\sin(a) \).

Using these identities helps transform the integral from a potentially intimidating problem to a manageable expression suitable for further calculations.

Definite Integrals

A definite integral allows you to find the aggregate value of a function over a specific interval. In simpler terms, it's like finding the total area under a curve between two points on the x-axis. For our function \( f(x) = \sin(x) - \cos(x) \), we calculate the definite integral over an interval \([a, a + \pi]\), which is of length \( \pi \).
The calculation proceeds by determining separate integrals for \( \sin(x) \) and \( \cos(x) \):

• \( \int \sin(x) \, dx = -\cos(x) + C \)
• \( \int \cos(x) \, dx = \sin(x) + C \)

Substituting these into the integral gives us the net change between the boundaries. It ultimately simplifies to an expression involving the trigonometric identities, which is ultimately solvable by plugging in known values and performing basic algebraic manipulation.

Average Value of a Function

The average value of a function over some interval tells you the "average" height of the curve between those two points. The formula to determine this average value across an interval \([a, b]\) is:
\[ \text{Average} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]In our specific problem, this formula becomes:
\[ \frac{1}{\pi} \int_{a}^{a+\pi} (\sin(x) - \cos(x)) \, dx \]After calculating, this average value becomes \( \frac{2}{\pi} (\cos(a) + \sin(a)) \). This expression helps us understand how the function behaves across the interval.
Knowing how to calculate the average value aids in various applications such as physics for charge distributions, or economics for analyzing consumer behaviors, making it a powerful analytical tool.

Maximizing Functions

Maximization in mathematics involves finding the largest value that a function can take. To maximize any expression like \( \cos(a) + \sin(a) \), we leverage our knowledge of trigonometric functions. Recall the identity \( R \sin(a + \phi) = \cos(a) + \sin(a) \).

• Here \( R = \sqrt{2} \), which comes from the amplitude of these combined sine and cosine terms.
• \( \sin(a + \phi) \) can be maximized when \( \sin(a + \phi) = 1 \).

This observation implies that the expression \( \cos(a) + \sin(a) \) achieves its maximum when \( a + \phi = \frac{\pi}{2} \). Solving \( a + \frac{\pi}{4} = \frac{\pi}{2} \) gives us \( a = \frac{\pi}{4} \), an assignment that maximizes our function's average value over the specified interval.
Maximizing functions is common in many domains such as economics, engineering, and operational research, since it determines optimal use of resources or highest achievable returns.