Problem 51 - Calculus: Early Transcendentals

Question

Describe the concavity of the functions below. $$ y=\cos ^{2} x-\sin ^{2} x $$

Step-by-Step Solution

Answer

The function is concave up on \((\frac{\pi}{4} + k\pi, \frac{3\pi}{4} + k\pi)\) and concave down on \(([0 + k\pi, \frac{\pi}{4} + k\pi) \cup (\frac{3\pi}{4} + k\pi, \pi + k\pi])\) for integer \(k\).

Step 1: Recall the formula for concavity

To determine the concavity of a function, you need the second derivative. A function is concave up if its second derivative is positive and concave down if its second derivative is negative.

Step 2: Find the first derivative

First, differentiate the function with respect to \(x\). The original function is \(y = \cos^2 x - \sin^2 x\). Use the chain rule: \(\frac{d}{dx}(\cos^2 x) = 2\cos x(-\sin x) = -2\cos x \sin x\) and \(\frac{d}{dx}(-\sin^2 x) = -2\sin x\cos x = -2\sin x \cos x\). So, \(y' = -4\sin x\cos x\).

Step 3: Simplify the first derivative using identities

Notice that \(-4\sin x\cos x\) can be written using the identity \(2\sin x\cos x = \sin 2x\), which gives us \(y' = -2\sin 2x\).

Step 4: Find the second derivative

Differentiate the first derivative to find the second derivative of the function. The first derivative is \(y' = -2\sin 2x\), so the second derivative is \(y'' = -2 \cdot (2\cos 2x) = -4\cos 2x\).

Step 5: Analyze the sign of the second derivative

The concavity of the function is determined by the sign of \(y'' = -4\cos 2x\). The function is concave up where \(-4\cos 2x > 0\), which simplifies to \(\cos 2x < 0\). The function is concave down where \(\cos 2x > 0\).

Step 6: Identify the intervals of concavity

Using the cosine function, we know \(\cos 2x < 0\) on the intervals \((\frac{\pi}{4} + k\pi, \frac{3\pi}{4} + k\pi)\) for integer \(k\), and \(\cos 2x > 0\) on the intervals \(([0 + k\pi, \frac{\pi}{4} + k\pi) \cup (\frac{3\pi}{4} + k\pi, \pi + k\pi])\).

Key Concepts

Second Derivative

When tackling the subject of concavity in calculus, the second derivative of a function is your best friend. It provides vital information on how the curve of the function behaves. Specifically, the second derivative helps us determine whether a function is concave up or concave down on a given interval. A positive second derivative, denoted as \( y'' \), implies that the graph of the function is concave up like a cup. This means the slope of the first derivative \( y' \) is increasing. On the other hand, a negative second derivative means the function is concave down, resembling a frown. Here, the slope of the first derivative \( y' \) is decreasing. Calculating \( y'' \) involves differentiating the first derivative \( y' \) once more. In the example function, after obtaining \( y' = -2\sin 2x \), taking another differentiation step yields \( y'' = -4\cos 2x \), focusing on how these changes describe the function's concavity.

Trigonometric Identities

Trigonometric identities are essential tools in calculus, especially when it comes to simplifying derivatives and integrals involving trigonometric functions. Recognizing and applying these identities can make your work much more manageable. In our exercise, the product \(-4\sin x\cos x\) can be simplified using the double-angle identity:

• \( 2\sin x\cos x = \sin 2x \)

Applying this identity, the expression for the first derivative becomes \( y' = -2\sin 2x \). This step not only simplifies further calculations, but it also helps in quickly reaching accurate conclusions about the function's behavior. Understanding and utilizing these identities effectively can thus transform a complex problem into a simpler one, making the analysis of functions much easier.

Chain Rule

The chain rule is a fundamental technique in calculus used for finding the derivative of composite functions — functions composed of other simpler functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. This rule is an indispensable part of differentiating complex expressions.In the problem at hand, we encounter the function components \( \cos^2 x \) and \( -\sin^2 x \). To differentiate these using the chain rule, we have:

• For \( \cos^2 x \), use \( \frac{d}{dx} (\cos^2 x) = 2\cos x (-\sin x) \)
• For \( -\sin^2 x \), use \( \frac{d}{dx} (-\sin^2 x) = -2\sin x \cos x \)

Mastering the chain rule enables you to efficiently handle derivatives of complex functions like those involving powers of trigonometric expressions.

Intervals of Concavity

Understanding the intervals of concavity involves determining where a function is concave up or down based on the sign of its second derivative. Once you've calculated the second derivative, the next step is to find where it is positive or negative. In our problem, we derived \( y'' = -4\cos 2x \).To assess concavity:- The function is concave up where \(-4\cos 2x > 0\), which simplifies to \(\cos 2x < 0\).- It is concave down where \(\cos 2x > 0\).Using our understanding of the cosine function, these intervals are determined in terms of \( x \):

• \( \cos 2x < 0 \) on \((\frac{\pi}{4} + k\pi, \frac{3\pi}{4} + k\pi)\)
• \( \cos 2x > 0 \) on \([0 + k\pi, \frac{\pi}{4} + k\pi) \cup (\frac{3\pi}{4} + k\pi, \pi + k\pi] \)

These intervals, defined for integer \( k \), help describe how the curve behaves and give us a clear picture of the function's geometry.