Problem 10 - Calculus

Question

the function continuous on the interval? $$\frac{e^{\sin \theta}}{\cos \theta} \text { on }\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$$

Step-by-Step Solution

Answer

The function \( \frac{e^{\sin \theta}}{\cos \theta} \) is continuous on \([-\frac{\pi}{4}, \frac{\pi}{4}]\).

Step 1: Definition of Continuity

A function is continuous on an interval if it is continuous at every point within the interval. A function is continuous at a point if the limit of the function as it approaches that point from either side is equal to the function's value at that point.

Step 2: Analyze the Function

Consider the function \( f(\theta) = \frac{e^{\sin \theta}}{\cos \theta} \). The key points to check for discontinuity in rational functions arise where the denominator equals zero, as these cause vertical asymptotes.

Step 3: Identify Potential Discontinuity Points

The function will be undefined and thus discontinuous at points where \( \cos \theta = 0 \). This occurs at \( \theta = \frac{\pi}{2} + k\pi \) for integer \(k\). Check within \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \).

Step 4: Check Critical Points in the Given Interval

Within \([-\frac{\pi}{4}, \frac{\pi}{4}]\), \( \cos \theta \) does not equal zero, as \( \cos(-\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \) and \( \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \). Since \( \cos \theta eq 0 \) throughout the interval, there are no discontinuities.

Step 5: Conclusion on Continuity

Since there are no points within the interval \([-\frac{\pi}{4}, \frac{\pi}{4}]\) where \( \cos \theta = 0 \), the function remains defined and continuous throughout the entire interval.

Key Concepts

Rational Functions

At the heart of many continuity problems in calculus lies the concept of rational functions. A rational function is simply the ratio of two polynomial functions. In the context of continuity, we usually express a rational function in the form \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \).Rational functions are particularly interesting because the main source of discontinuities comes from their denominators. When the denominator becomes zero, the function is not defined. As a result, the function is discontinuous at these points. For instance, the function \( f(\theta) = \frac{e^{\sin \theta}}{\cos \theta} \) becomes undefined wherever \( \cos \theta = 0 \).In analyzing rational functions, our primary job is to identify where those dreaded zeros of the denominator occur, as those are the spots where discontinuities might happen. Once identified, we can better understand and delineate the interval in which the function is continuous.

Discontinuity

Discontinuity is a critical concept in calculus, defining where a function fails to be continuous. For rational functions, discontinuity happens at points where the denominator is zero since it leads to an undefined condition. Take the function from the exercise, \( f(\theta) = \frac{e^{\sin \theta}}{\cos \theta} \). Here, we face potential discontinuities wherever \( \cos \theta = 0 \).Understanding discontinuity involves:

• Identifying the zero points of the denominator.
• Determining if these points lie within the interval of interest.

In the interval \([-\frac{\pi}{4}, \frac{\pi}{4}]\), \( \cos \theta \) does not hit zero because all angles within this range have a positive cosine value. The function is thus continuous in this interval. This continuity is confirmed not just by avoiding zero denominators, but also by observing that no piece of the original function leads to any other kind of undefined behavior in this domain.

Interval Analysis

Interval analysis is an essential tool used to study the behavior of functions over specific domains. In the exercise given, we are tasked to determine if the function \( f(\theta) = \frac{e^{\sin \theta}}{\cos \theta} \) remains continuous over the interval \([-\frac{\pi}{4}, \frac{\pi}{4}]\).The process usually involves these steps:

• Evaluate the endpoints of the interval to check for any abrupt changes.
• Analyze the function within the interval for values or conditions that indicate discontinuity.
• Check the continuity of any sub-functions, such as \( \sin \theta \) and \( \cos \theta \), relevant to the main function within the interval.

In our specific range, we observed that \( \cos \theta \) never hits zero, which means our function stays defined and smooth across the whole interval. Interval analysis allows us to conclude with confidence that \( f(\theta) \) is continuous from \(-\frac{\pi}{4}\) to \(\frac{\pi}{4}\), illustrating how interval analysis combines with other calculus concepts to provide a thorough understanding of function behavior.