Problem 65 - Essential Calculus

Question

In Exercises 65 and 66, apply the Extended Mean Value Theorem to the functions \(f\) and \(g\) on the given interval. Find all values \(c\) in the interval \((a, b)\) such that \(\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\) \(\begin{array}{l} \underline{\text { Functions }} \\ f(x)=\sin x, \quad g(x)=\cos x \end{array} \quad \frac{\text { Interval }}{\left[0, \frac{\pi}{2}\right]}\)

Step-by-Step Solution

Answer

The value of \(c\) that satisfies the given conditions of the Extended Mean Value Theorem for the given functions and interval is \(c = \frac{\pi}{4}\).

Step 1: Calculate the derivatives

First, the derivatives of the functions \(f(x) = \sin x\) and \(g(x) = \cos x\) are needed. The derivative of \(f(x) = \sin x\) is \(f^{\prime}(x) = \cos x\) and the derivative of \(g(x) = \cos x\) is \(g^{\prime}(x) = -\sin x\).

Step 2: Calculate \(f(b) - f(a)\) and \(g(b) - g(a)\)

In the next step, calculate the differences in the values of \(f\) and \(g\) at \(b\) and \(a\) respectively. Here, \(f(b) - f(a) = f\(\frac{\pi}{2}\) - f(0)= 1 - 0 = 1\), and \(g(b) - g(a) = g\(\frac{\pi}{2}\) - g(0) = 0 - 1 = -1\).

Step 3: Apply the Extended Mean Value Theorem

According to the EMVT, there exists a \(c\) in the interval \((0, \frac{\pi}{2})\) where \(\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\). Substituting the given values, we get: \(\frac{\cos c}{-\sin c} = \frac{1}{-1}\). Solving this, we get: \(\cos c = \sin c\).

Step 4: Find \(c\) satisfying the equation.

The equation \(\sin c = \cos c\) is satisfied when \(c = \frac{\pi}{4}\). However, it's important to note other solutions for this equation exist, but they fall outside the interval \((0, \frac{\pi}{2})\). Therefore, \(c = \frac{\pi}{4}\) is the only solution within this interval.

Key Concepts

Derivatives Explained

In mathematics, derivatives represent the rate at which a function is changing at any given point. When you have a function, its derivative will tell you how steep its curve is at any particular moment. This is particularly useful when examining behavior in physics or economics, for instance. For the trigonometric functions in our exercise, the derivative of \( f(x) = \sin x \) is \( f'(x) = \cos x \). This tells us that the rate of change of the sine function at any point is given by the cosine of that point.

• The derivative of sine is directly found as the cosine function.
• Similarly, for \( g(x) = \cos x \), the derivative \( g'(x) = -\sin x \) indicates the rate at which the cosine decreases.

These derivatives are essential when applying the Extended Mean Value Theorem, as they help establish the relationship defining the value \( c \) within the interval.

Understanding Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in mathematics. They describe the relationship between angles and sides of triangles, but are also crucial in modeling wave patterns, such as sound waves.

• \( f(x) = \sin x \) gives the opposite side over the hypotenuse in a right-angle triangle.
• \( g(x) = \cos x \) gives the adjacent side over the hypotenuse.

During the solution, understanding what \( \sin \) and \( \cos \) represent helps visualize the changes these curves undergo over specific intervals.
The sine function begins at 0 at \( x = 0 \), rises to 1 at \( x = \frac{\pi}{2} \), which signifies where the angle reaches its peak. Meanwhile, the cosine function starts at 1 at \( x = 0 \) and falls to 0 at \( x = \frac{\pi}{2} \). Knowing this, we can better appreciate why \( \cos c = \sin c \) leads to \( c = \frac{\pi}{4} \). Both functions are equal at this angle because they have traveled symmetrically along their curves by equal amounts.

The Importance of Interval Calculations

Intervals define the range within which we are concerned about the behavior of a function. It's crucial to restrict solutions to predefined intervals to make rational conclusions in applications such as the Extended Mean Value Theorem (EMVT).

• In our problem, the interval is \((0, \frac{\pi}{2})\).
• This means any solution for \( c \) must lie within these bounds.

The calculation of \( c \) involves ensuring that it satisfies the given functional equation derived from EMVT in this specific range. The importance can't be overstated, as it ensures that conclusions apply logically to the scenario outlined by the interval.
In practical applications, ignoring the interval could lead to incorrect assumptions or interpretations, especially in fields that heavily rely upon precise angular or periodic behavior analysis. Identifying \( c = \frac{\pi}{4} \) is a direct result of checking which solutions fit the constraints provided by the interval, ensuring our results' relevance and accuracy.