Problem 19 - Calculus

Question

Where does \(y=\sin x+\cos x\) have zero slope?

Step-by-Step Solution

Answer

The slope is zero at \(x = \frac{\pi}{4} + k\pi\) where \(k\in\mathbb{Z}\).

Step 1: Find the derivative

To find where the slope of the function is zero, we first need to compute the derivative of the function. The function given is \(y = \sin x + \cos x\). The derivative, \(y'\), can be found using the derivatives of sine and cosine functions: \(\frac{d}{dx}(\sin x) = \cos x\) and \(\frac{d}{dx}(\cos x) = -\sin x\). Therefore, the derivative is \(y' = \cos x - \sin x\).

Step 2: Set the derivative to zero

To find where the slope is zero, we set the derivative equal to zero: \(\cos x - \sin x = 0\). This equation implies \(\cos x = \sin x\).

Step 3: Solve for x

When \(\cos x = \sin x\), we can divide both sides by \(\cos x\) (as long as \(\cos x eq 0\)) to obtain \(1 = \tan x\). This implies \(\tan x = 1\). The general solution for \(\tan x = 1\) is \(x = \frac{\pi}{4} + k\pi\), where \(k\) is any integer. These are the points where the slope of the function is zero.

Step 4: Confirm the solutions

Values of \(x\) that satisfy \(x = \frac{\pi}{4} + k\pi\) correspond to points where the slope of \(y = \sin x + \cos x\) is zero. Verifying by substituting back into the derivative, \(y' = \cos x - \sin x\), confirms that these values lead to a derivative of zero.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, focusing on how a function changes as its input changes. Imagine you have a curve representing a function on a graph. Differentiation helps us find the slope of the tangent line at any point on this curve. This slope tells us the rate at which the function is changing at that particular point. For trigonometric functions like sine and cosine, the process of differentiation follows specific rules:

• The derivative of the sine function, \( \frac{d}{dx}(\sin x) \), is \( \cos x \).
• The derivative of the cosine function, \( \frac{d}{dx}(\cos x) \), is \( -\sin x \).

By applying these rules, we can differentiate complex functions, such as \( y = \sin x + \cos x \), by treating each component separately and then combining the results. The goal in differentiation is often to find points on a function where certain conditions are met, such as where the slope is zero, indicating a potential maximum, minimum, or saddle point.

Trigonometric Functions

Trigonometric functions like \( \sin x \) and \( \cos x \) are fundamental in many areas of mathematics and occur frequently in calculus. They describe relationships in right-angled triangles but also extend to more complex periodic phenomena. When we differentiate trigonometric functions:

• The derivative of \( \sin x \) is \( \cos x \), which means that the rate of change of the sine function is described by the cosine function.
• The derivative of \( \cos x \) is \( -\sin x \), indicating that the change in the cosine function is captured by the negative sine function.

These derivatives are pivotal in solving calculus problems involving waves, oscillations, and circular motion. In our task to find where the function \( y = \sin x + \cos x \) has a zero slope, we deal with both \( \sin x \) and \( \cos x \). Their interplay is crucial, as setting their combined derivative to zero allows us to find those points of interest where the function does not rise or fall.

Critical Points

Critical points are points on a graph where the derivative (or slope) of a function is zero or undefined. These points are significant because they often indicate local maxima, minima, or points of inflection. For the function \( y = \sin x + \cos x \), critical points occur when the derivative, \( y' = \cos x - \sin x \), equals zero.Setting \( \cos x - \sin x = 0 \), we find that \( \cos x = \sin x \). By solving the equation \( 1 = \tan x \), it becomes clear that for these trigonometric functions, the tangent having a value of 1 points to specific angles on the unit circle. The general solution \( x = \frac{\pi}{4} + k\pi \) (where \( k \) is an integer) reveals where the function levels out, achieving a critical point.Understanding critical points helps us describe the behavior of functions and predict how they change, a crucial element in fields ranging from engineering to physics. By identifying these points in the context of the exercise, students gain insight into how calculus helps map out the nuanced behavior of functions.